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# Dividing Complex Numbers - Problem 2

###### Norm Prokup

###### Norm Prokup

**Cornell University**

PhD. in Mathematics

Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.

We're dividing complex numbers in trigonometric form. I have a problem that asks me to express z1, and z2 these two numbers, and their quotient in trigonometric form. These guys are actually in rectangular form, so I first need to put them in trig form, and then divide and I'll express the answer in trig form.

First let's start with z1. Before I do anything else, I'm going to need to find the modulus of z1, I'll call that r. R is the square root of this number squared plus this number squared. 6² plus, and I would count this as -6. That's the imaginary coefficient -6². So I get root 36 plus 36, root 72. That's going to simplify to 6 root 2. That's my modulus value. Remember modulus tells you how far the number is from the origin.

Now the argument is going to be cosine of theta, theta is going to be the argument, equals 6 over 6 root 2. The 6's cancel giving me 1 over root 2, which is the same as root 2 over 2.What about the sine of theta?

Sine of theta is -6 over 6 root 2. It's exactly the opposite of what the cosine was. So this will simplify to negative root 2 over 2. So I need to find an angle whose cosine is root 2 over 2, and whose sine is negative root 2 over 2. Let me draw a little picture.

Cosine is positive, so over here sine is negative, so down here. The root 2 over 2's tell me, that my reference angle is going to be pi over 4. So this angle will be pi over 4. That means that my modulus is going to be 7 pi over 4. So theta is 7 pi over 4. That means that z1 can be written 6 root 2, the modulus, times cosine 7 pi over 4 plus i sine 7 pi over 4.

What about z2? Well, for z2, I also want to start with its modulus, and I'll call its modulus s so I don't confuse it with r. S is going to be the square root of 3² plus 3². That's going to be square root of 9 plus 9, root 18, which is 3 root 2.

What about its argument? I'll call its argument phi, so I don't confuse it with theta. Cosine of phi is going to be the real part divided by the modulus. So 3 over 3 root 2 which is also root 2 over 2. The sine of phi is going to be the imaginary part 3 over 3 root 2. It's the same thing also root 2 over 2.

So what angle has a cosine, and a sine of root 2 over 2? It's got to be pi over 4. So phi is pi over 4. That means that I can write z2 as 3 root 2 cosine pi over 4 plus i sine pi over 4.

What about the quotient; Z1 over z2? Remember when you divide two numbers in trig form, you divide their moduli. So I'm going to divide 6 root 2 by 3 root 2, and then you subtract their arguments. I have 7 pi over 4, and pi over 4. 7 pi over 4 minus pi over is 6 pi over 4, which is the same as 3 pi over 2. So all I have to do is simplify this. This simplifies to 2 cosine 3 pi over 2 plus i sine 3 pi over 2. That's all we have to do if we're asked to express our answer in trig form.

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###### Norm Prokup

PhD. in Mathematics, University of Rhode Island

B.S. in Mechanical Engineering, Cornell University

He uses really creative examples for explaining tough concepts and illustrates them perfectly on the whiteboard. It's impossible to get lost during his lessons.

Thiswas EXCELLENT! I am a math teacher and have been looking for an easy/logical way to explain the lateral area of a cone to my students and this was incredibly helpful, thank you very much!”

I just learned more In 3 minutes of polygons here than I do in 3 weeks in my math class”

Hahaha, his examples are the same problems of my math HW!”

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