How do I solve this?

[Matthew Ko]

I am a junior. I tried to solve #19, and what's written is how far I've got. How do I continue on? Or am I even on the right track? What does "both roots" mean in the parentheses? (P.S. Ignore "Multiply . . . the" at the bottom; it's the next question)
Thanks!

 

[Harry_the_cat]

Can you answer these two questions
1. When two complex numbers are multiplied, how is the new modulus found?
2. When two complex numbers are multiplied, how is the new argument found?

 

[ksdhart2]

I'm not even sure what the question is, much less how to solve it. I can guess from context that you're meant to find both roots of the given expression. That is, solve:

\(\displaystyle \left( 2 - 2\sqrt{3} i \right)^{\frac{1}{2}} = 0\)

Is that correct? Assuming that it is, you've correctly identified one of the roots, and but what about the other one? You've also correctly identified that \(\displaystyle \sin(\theta) = -\frac{\sqrt{3}}{2}\) so either use a unit circle to help you, or consider the fact that \(\displaystyle \sin(\theta) = \sin(\pi - \theta)\).

 

[pka]

Suppose that \(\displaystyle x\cdot y\ne 0 \) then \(\displaystyle \arg(x + yi) = \left\{ {\begin{array}{{rl}} {\arctan \left( {\frac{y}{x}} \right),}&{x > 0} \\ {\arctan \left( {\frac{y}{x}} \right) + \pi ,}&{x < 0\;\& \;y > 0} \\ {\arctan \left( {\frac{y}{x}} \right) - \pi ,}&{x < 0\;\& \;y < 0} \end{array}} \right. \)
The above can be programmed into a calculator.
So \(\displaystyle \theta=\arg(2-2\sqrt 3 {\bf{i}})=\arctan\left(\frac{-2\sqrt 3}{2}\right)\)
Can you finish?

 

[lookagain]

I can guess from context that you're meant to find both roots of the given expression. That is, solve:

\(\displaystyle \left( 2 - 2\sqrt{3 \ } i \right)^{\frac{1}{2}} = 0\)

That is not an equation to solve. The left-hand side is not equal to zero. The
square root of the left-hand side could be set equal to a + bi. Then square
each side.

\(\displaystyle 2 - 2\sqrt{3 \ }i \ = \ a^2 + 2abi - b^2\)

Set corresponding parts equal to each other:

\(\displaystyle a^2 - b^2 \ = \ 2 \ \ \ \ \ \ \ \ \ \) (1)

\(\displaystyle 2abi\ = \ -2\sqrt{3 \ }i\)

Simplify the second equation to get:

\(\displaystyle ab \ = \ -\sqrt{3} \ \ \ \ \ \ \ \ \ \) (2)

Use the first equation and the new second equation to solve for a and b.
a must be a real number. Reject any imaginary values that would come up
for a.

The two square roots should be conjugates of each other.


Note: This optional method might not be allowed in any unseen-to-us
instructions.

 

[Matthew Ko]

Thanks a lot! I solve the problem!