Question

(2-2i)^5

Answer 1

write in trig form would be easiest

Answer 2

you know how to do that?

Answer 3

I dont think so

Answer 4

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

Answer 5

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

Answer 6

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

Answer 7

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

Answer 8

What is next?

Answer 9

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

Answer 10

then multiply the angle by 5

Answer 11

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

Answer 12

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

Answer 13

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

Answer 14

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

Answer 15

then convert back

Answer 16

once you convert back you will no doubt get answer D

Answer 17

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

Answer 18

Thank you

Answer 19

yw