8.3+Content+-+Complex+Numbers

toc Add Complex Numbers content to this page. See specific details on the home page. This unit has been split into parts A & B. Content between the two must be cohesive as it is one continuous, connected material. They are split simply to better organize the page.
 * Quick Instructions**


 * Complex Numbers (Unit 8B) Learning Targets**
 * Write complex numbers in trigonometric form and find products and quotients.
 * Use DeMoivre’s Theorem to find powers and roots of complex numbers.
 * Solve equations with complex numbers in polar form.

=How to Deal with Complex Numbers=

Complex numbers are numbers written in the form of a + ib, where i is the imaginary component of the number (the square root of -1) and a and b are both real numbers.

Graphing Complex Numbers
These complex numbers can be expressed in a polar form as well as the rectangular form mentioned above.



This is a graphical representation of a complex number across the real (measured by a) and imaginary (measured by b) planes.

Here, you can see how the components come together.

In order to express the complex number in polar form, you must first write an equation that expresses a in terms of r (the distance from the origin) and theta (the angle between the x axis and the line).

math a=rcos\theta math

Then, write an equation that expresses b in terms of r and theta

math b=rsin\theta math

Now, given the complex number z=a + bi, rewrite z in polar form! math z= rcos\theta+ irsin\theta math

math z=(r(cos\theta))+i(sin\theta) math

In order to better represent this form, we use the CiS notation: math z=rcis\theta math

Now, graphing complex numbers is easy! By using the same graphing techniques used earlier in the chapter, you can graph complex numbers in the polar plane.
 * It is important to note that graphing complex numbers is simply plotting one point on a plane and not a function!!!**

‍Now, some examples: ‍
Convert the following to Trigonometric (Polar) Form: math -1+\sqrt{3}i math

By using inverse sine and cosine laws as well as the pythagorean theorem, you can find the value of theta and r.

math z=2(cos\frac{2\pi}{3}+isin\frac{2\pi}{3}) math

math 2cis(\frac{2\pi}{3}) math However, an alternate way to do this problem(and other problems with special right triangles or triples is to use the stuff we learned back in geometry. We know that in a 30-60-90 triangle the ratios of the sides is x-xsqrt{3}-2x. If you see the any 30-60-90 or 45-45-90, you can save a lot of time by using those ratios.

=**Multiplying Complex numbers:**=

Complex numbers can be multiplied or divided by other complex numbers, and the resulting numbers can be expressed by using their polar forms.

"Normal" multiplication of Complex numbers:

math (a+bi)(c+di) math

math (ac+adi+bci-bd) math

Now, using their polar forms, we can multiply their Polar forms as well! math r_{1}(cis\theta_{1}) * r_{2}(cis\theta_{2}) math

math r_1(cos(\theta_{1} )+isin(\theta_{1})r_2(cos(\theta_{2} )+isin(\theta_{2})\\

r_1r_2 (cos(\theta_1)cos(\theta_2)-sin(\theta_1)sin(\theta_2))+(cos(\theta_1)isin(\theta_2)+isin(\theta_1)cos(\theta_2)) math Using the trig identities.... math r_{1}* r_{2}(cos(\theta_{1}+\theta_{2})+ isin(\theta_{1}+\theta_{2}) ) math math r_{1}* r_{2}(cis(\theta_{1}+\theta_{2})) math

We can also use a similar formula to divide complex numbers by other complex numbers: math \[\frac{r_1(cis\theta _1)}{r_2(cis\theta_2)}\] math

Multiplying Complex Numbers Example
Ex1(Final Exam Review #1)

5) Given math w = 3-3\sqrt{3}i math math z = -5+5i, find math math a) w*z \\ \indent \indent 1. \text{Convert to polar form} \\ \indent \indent 2. w=6cis(\frac{4\pi}{3}) and \indent z=5\sqrt{2}cis(\frac{3\pi}{4}) \\ \indent \indent 3. \text{To multiply complex numbers, you multiply the coefficients and add the angles} \\ \indent \indent 4. \text{The result of} w*z= 6*5\sqrt{2}cis(\frac{4\pi}{3}+ \frac {3\pi} {4}) \\ \indent \indent 5. w*z=30\sqrt{2}cis(\frac{5\pi}{12}) \\ math

Dividing Complex Numbers Example
Given math w = 3-3\sqrt{3}i math math z = -5+5i, find math math b) w/z\\ \indent \indent 1. \text{ Again, convert to polar form} \\ \indent \indent 2. w=6cis(\frac{4\pi}{3}) and \indent z=5\sqrt{2}cis(\frac{3\pi}{4}) \\ \indent \indent 3.\text{To divide complex numbers, you divide the respective coefficients and subtract the angles} \\ \indent \indent 4.\text{The result of } \frac{w}{z}= \frac {6} {5\sqrt{2}} cis(\frac{4\pi}{3}- \frac {3\pi} {4}) \\ \indent \indent 5. \frac{w}{z} = \frac {6} {5\sqrt{2}} cis(\frac{7\pi} {12} ) \\ math

= Using DeMoivre's Theorem: =

DeMoivre's Theorem deals with taking polar forms to certain exponential powers and vice versa. DeMoivre's is an extension of multiplying/dividing complex numbers in polar form, since taking something to the nth power is multiplying the number by itself n times.

math \text{If z=a+bi is any complex number with trig form}\, rcis\theta \,\text{and n is any positive integer then the nth power of z is given by:}\\ z^{n} = r^{n} cis (n\theta) math

DeMoivre's Theorem Example:
To show how DeMoivre's theorem is an extension of multiplying/dividing complex numbers in polar form, let us say that math \[Z= 6cis(\frac \pi 3)\] math Suppose we wanted to square the above complex number math \[Z^{2}= 6cis(\frac \pi 3)^{2}\] math We know that the coefficient gets squared **but** the radian angle value is multiplied by the exponential value. Therefore, math \[Z^{2}= 36cis((2\pi) /3)\] math

Going Backwards is a bit trickier
Let us say that:

math \[Z^{3}= 1+i\] math Firstly, we must convert the above expression into polar form we get math \[Z^{3}= (\sqrt 2 cis \frac{\pi }{4})\] math Then, math \[Z= (\sqrt 2 cis \frac{\pi }{4})^{\frac{1}{3}}\] math math \[(2)^{\frac{1}{6}}\] math is the coefficient because math \[\sqrt{2}^{\frac{1}{3}} = 2^{\frac{1}{6}}\] math

Getting the angle is a bit trickier You know that math \[\[\frac{\pi}{4} *{\frac{1}{3}}\]\] math but you also know that angle can differ by math \[\[\[\frac{2\pi}{3}\]\] math because solutions in the complex plane are evenly spaced. If there are three solutions (cubed) over 2pi, then each solution would naturally be separated by 2pi/3. So, the angles are math \[\[\frac{\pi}{12}, \frac{3\pi}{4}, \frac{4\pi} {3}\]\] math


 * REMEMBER: Convert the complex number from its real form to its polar form to perform these functions**

Extensions of DeMoivre's theorem
Sometimes, seemingly simple equations have more solutions than meets the eye For example, we can use the equation x^3=8

math 1.\indent x^{3} = 8 math

math 2. \indent x^{3}= 8+ 0i math

math 3. x= 8^{\frac{1}{3}} cis( 0), \indent 8^{\frac{1}{3}} cis( 0+ \frac {2\pi} {3} ), \indent 8^{\frac{1}{3}} cis( 0 + \frac {4\pi} {3} ) math

math 4.x=2cis(0), \indent 2cis( \frac {2\pi} {3} ),\indent 2cis(\frac{4\pi} {3} ) math

math 5. x= 2, \indent x= -1+ \sqrt {3} * i , \indent x=-1- \sqrt {3} * i math

math 1.\indent x^{5}= \sqrt{3}+ 1i math math 2. \indent x^{5} = 2cis(\frac{\pi}{3}) math math 3. x=2^{\frac{1}{5}}cis(\frac{\pi}{3}), x=2^{\frac{1}{5}}cis(\frac{11\pi}{15}), x=2^{\frac{1}{5}}cis(\frac{17\pi}{15}), x=2^{\frac{1}{5}}cis(\frac{23\pi}{15}), x=2^{\frac{1}{5}}cis(\frac{29\pi}{15}) math
 * Example 2:**

DeMoivre's Theorem involving large exponents
Sometimes, we encounter problems when we have to use DeMoivre's Theorem for high exponential powers. Let us say that math z= 4cis(\frac{2\pi} {3}) math Suppose we wanted to find what z^101 evaluates to. We have to use the assumptions of DeMoivre's theorem in order to solve this problem.

Firstly, the radius is simply taking to the respective power so the radius becomes math 4^{101} math Secondly, the angle becomes math 101* \frac {2\pi} {3} = \frac{202\pi} {3} math As a result, we get that math z^{101}= 4^{101}cis(\frac{202\pi}{3}) math
 * HOWEVER**
 * we are not done**

We know from earlier units that the unit circle repeats every math 2\pi or \frac {6\pi} { 3} math So, math \frac {202\pi} {3} = \frac {4\pi} {3} math

We can also get this through doing the first few powers

For 2nd power, math angle = \frac{4\pi} {3} math For 3rd power, math angle= 2\pi math

For 4th power, math angle = \frac{2\pi} {3} math

and this cycle repeats

As a result, we get the final result to be math 4^{101}cis(\frac {4\pi} {3}) math

It is even possible to convert the answer back into rectangular form by expanding CiS: math 4^{101}Cos(\frac {4\pi} {3}) + 4^{101}Sin(\frac {4\pi} {3}) math

math (4^{101})(\frac {-1} {2}) + (4^{101})(-\frac {\sqrt {3}} {2}) math

=Authors=
 * Primary authors of this page (as of 06/02/12):**
 * Chandru Rasendran**
 * Kyle Friedrichs**
 * Jay Wang**

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