Complex number plane (eulers formula)

[Quasimoto]

Dear forum users,

I have been struggling with a problem for a while.
I may use Euler's formula but I have to write it in the (x+yi) form



I know how to solve it in the form of something like;



which would be the following;



But for some reason I keep getting the wrong answer for my initial problem, which should be

-2 471 241 472 - 378 107 904 i

If there is something that is not clear or if I forgot something to mention please do say so.
I really hope someone can help me out.

Thanks in advance.

 

[topsquark]

We can't figure out what you did wrong unitl you show us what you did! How did you get your answer?

-Dan

 

[Quasimoto]

I got my answer from my book, and also I figured it out just a few minutes ago. so Ill try to delete this thread thanks anyway!

 

[MarkFL]

I am assuming you meant for the expression to be:

[MATH](-3+4i)^6(-2-4i)^8[/MATH]
I would begin by writing:

[MATH](-3+4i)^6(2+4i)^8=2^8(1+2i)^{20}[/MATH]

 

[pka]

I have been struggling with a problem for a while.
I may use Euler's formula but I have to write it in the (x+yi) form
View attachment 17184

I would rewrite \((-3+4i)^6\cdot(-2-4i)^8=2^8(-3+4i)^6\cdot(1+2i)^8\)
Using Euler's formula we need the arguments: \(\arg(-3+4i)=\pi-\arctan\left(\dfrac{3}{4}\right)=\theta~\&~\arg(1+2i)=\arctan\left(2\right)=\gamma\)
Now we write \((-3+4i)=5\exp(i\theta)~\&~(1+2i)=\sqrt{5}\exp(i\gamma)\)
Thus \((-3+4i)^6=5^6\exp(5i\theta)~\&~(1+2i)^8=(\sqrt{5})^8\exp(8 i\gamma)\)

Some people will prefer this