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??

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??

anonymous · Answer

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

anonymous · Answer

I got, -500-500isaqrt(3)

anonymous · Answer

That's correct.

anonymous · Answer

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

anonymous · Answer

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.

anonymous · Answer

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.

anonymous · Answer

Sure

anonymous · Answer

yw