Question

Use De Moivre's theorem to write the complex number in trigonometric form.
(√2 + i√2)^3

Answer 1

Ooo De'Moivre's!
These are usually pretty fun :D

Answer 2

\[\large\rm \left(\sqrt2+i\sqrt2\right)^3\]So what happens if we factor a 2 out of each term? Hmm...\[\large\rm =\frac{1}{2^3}\left(\frac{\sqrt2}{2}+i\frac{\sqrt2}{2}\right)^3\]Do these values sqrt2/2 look familiar?
Do you remember how they relate to our trig functions for some of the special angles? :o

Answer 3

Woops that line should read like this\[\large\rm =2^3\left(\frac{\sqrt2}{2}+i\frac{\sqrt2}{2}\right)^3\]

Answer 4

Is that process too confusing?
You can do it the more traditional way if preferred,\[\large\rm r=\sqrt{a^2+b^2}\]\[\large\rm \arctan\left(\frac{b}{a}\right)=\theta\]

Answer 5

so for r I got 2, and then tan theta=1. I'm not really sure about the special angles.

Answer 6

Recall that tangent = sine/cosine

So our tangent = 1 only when sine and cosine are the same.
(They divide evenly giving us 1).

So at what angle are sine and cosine the same, do you remember? :o

Answer 7

No I'm not sure D: