###### 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.

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# Dividing Complex Numbers - Problem 3

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.

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We're dividing complex numbers in trig form. Here I'm asked to express z1, z2, and z1 over z2 in trig form. I'm given these two complex numbers in rectangular form. So let's start by converting z1 to trig form.

First let's find the modulus. The modulus r, is going to be the square root of 10 root 3² plus 30². This is going to be 10² times root 3², 100 times 3, 300, plus 30² is 900. So that's root 1200. Now 1200 is 3 times 400, and 400 is a perfect square, the square root is 20. So this is going to give me 20 root 3. That's my modulus. That's how far this point is away from the origin.

Now let's find the argument. Cosine of the argument is going to be 10 root 3 over the modulus, 20 root 3. 10 root 3 over 20 root 3 is 1/2. So we're looking for cosine of theta to be a half. What about the sine of theta? 30 over the modulus. 30 over 20 root 3. Now if I multiply by root 3 over root 3, I get 30 root 3, over 20 times 3, 60. This is the same as root 3 over 2. So I have sine theta is root 3 over 2, cosine theta is 1/2. Theta has to be pi over 3 because I want theta to be between 0 and 2 pi. So my complex number z1 is going to be 20 root 3 cosine pi over 3 plus i sine pi over 3.

Next I need to get z2 in trig form. I'll start with its modulus. I'll call its modulus s so I don't confuse it with r. I have to take these numbers -1 and root 3 and square them. So s is the square root of -1² and root 3². I get the square root of 1 plus 3, root 4 and that's 2.

Next I need to find the argument of z2. I'll call the argument phi. Cosine of the argument is -1 over 2, negative 1/2, and the sine of the argument is root 3 over 2. So the cosine is -1/2, the sine is root 3 over 2, so phi looks like it could be 2 pi over 3. That means that z2 can be written 2 times cosine 2 pi over 3 plus i sine 2 pi over 3.

Now it's time to divide. This is the fun part, because dividing numbers in trig form is really easy. Z1 divided by z2.

First, you divide the moduli. You divide 20 root 3 over 2. Then, you subtract the arguments; pi over 3 minus 2 pi over 3 plus i sine pi over 3 minus 2 pi over 3. Let me simplify this a little bit.

First of all 20 root 3 over 2 is 10 root 3. Second pi over 3 minus 2 pi over 3 is negative pi over 3. Now I can't leave my answer this way. Just in case your teacher wants you to give an argument between 0 and 2 pi, we have to be prepared to switch this for another angle.

Now, you have to think about what angle between 0 and 2 pi is co-terminal with negative pi over 3. It's 5 pi over 3. So I have to switch to 10 root 3 cosine of 5 pi over 3 plus i sine 5 pi over 3. That's my final answer.