Question

Am I right?

[sqrt(10)(cos(40)+i sin(40)]^6

I need to put it in a+bi format

but I just keep getting 1000??

Answer 1

From DeMoivre's theorem, you get
\[\left(\sqrt{10}(\cos40+i\sin40)\right)^6=10^3(\cos240+i\sin240)\]
Check your sine/cosine again.

Answer 2

I got, -500-500isaqrt(3)

Answer 3

That's correct.

Answer 4

that's the answer? It doesn't have to be positive then?

Answer 5

That's right, and no, if you recall the unit circle, an angle of \(240^\circ\) will land you in the third quadrant, which contains points with negative \(x\) and \(y\) components.

Answer 6

Thank you, do you think you can also help me in another question? I'll open a new thread so I can give you another medal.

Answer 7

Sure

Answer 8

yw