Question

write \(\large (1-i)^{2i}\) as of the form x+iy

Answer 1

@ganeshie8 wanna try ?

Answer 2

@Loser66

Answer 3

\[\large (1-i)^{2i} = e^{2i \ln (1-i)}\]

Answer 4

how ?and sure :O
when there is i im not sure of log or ln

Answer 5

lets hope it works... :) we can check wid wolfram in the end maybe

Answer 6

log z = ln r +i ( theta +2k pi)

Answer 7

ok

Answer 8

i never took complex analysis so idk from where u pulled up that formula

Answer 9

oh ok got it :)

Answer 10

from the notes there is box full of formulas xD

Answer 11

okay looks like the log thingy works :
http://www.wolframalpha.com/input/?i=%281-i%29%5E%282i%29++%3D+e%5E%282i*ln%281-i%29%29

Answer 12

ok find somthing
log z = ln |z| + i arg z

Answer 13

ok go ahead
im doing this with music
https://www.youtube.com/watch?v=fSKCJrh_VRc

Answer 14

\[\large (1-i)^{2i} = e^{2i \ln (1-i)} = e^{2i \ln \left(\sqrt{2}e^{i(\frac{-\pi}{4})}\right)}\]

Answer 15

fine so far ?

Answer 16

no , log e^z \(\neq\) z :\

Answer 17

but
e^log z = z

Answer 18

why not

Answer 19

one of the properties

Answer 20

log e^z =z + i2k pi

Answer 21

well we're using the second property, so we should be fine ?

Answer 22

wolfram is fine wid this

http://www.wolframalpha.com/input/?i=%281-i%29%5E%7B2i%7D+%3D+e%5E%7B2i+%5Cln+%281-i%29%7D++%3D++e%5E%7B2i+%5Cln+%5Cleft%28%5Csqrt%7B2%7De%5E%7Bi%28%5Cfrac%7B-%5Cpi%7D%7B4%7D%29%7D%5Cright%29%7D

Answer 23

ok keep going :O
but i dont trust walfram sometimes

Answer 24

\[\large (1-i)^{2i} = e^{2i \ln (1-i)} = e^{2i \ln \left(\sqrt{2}e^{i(\frac{-\pi}{4})}\right)} = e^{2i \left[\ln \sqrt{2} - \frac{i\pi}{4}\right] }\]

Answer 25

nw we have to convert to x+iy

Answer 26

yes break them

Answer 27

r=2
theta=the rest ?

Answer 28

break it into two using product property of exponents : \(a^{m+n} = a^m.a^n\)

Answer 29

O.O

Answer 30

\[\large (1-i)^{2i} = e^{2i \ln (1-i)} = e^{2i \ln \left(\sqrt{2}e^{i(\frac{-\pi}{4})}\right)} = e^{2i \left[\ln \sqrt{2} - \frac{i\pi}{4}\right] } \]
\[\large = e^{2i \ln \sqrt{2} } . e^ {2i\frac{-i\pi}{4}} = e^{i \ln 2 } . e^ {\frac{\pi}{2}} \]

Answer 31

ok then ?

Answer 32

we're done

Answer 33

\[\large (1-i)^{2i} = e^{2i \ln (1-i)} = e^{2i \ln \left(\sqrt{2}e^{i(\frac{-\pi}{4})}\right)} = e^{2i \left[\ln \sqrt{2} - \frac{i\pi}{4}\right] } \]
\[\large = e^{2i \ln \sqrt{2} } . e^ {2i\frac{-i\pi}{4}} = e^{i \ln 2 } . e^ {\frac{\pi}{2}} \]
\[ = e^ {\frac{\pi}{2}} \left[ \cos(\ln 2 ) + i\sin (\ln 2) \right] \]

Answer 34

ok lol
i was confused , but nw only i got it xD
ty !