OpenStudy (anonymous):

(2-2i)^5

2 years ago
OpenStudy (anonymous):

write in trig form would be easiest

2 years ago
OpenStudy (anonymous):

you know how to do that?

2 years ago
OpenStudy (anonymous):

I dont think so

2 years ago
OpenStudy (anonymous):

you need two number \[r=\sqrt{a^2+b^2}\] and also \(\theta\)

2 years ago
OpenStudy (anonymous):

in your case \[r=\sqrt{2^2+2^2}=\sqrt8=2\sqrt2\]

2 years ago
OpenStudy (anonymous):

|dw:1416365862786:dw|

2 years ago
OpenStudy (anonymous):

So \[\sqrt{8}\] and then if we are working in radians would the angle be 7pi/4 ?

2 years ago
OpenStudy (anonymous):

\(\theta\) you have choices if you are working in radians it would \(\frac{7\pi}{4}\) right

2 years ago
OpenStudy (anonymous):

What is next?

2 years ago
OpenStudy (anonymous):

then it is \[2\sqrt2\left(\cos(\frac{7\pi}{4})+i\sin(\frac{7\pi}{4})\right)\]

2 years ago
OpenStudy (anonymous):

then multiply the angle by 5

2 years ago
OpenStudy (anonymous):

and also find \((2\sqrt2)^5\)

2 years ago
OpenStudy (anonymous):

\[(2\sqrt2)^5=128\sqrt2\]

2 years ago
OpenStudy (anonymous):

\[2\sqrt2\left(\cos(\frac{7\pi}{4})+i\sin(\frac{7\pi}{4})\right)]^5=128\sqrt2\left(\cos(\frac{35\pi}{4})+i\sin(\frac{35\pi}{4})\right)\]

2 years ago
OpenStudy (anonymous):

right, but my answers are A) 10-10i B) 32-32i C)32-10i D) -128+128i

2 years ago
OpenStudy (anonymous):

then convert back

2 years ago
OpenStudy (anonymous):

once you convert back you will no doubt get answer D

2 years ago
OpenStudy (anonymous):

since \(\cos(\frac{35\pi}{4})=-\frac{1}{\sqrt2}\) etc

2 years ago
OpenStudy (anonymous):

Thank you

2 years ago
OpenStudy (anonymous):

yw

2 years ago