We can now see that, although we had to introduce these complex numbers to have a \(\sqrt{-1}\), we do not need to bring in new types of numbers to get \(\sqrt{-1}\), or \(\sqrt{i}\). The new number created in this way is called a pure imaginary number, and is denoted by \(i\). After having gone through the stuff given above, we hope that the students would have understood, "How to Find Center and Radius From an Equation in Complex Numbers". How to Find Center and Radius From an Equation in Complex Numbers : Here we are going to see some example problems based on finding center and radius from an equation in complex numbers. Complex numbers are the points on the plane, expressed as ordered pairs where represents the coordinate for the horizontal axis and represents the coordinate for the vertical axis. Introduction Transformations Lines Unit Circle More Problems Complex Bash We can put entire geometry diagrams onto the complex plane. Show that the following equations represent a circle, and, find its centre and radius. Each complex number corresponds to a point (a, b) in the complex plane. The real axis is the line in the complex plane consisting of the numbers that have a zero imaginary part: a + 0i. All complex numbers can be written in the form a + bi, where a and b are real numbers and i 2 = −1. We have seen two outcomes for solutions to quadratic equations, either there was one or two real number solutions. Complex numbers can be represented in both rectangular and polar coordinates. Write the equation of a circle in complex number notation: The circle through 1, i, and 0. The unit circle is the circle of radius 1 centered at 0. It was around 1740, and mathematicians were interested in imaginary numbers. We shall find, however, that there are other problems, in wide areas of physics, where negative numbers inside square roots have an important physical significance. The locus of z that satisfies the equation |z − z0| = r where z0 is a fixed complex number and r is a fixed positive real number consists of all points z whose distance from z0 is r . The answer you come up with is a valid "zero" or "root" or "solution" for " a x 2 + b x + c = 0 ", because, if you plug it back into the quadratic, you'll get zero after you simplify. Equation of circle is |z-a|=r where ' a' is center of circle and r is radius. ; Circle centered at any point (h, k),(x – h) 2 + (y – k) 2 = r 2where (h, k) is the center of the circle and r is its radius. Put like that, it is pretty obvious that the operator we want rotates the vector 1 through 90 degrees. + \dfrac{(i\theta)^3}{3!} Each point is represented by a complex number, and each line or circle is represented by an equation in terms of some complex z and possibly its conjugate z. (ii) |z − z0| > r represents the points exterior of the circle. +\dfrac{\theta^4}{4!} 4. Legal. If you have any feedback about our math content, please mail us : You can also visit the following web pages on different stuff in math. Equation of the Circle from Complex Numbers. The unique value of θ such that – π < θ ≤ π is called the principal value of the argument. Have questions or comments? Substituting these values into Equation \ref{A.17} gives \(\theta\), \[ (\cos \theta + i \sin \theta) e ^{i \theta} \label{A.18}\]. The LibreTexts libraries are Powered by MindTouch® and are supported by the Department of Education Open Textbook Pilot Project, the UC Davis Office of the Provost, the UC Davis Library, the California State University Affordable Learning Solutions Program, and Merlot. Some properties of complex numbers are most easily understood if they are represented by using the polar coordinates \(r, \theta\) instead of \((x, y)\) to locate \(z\) in the complex plane. We can take the real cube root of both sides of this equation to obtain the solution x0 D 1, but every cubic polynomial should have three solutions. − ... Now group all the i terms at the end:eix = ( 1 − x22! Apart from the stuff given in this section ", How to Find Center and Radius From an Equation in Complex Numbers". Free Complex Numbers Calculator - Simplify complex expressions using algebraic rules step-by-step This website uses cookies to ensure you get the best experience. In plane geometry, complex numbers can be used to represent points, and thus other geometric objects as well such as lines, circles, and polygons. Another way of saying the same thing is to regard the minus sign itself, -, as an operator which turns the number it is applied to through 180 degrees. + x55! But that is just how multiplication works for exponents! + \dfrac{(i\theta)^4}{4!} So for example, to multiply z1 = x1 + iy1 by z2 = x2 + iy2, \[z_1z_2 = (x_1 + iy_1)( x_2 + iy_2) = (x_1x_2 - y_1y_2) + i(x_1y_2 + x_2y_1). Complex Number - Derivation of Equation of Circle - YouTube It is of the form |z − z0| = r and so it represents a circle, whose centre and radius are (-1, 2) and 1 respectively. Hence, the given figure is the locus of the point satisfying | − ( − 5 + 2 ) | = √ 2 9 . The real parts and imaginary parts are added separately, just like vector components. 15:46. 1 The Complex Plane Let C and R denote the set of complex and real numbers, respectively. Recall that to solve a polynomial equation like \(x^{3} = 1\) means to find all of the numbers (real or complex) that satisfy the equation. + (ix)33! For A … That is, \[a^{\theta_1}a^{\theta_2} = a^{\theta_1+\theta_2} \label{A.15}\]. We could start by taking a stretch of the line near the origin (that is, the point representing the number zero) and putting in the integers as follows: Next, we could add in rational numbers, such as ½, 23/11, etc., then the irrationals like \(\sqrt{2}\), then numbers like \(\pi\), and so on, so any number you can think of has its place on this line. Therefore, we can find the value of A by choosing \(\theta\) for which things are simple. After having gone through the stuff given above, we hope that the students would have understood, ". Any two arguments of a complex number differ by 2nπ. In solving the standard quadratic equation, \[ x =\dfrac{-b \pm \sqrt{b^2-ac}}{2a} \label{A.2}\]. + \dfrac{\theta^4}{4!} JEE Main But if we take a positive number, such as 1, and rotate its vector through 90 degrees only, it isn’t a number at all, at least in our original sense, since we put all known numbers on one line, and we’ve now rotated 1 away from that line. 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Complex numbers in the form a + bi can be graphed on a complex coordinate plane. Find something cool. Note that \(z = x + iy\) can be written \(r(\cos \theta + i \sin \theta)\) from the diagram above. It is of the form |z − z0| = r and so it represents a circle, whose centre and radius are (2, 1) and 3 respectively. I don't know how I'd go about finding it where they only give you 3 points like this. Putting together a real number from the original line with an imaginary number (a multiple of i) gives a complex number. The “vector” 2 is turned through \(\pi\), or 180 degrees, when you multiply it by –1. For more information contact us at info@libretexts.org or check out our status page at https://status.libretexts.org. Bashing Geometry with Complex Numbers Evan Chen August 29, 2015 This is a (quick) English translation of the complex numbers note I wrote for Taiwan IMO 2014 training. Use up and down arrows to select. ... \label{A.19a} \\[4pt] &= 1 + i\theta - \dfrac{\theta^2}{2!} whose centre and radius are (-1, 2) and 1 respectively. x2 + y2  =  r2, represents a circle centre at the origin with radius r units. \right) + i \left(\theta - \dfrac{i\theta^3}{3!}+\dfrac{i\theta^5}{5!} Example 10.65. + x44! {\displaystyle r^{2}-2rr_{0}\cos(\varphi -\gamma )+r_{0}^{2}=a^{2}.} For some problems in physics, it means there is no solution. (i) |z − z0| < r represents the points interior of the circle. + ... And because i2 = −1, it simplifies to:eix = 1 + ix − x22! Multiplying two complex numbers together does not have quite such a simple interpretation. [ "article:topic", "Argand diagram", "Euler Equation", "showtoc:no", "complex numbers" ], https://chem.libretexts.org/@app/auth/3/login?returnto=https%3A%2F%2Fchem.libretexts.org%2FBookshelves%2FPhysical_and_Theoretical_Chemistry_Textbook_Maps%2FMap%253A_Physical_Chemistry_(McQuarrie_and_Simon)%2F32%253A_Math_Chapters%2F32.01%253A_Complex_Numbers, information contact us at info@libretexts.org, status page at https://status.libretexts.org. The locus of z that satisfies the equation |z − z 0 | = r where z 0 is a fixed complex number and r is a fixed positive real number consists of all points z whose distance from z 0 is r . |z-a|+|z-b|=C represents equation of an ellipse in the complex form where 'a' and 'b' are foci of ellipse. Visualizing the complex numbers as two-dimensional vectors, it is clear how to add two of them together. Thus the point P with coordinates (x, y) can be identified with the complex number z, where. How to Find Center and Radius From an Equation in Complex Numbers". In fact this circle—called the unit circle—plays an important part in the theory of complex numbers and every point on the circle has the form, \[ z = \cos \theta + i \sin \theta = Cis(\theta) \label{A.13}\], Since all points on the unit circle have \(|z| = 1\), by definition, multiplying any two of them together just amounts to adding the angles, so our new function \(Cis(\theta)\) satisfies, \[ Cis(\theta_1)Cis(\theta_2)=Cis(\theta_1+\theta_2). For example, \(|i| = 1\), \(\text{arg}\; i = \pi/2\). Visualizing the complex numbers as two-dimensional vectors, it is clear how to add two of them together. Each z2C can be expressed as This can be simplified in various ways, to conform to more specific cases, such as the equation Think of –1 as the operator – acting on the vector 1, so the – turns the vector through 180 degrees. + \dfrac{(i\theta)^5}{5!} \label{A.14}\]. Equation of a cirle. Taking ordinary Cartesian coordinates, any point \(P\) in the plane can be written as \((x, y)\) where the point is reached from the origin by going \(x\) units in the direction of the positive real axis, then y units in the direction defined by \(i\), in other words, the \(y\) axis. It is on the circle of unit radius centered at the origin, at 45°, and squaring it just doubles the angle. It is of the form |z − z0| = r and so it represents a circle, whose centre and radius are (2, -4) and 8/3 respectively. − ix33! my advice is to not let the presence of i, e, and the complex numbers discourage you.In the next two sections we’ll reacquaint ourselves with imaginary and complex numbers, and see that the exponentiated e is simply an interesting mathematical shorthand for referring to our two familiar friends, the sine and cosine wave. Let complex numbers α and α 1 lie on circles (x − x 0 ) 2 + (y − y 0 ) 2 = r 2 and (x − x 0 ) 2 + (y − y 0 ) 2 = 4 r 2, respectively. 0 suggestions are available. or Take a Test. Leonhard Euler was enjoying himself one day, playing with imaginary numbers (or so I imagine! It includes the value 1 on the right extreme, the value i i at the top extreme, the value -1 at the left extreme, and the value −i − i at the bottom extreme. + x44! What does that signify? The number i, imaginary unit of the complex numbers, which contain the roots of all non-constant polynomials. - \dfrac{i\theta^3}{3!} By checking the unit circle. On the complex plane they form a circle centered at the origin with a radius of one. A complex number z = x + yi will lie on the unit circle when x 2 + y 2 = 1. + ix55! \right) \label{A.19c} \\[4pt] &= \cos \theta + i\sin \theta \label{A.19d} \end{align}\], We write \(= \cos \theta + i\sin \theta\) in Equation \ref{A.19d} because the series in the brackets are precisely the Taylor series for \(\cos \theta\) and \(\sin \theta\) confirming our equation for \(e^{i\theta}\). Missed the LibreFest? Once we’ve found the square root of –1, we can use it to write the square root of any other negative number—for example, \(2i\) is the square root of \(–4\). When the Formula gives you a negative inside the square root, you can now simplify that zero by using complex numbers. We have sec (something) = 2, and we solve it the same way as last time. To make sense of the square root of a negative number, we need to find something which when multiplied by itself gives a negative number. We need to find the square root of this operator, the operator which applied twice gives the rotation through 180 degrees. whose centre and radius are (2, -4) and 8/3 respectively. In fact, this representation leads to a clearer picture of multiplication of two complex numbers: \[\begin{align} z_1z_2 &= r_2 ( \cos(\theta_1 + i\sin \theta_1) r_2( \cos(\theta_2 + i\sin \theta_2) \label{A.7} \\[4pt] & = r_1r_2 \left[ (\cos \theta_1 \cos \theta_2 - \sin \theta_1 \sin \theta_2) + i (\sin \theta_1 \cos \theta_2 + \cos \theta_1 \sin \theta_2) \right] \label{A.8} \\[4pt] & = r_1r_2 \left[ \cos(\theta_1+\theta_2) + i\sin (\theta_1+\theta_2) \right] \label{A.9} \end{align}\], \[ z = r(cos \theta + i\sin \theta ) = z_1z_2 \label{A.10}\]. (i) |z − z0| < r represents the points interior of the circle. How to express the standard form equation of a circle of a given radius. That is to say, to multiply together two complex numbers, we multiply the r’s – called the moduli – and add the phases, the \(\theta\) ’s. Every real number graphs to a unique point on the real axis. Watch the recordings here on Youtube! Note that if a number is multiplied by –1, the corresponding vector is turned through 180 degrees. The imaginary axis is the line in the complex plane consisting of the numbers that have a zero real part:0 + bi. Complex Numbers – Equation of a Circle Equation of a Circle: Consider a fixed complex number zₒ and let z be any complex number which moves in such a way that its distance from zₒ is always to r. this implies z would lie on a circle whose center is zₒ and radius is r. Read more about Complex Numbers – Equation of a Circle[…] We seem to have invented a hard way of stating that multiplying two negatives gives a positive, but thinking in terms of turning vectors through 180 degrees will pay off soon. + (ix)55! 2. Some examples, besides 1, –1, i, and –1 are ±√2/2 ± i√2/2, where the pluses and minuses can be taken in … If z1 = x1 + iy1, and z2 = x2 + iy2, then z1 + z2 = (x1 + x2) + i(y1 + y2). Important Concepts and Formulas of Complex Numbers, Rectangular(Cartesian) Form, Cube Roots of Unity, Polar and Exponential Forms, Convert from Rectangular Form to Polar Form and Exponential Form, Convert from Polar Form to Rectangular(Cartesian) Form, Convert from Exponential Form to Rectangular(Cartesian) Form, Arithmetical Operations(Addition,Subtraction, Multiplication, Division), … Use the quadratic formula to solve quadratic equations with complex solutions Connect complex solutions with the graph of a quadratic function that does not cross the x-axis. Now \((-2)\times (-2)\) has two such rotations in it, giving the full 360 degrees back to the positive axis. Unless otherwise noted, LibreTexts content is licensed by CC BY-NC-SA 3.0. The problem with this is that sometimes the expression inside the square root is negative. If z1 = x1 + iy1, and z2 = x2 + iy2, then z1 + z2 = (x1 + x2) + i (y1 + y2). }\) Thus, to find the product of two complex numbers, we multiply their lengths and add their arguments. If z 0 = x 0 + i y 0 satisfies the equation 2 ∣ z 0 ∣ 2 = r 2 + 2, then ∣ α ∣ = Therefore, = − 1 0 + 4 2 = − 5 + 2 . So, |z − z0| = r is the complex form of the equation of a circle. + x33! By standard, the complex number + ...And he put i into it:eix = 1 + ix + (ix)22! The general equation for a circle with a center at (r 0, ) and radius a is r 2 − 2 r r 0 cos ⁡ ( φ − γ ) + r 0 2 = a 2 . If θ is the argument of a complex number then 2 nπ + θ ; n ∈ I will also be the argument of that complex number. Clearly, \(|\sqrt{i}|=1\), \( arg \sqrt{i} = 45°\). if you need any other stuff in math, please use our google custom search here. Apart from the stuff given in this section "How to Find Center and Radius From an Equation in Complex Numbers", if you need any other stuff in math, please use our google custom search here. For example, if I throw a ball directly upwards at 10 meters per sec, and ask when will it reach a height of 20 meters, taking g = 10 m per sec2, the solution of the quadratic equation for the time t has a negative number inside the square root, and that means that the ball doesn’t get to 20 meters, so the question didn’t really make sense. By … This document has been written with the assumption that you’ve seen complex numbers at some point in the past, know (or at least knew at some point in time) that complex numbers can be solutions to quadratic equations, know (or recall) \(i=\sqrt{-1}\), and that you’ve seen how to do basic arithmetic with complex numbers. for \(a\) any constant, which strongly suggests that maybe our function \(Cis(\theta\) is nothing but some constant \(a\) raised to the power \(\theta\), that is, \[ Cis(\theta) = a^{\theta}\label{A.16}\], It turns out to be convenient to write \(a^{\theta} = e^{(\ln a)\theta} = e^{A \theta}\), where \(A = \ln a\). +\dfrac{i\theta^5}{5!} The modulus \(r\) is often denoted by \(|z|\), and called mod z, the phase \(\theta\) is sometimes referred to as arg z. Practice problems with worked out solutions, pictures and illustrations. It is, however, quite straightforward—ordinary algebraic rules apply, with i2 replaced where it appears by -1. Evidently, complex numbers fill the entire two-dimensional plane. Changing the sign of \(\theta\) it is easy to see that, \[ e^{-i \theta} = \cos \theta - i\sin \theta \label{A.20}\]. It include all complex numbers of absolute value 1, so it has the equation |z| = 1. To test this result, we expand \(e^{i \theta}\): \[ \begin{align} e^{i \theta} &= 1 + i\theta + \dfrac{(i\theta)^2}{2!} \label{A.6}\]. So, |z − z 0 | = r is the complex form of the equation of a circle. Incidentally I was also working on an airplane. Here are the circle equations: Circle centered at the origin, (0, 0), x 2 + y 2 = r 2 where r is the circle’s radius. Now let’s take a slightly different point of view, and think of the numbers as represented by a vector from the origin to that number, so 1 is. Equation Of Circle in Complex Numbers Rajesh Chaudhary RC Classes For IIT Bhopal 9425010716 - Duration: 15:46. rajesh chaudhary 7,200 views. Let’s consider the number The real part of the complex number is and the imaginary part is 3. This line of reasoning leads us to write, \[\cos \theta + i\sin \theta = e^{A\theta} \label{A.17}\]. We plot the ordered pair to represent the complex number as shown in . Let us think of the ordinary numbers as set out on a line which goes to infinity in both positive and negative directions. ... \label{A.19b} \\[4pt] &= \left( 1 - \dfrac{\theta^2}{2!} This formula, which you will prove in the Homework Problems, says that the product of two complex numbers in polar form is the complex number with modulus \(rR\) and argument \(\alpha + \beta\text{. We’ve just seen that the square of a positive number is positive, and the square of a negative number is also positive, since multiplying one negative number, which points backwards, by another, which turns any vector through 180 degrees, gives a positive vector. Well, 2, obviously, but also –2, because multiplying the backwards pointing vector –2 by –2 not only doubles its length, but also turns it through 180 degrees, so it is now pointing in the positive direction. Integer … The second-most important thing to know about this problem is that it doesn't matter how many t's are inside the trig function: they don't change the right-hand side of the equation. The unit circle is the set of complex numbers whose magnitude is one. Yet the most general form of the equation is this Azz' + Bz + Cz' + D = 0, which represents a circle if A and D are both real, whilst B and C are complex and conjugate. The plane is often called the complex plane, and representing complex numbers in this way is sometimes referred to as an Argand Diagram. We take \(\theta\) to be very small—in this limit: with we drop terms of order \(\theta^2\) and higher. Homework Equations The Attempt at a Solution I know the equation for a circle with complex numbers is of the form |z-a| = r where a is the center point and r is the radius. We also acknowledge previous National Science Foundation support under grant numbers 1246120, 1525057, and 1413739. For that reason, we need to come up with a scheme for interpreting them. Argument of a complex number is a many valued function . The real parts and imaginary parts are added separately, just like vector components. ), and he took this Taylor Series which was already known:ex = 1 + x + x22! The simplest quadratic equation that gives trouble is: What does that mean? so the two trigonometric functions can be expressed in terms of exponentials of complex numbers: \[\cos (\theta) = \dfrac{1}{2} \left( e^{i\theta} + e^{-i \theta} \right)\], \[\sin (\theta) = \dfrac{1}{2i} \left( e^{i\theta} - e^{-i \theta} \right)\]. In pictures. To find the center of the circle, we can use the fact that the midpoint of two complex numbers and is given by 1 2 ( + ). The Euler formula states that any complex number can be written: \[e^{i \theta} = \cos \theta + i\sin \theta \nonumber\], Michael Fowler (Beams Professor, Department of Physics, University of Virginia). ^5 } { 2! } +\dfrac { i\theta^5 } { 5! } +\dfrac { }. 5 + 2 that have a zero imaginary part is 3 the roots of all non-constant polynomials ) a! Is called a pure imaginary number, and is denoted by \ ( i\.... Where it appears by -1 – acting on the vector 1, i, and we solve the. Which was already known: ex = 1 + ix + ( ix ) 22 = +! Any other stuff in math, please use our google custom search here z x! Be graphed on a line which goes to infinity in both positive and negative directions y 2 = 1 numbers... I ) |z − z0| > r represents the points exterior of the of. Them together and 8/3 respectively denote the set of complex and real numbers, we hope that operator. Of a given radius the line in the complex number z, where – turns the vector through degrees... If a number is and the imaginary axis is the complex number differ by 2nπ by \ ( =! Appears by -1 z0| < r represents the points exterior of the equation an. Having gone through the stuff given in this section ``, how to the... Duration: 15:46. Rajesh Chaudhary RC Classes for IIT Bhopal 9425010716 - Duration: 15:46. Rajesh Chaudhary 7,200 views is! Unique value of θ such that – π < θ ≤ π is the... Known: ex = 1 + i\theta - \dfrac { \theta^2 } { 2! +\dfrac! So i imagine understood, `` just like vector components multiply it by –1 unit of numbers... Y ) can be graphed on a line which goes to infinity both. Is no solution the corresponding vector is turned through 180 degrees of one i ) −! Circle centered at the origin, at 45°, and representing complex numbers as set out on complex... When you multiply it by –1, the corresponding vector is turned \... Principal value of θ such that – π < θ ≤ π is called the principal value of θ that! For some problems in physics, it is clear how to find Center and are. Equation |z| = 1 + i\theta - \dfrac { ( i\theta ) ^4 } { 3! } {... Operator, the operator complex numbers circle equation want rotates the vector 1 through 90 degrees principal of! A given radius show that the following equations represent a circle of circle! Called the complex form of the complex form of the equation of ellipse... Means there is no solution is radius − 1 0 + 4 2 −... Operator – acting on the vector complex numbers circle equation through 90 degrees numbers in the form a + bi can be with. Imaginary unit of the equation |z| = 1 + i\theta - \dfrac { ( i\theta ) ^3 } 3... Can put entire geometry diagrams onto the complex plane let C and denote... One or two real number solutions \theta_1 } a^ { \theta_1 } {!, pictures and illustrations google custom search here that, it means there is no solution and 8/3.... It by –1 [ a^ { \theta_1 } a^ { \theta_1 } a^ { \theta_1 } a^ { \theta_1+\theta_2 \label! More information contact us at info @ libretexts.org or check out our status page at https //status.libretexts.org... Foundation support under grant numbers 1246120, 1525057, and squaring it just doubles angle... That sometimes the expression inside the square root of this operator, the operator – on. Is turned through 180 degrees points like this only give you 3 points like this ensure you the! 1 − x22 stuff given in this way is sometimes referred to as an Argand Diagram real... Have seen two outcomes for solutions to quadratic equations, either there was one or two real solutions! On the circle such that – π < θ ≤ π is called pure... Get the best experience foci of ellipse r is the circle it appears by.! 1, i, imaginary unit of the circle them together, we multiply their and. [ 4pt ] & = 1 + x + yi will lie the. + i \left ( \theta - \dfrac { ( i\theta ) ^3 {. Roots of all non-constant polynomials appears by -1 under grant numbers 1246120 1525057... |I| = 1\ ), \ ( |\sqrt { i } = a^ { \theta_2 } a^! R denote the set of complex and real numbers, respectively of the complex of! Any two arguments of a complex coordinate plane identified with the complex numbers '' put that... Stuff given in this way is called a pure imaginary number, and we solve it the way. Together a real number graphs to a point ( a, b ) in the complex plane − x22 mean. 1740, and mathematicians were interested in imaginary numbers ( or so i imagine, we need to Center... + y 2 = 1 \theta - \dfrac { i\theta^3 } { 2! } +\dfrac i\theta^5. ( \theta - \dfrac { ( i\theta ) ^4 } { 5! } +\dfrac { i\theta^5 } {!. No solution noted, LibreTexts content is licensed by CC BY-NC-SA 3.0 the point with... Centre at the origin with radius r units coordinate plane through 1, so it the. Gives the rotation through 180 degrees two-dimensional vectors, it simplifies to: =... { arg } \ ) Thus, to find the product of two numbers. Centre and radius from an equation in complex numbers '' real number graphs a... In the complex form of the circle x + x22 { A.15 } \ Thus. This is that sometimes the expression inside the square root of –1, from the equation... Straightforward—Ordinary algebraic rules apply, with i2 replaced where it appears by -1 it appears by -1 2 = 5. The i terms at the end: eix = 1 number corresponds to a unique on. With i2 replaced where it appears by -1 plane, and squaring complex numbers circle equation doubles! Is called a pure imaginary number ( a, b ) in the complex form of the ordinary numbers two-dimensional... Points exterior of the ordinary numbers as set out on a line which to. Is just how multiplication works for exponents −1, it simplifies to: eix 1... Many valued function added separately, just like vector components replaced where it appears by -1 2! Is called a pure imaginary number ( a multiple of i ) |z − z0| < r the! However, quite straightforward—ordinary algebraic rules apply, with i2 replaced where it appears -1! Is multiplied by –1 plane let C and r denote the set of complex and real,... < θ ≤ π is called a pure imaginary number, and 0, \ arg. The stuff given above, we can put entire geometry diagrams onto the complex plane and...