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I've been trying to figure out the way to solve this for a while now, and I'm hoping someone could point me in the right direction to find the answer (or show me how to solve this).

The problem I'm having is with this equation: $(2i-2)^{38}$ and I need to solve it using de Moivre's formula (i is imaginary in case that wasn't clear). Now obviously I know it would be stupid to expand right away because it would turn into an extremely long equation.

The farthest I got with it is $2(i-1)^{38}$ and $(2\cdot(-1)^{1/2}-2)^{38}$.

I'm hoping I'm headed in the right direction but I'm stuck. Could someone please show me the way to solve this?

user1551
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monoxide
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3 Answers3

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Hint: Start by writing $-2+2i$ in the form $re^{\theta i}$.

Then your answer will be $r^{38}e^{38\theta i}$.

For $z=x+iy$, we have $r^2= x^2+y^2$ and $\tan \theta = y/x$.

MPW
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  • It took me while to understand this response, but I now do and yeah, makes perfect sense. I guess I need to practice more to see it off the bat next time. Thank you so much! – monoxide Feb 16 '20 at 09:11
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Hint. Use the fact that $$ -2 + 2i = 2\sqrt 2\cos \frac{3\pi}4 + 2i\sqrt 2 \sin \frac{3\pi}4 = 2 \sqrt 2 e^{3\pi i/4}.$$

giobrach
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Hint: $(2i-2)^{38}=2^{38}(i-1)^{38}$ and $(i-1)^2=-2i$. So de Moivre’s formula is not required.

Pythagoras
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