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# Multiplying 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.

I'm multiplying complex numbers in trig form. Here I have two numbers in rectangular form, and I'm asked to express those two numbers, and their product in trigonometric form. So let me start by writing z1 in trig form.

The first thing I want to do is find r, its modulus. R is the square root of the sum of the squares of the a, and b values; the real and imaginary coefficients. So I'm going to get 2² plus 2 root 3 squared. Now 2² is 4, and 2 root 3 squared is 4 times 3, 12. So that's root 16 which is 4. So that's our modulus. That's how far away this number is from the origin.

Second let's find the argument. The cosine of the argument is going to be a over r, where a is this value, the real coefficient. So 2 over 4, 1/2. Sine theta is going to be b over r, where b is this value; 2 root 3 over 4 so root 3 over 2. Let's see cosine theta is 1/2, sine theta is root 3 over 2, theta is going to be pi over 3. So finally my z1 is going to be 4 times cosine pi over 3 plus i sine pi over 3. So that's z1 in trig form.

What about z2? Start with the modulus. So the modulus occur at s, so I don't confuse it with the modulus of z1. S is going to be root 3² plus -1², so plus -1². That's going to be root 3 plus 1, root 4, which is 2. So the modulus is 2, that means this number is two units away from the origin.

The argument I'll call phi. Cosine of phi is root 3 over 2, and the sine of phi is -1. Remember this coefficient is going to be -1, -1 times i, -1 over 2. So what angle has a cosine of root 3 over 2, and a sine of -1/2? You can draw a picture if you like. Sine is -1/2, cosine is root 3 over 2, we're down here. So I could give the answer, negative pi over 6. But, sometimes your teacher will want you to give your arguments between 0 and 2 pi, in which case you've got to go with 11 pi over 6. So phi will be 11 pi over 6. That means z2 is 2 times the cosine of 11 pi over 6 plus i sine 11 pi over 6.

Now we multiply them; Z1 times z2. Now, remember when you multiply two complex numbers, you multiply their moduli, 4 and 2. So 4 times 2 is going to be 8, that's the first thing I'll write. Then you add their arguments. So we have pi over 3, and 11 pi over 6. I want to add these two.

Now pi over 3 is the same as 2 pi over 6, plus 11 pi over 6 gives me 13 pi over 6. So cosine of 13 pi over 6 plus i sine 13 pi over 6. Now that would almost be an answer, except, again, I want my arguments to all be between 0 and 2 pi. This is not, this is actually pi over 6 greater than 2 pi. So think about 13 pi over 6. It's 12 pi over 6, plus pi over 6, and that's 2 pi.

Now because both of these functions are periodic with period 2 pi, cosine of 13 pi over 6, and sine of 13 pi over 6, will be the same as the cosine, and sine of pi over 6. So I write my final answer as 8 times cosine pi over 6 plus i sine pi over 6. That's my answer for the product.

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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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