Question

(1-i)^17/(1+isqur3)^9

Answer 1

-1.81161

Answer 2

its an ugly number

Answer 3

write the complex numbers into its trigonometric form and use DeMoivre's law or use the binomial theorem on both

Answer 4

it becomes squr2(cos 315+isin315)^17/2(cos60+isin60)

Answer 5

^9

Answer 6

2^7.5(cos(4815)+isin(4815))=0 ?

Answer 7

does anything seem wrong?

Answer 8

let's see

Answer 9

\[ \large 1-i=\sqrt{2}(\cos(7\pi/4)+i\sin(7\pi/4)) \]
\[ \large 1+i\sqrt{3}=2(\cos(\pi/3)+i\sin(\pi/3)) \]
right?

Answer 10

Correct

Answer 11

2=2^1 squr2=2^0.5

Answer 12

17-9=8

Answer 13

8-0.5=7.5

Answer 14

thats how I thought

Answer 15

then
\[ \large (1-i)^{17}=\sqrt{2}^{17}[\cos(119\pi/4)+i\sin(119\pi/4)] \]
\[ \large =2^8\cdot\sqrt{2}[\cos(14\cdot2\pi+7\pi/4)+i\sin(14\cdot2\pi+7\pi/4)] \]
\[ \large =2^8\cdot\sqrt{2}[\cos(7\pi/4)+i\sin(7\pi/4)] \]

Answer 16

I dont really understand...

Answer 17

\[ \large (1+i\sqrt{3})^9=2^9[\cos(9\pi/3)+i\sin(9\pi/3)] \]
\[ \large =2^9[\cos(3\pi)+i\sin(3\pi)]=2^9[\cos\pi+i\sin\pi] \]

Answer 18

what don't u understand?

Answer 19

the 119pi part

Answer 20

do u know DeMoivre's theorem
\[ \large (\rho(\cos\phi+i\sin\phi))^n=\rho^n(\cos(n\phi)+i\sin(n\phi)) \]
??

Answer 21

Yes but how does that make it 119pi. I got 5355degrees and its not the same

Answer 22

for \(z=1-i\) we have \(\rho=\sqrt{2}\) and \(\phi=7\pi/4\)
right?

Answer 23

So its wrong becouse I wrote in deggres and not pi? becouse 7pi/4=315 wich is what i got

Answer 24

yes

Answer 25

Why do I need to write in pi? Thanks so much for helping me!

Answer 26

it is usually better (actually customary) to use radians instead of degress when dealing with complex numbers

Answer 27

u r welcome