 ###### 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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# Multiplying 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.

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We're multiplying complex numbers in trigonometric form. I have here two numbers in rectangular form. I first want to convert these two guys to trig form, and then I'm going to multiply them. I'll leave my answer in trig form.

So first, let's convert z1 into trig form. Z1, I have to find the modulus which is the distance of this number from 0, and that's going to be r equals the square root of a² plus b² or a and b, are these numbers. So we have square root of 3² plus 3². That's the square root of 9 plus 9 root 18. Root 18 is 3 root 2.

Now I need to find theta, the argument. So cosine theta is going to equal this value, the a value, over r. So it's 3 over 3 root 2, and that's the same as, 1 over root 2, which I can rationalize. Root 2, and I get root 2 over 2. Sine theta is b over r, where b is this value. It's the same value, so I'm going to get root 2 over 2 again.

Now, what angle has a cosine value and a sine value both of root 2 over 2? That would be pi over 4. So that's our argument, pi over 4. So I write my z1, as 3 root 2 cosine pi over 4 plus i sine pi over 4.

Now what about z2? Z2 first I'll find its modulus. I want to give it another letter name so I'll call it s. S equals the square root of a² plus b², a is 1, b is root 3. So 1² plus root 3². So that's going to be 1 plus 3, root 4, which is 2. Then I need to find the argument, and I should give it a different name. I'll call it phi. That's not a very good phi. Phi is the Greek letter 'ph' like in Philosophy, or Physics.

Cosine of phi equals a over s. So it's going to be 1/2, and the sine of phi equals root 3 over 2. Now what angle has a cosine of 1/2, and a sine of root 3 over 2? That sounds like pi over 3. So my number is z2 equals 2 times cosine pi over 3 plus i sine pi over 3.

Now let's multiply these guys; z1 times z2. It's much easier to multiply them in trig form. Now for z1 we had 3 root 2, times the cosine pi over 4 plus i sine pi over 4, times 2 cosine pi over 3, plus i sine pi over 3. So when you multiply complex numbers, you multiply their moduli. 3 root 2 times 2, 6 root 2. You add their arguments, so I'll get cosine of pi over 4 plus pi over 3, plus i sine pi over 4, plus pi over 3. This is an answer for the product, but it's not simplified.

So let me just simplify this. I need to get a common denominator which in this case is 12. This would be 3 pi over 12, and 4 pi over 12. So I get 7 pi over 12, and my final answer is 6 root 2, cosine 7 pi over 12, plus i sine 7 pi over 12. That's my product z1 times z2.