## anonymous one year ago Use DeMoivre's Theorem to find the indicated power of the following complex number. (-3 + 3i)⁴ My work r = √18 and tan θ = -1 so θ = 135° because the complex number is in Quadrant II. √18 [cos(135°) + i sin(135°)] (√18)⁴ [cos(4 × 135°) + i sin(4 × 135°)] 324[cos(540°) + i sin(540°)] 324(-1 + i × 0) 324(-1) -324

1. anonymous

I just need to know if I am correct

2. zepdrix

Hmm I took a very different approach, got the same result though. Yay good job, looks correct! :)

3. zepdrix

https://www.wolframalpha.com/input/?i=%28-3%2B3i%29%5E4 Just in case you need to verify ^

4. anonymous

If you dont mind, can you show me how you did it?

5. anonymous

either way thanks a lot :)

6. zepdrix

My method is a little strange. I factor a 3 out to start.$\large\rm 3^4(-1+i)^4$I recognize that the magnitude of the real and imaginary parts are the same, that only happens at the pi/4 angles. So we need to turn these into sqrt(2)/2's. So I'll factor a sqrt(2) out of each term.$\large\rm 3^4(\sqrt{2})^4\left(-\frac{\sqrt2}{2}+\frac{\sqrt2}{2}i\right)^4$And as you said, this is in quadrant 2, so it's 3pi/4.$\large\rm 3^4(\sqrt{2})^4\left(\cos\frac{3\pi}{4}+i \sin\frac{3\pi}{4}\right)^4$Then, De'Moivre gives us,$\large\rm 3^4(\sqrt{2})^4\left(\cos3\pi+i \sin3\pi\right)$Sine is 0 at the 3pi angle, cosine -1,$\large\rm 3^4(\sqrt{2})^4\left(-1+0\right)$And then just simplify a little further to get the same result. I know, I know, my method is a little goofy :) But I like it lol

7. anonymous

Hahaha thanks a lot man, whatever gets the job done :P