Question

find real and imaginary parts of (1-i)^i
Please, help

Answer 1

consider Z=(1-i)^i
take log

Answer 2

@SithsAndGiggles if we do that way, we don't have imaginary part, right?
@rvc I did and got stuck

Answer 3

Hold on, those rules don't hold up....

Answer 4

we know that \(z^a = e^{alogz}\)

Answer 5

\((1-i)^i = e^{ilog(1-i)}\)

Answer 6

and log (1-i) = ln|1-i| + i arg (-pi/4)

Answer 7

\[\rm LogZ=i(\log(1-i))\\~~~~~=i(1/2\log2+i\pi/4)\]

Answer 8

have you tried writing 1-i in polar form ?
r*e^(i theta) form is what I'm talking about

Answer 9

oops its negative pi/4

Answer 10

THen multiply the i

Answer 11

but ln|1-i| = ln (sqrt 2), right?

Answer 12

then, \(e^{ilog(1-i)}= e^{i (ln\sqrt 2+i(-\pi/4))}\)
expand it, I got a complicated answer :(

Answer 13

\[\rm ~Z=(1-i)^i \\ logZ=i(\log[1-i]) \\ logZ=i(\log \sqrt2-i\pi/4) \\logZ=\pi/4+ilog \sqrt2\\Z=e^{\pi/4} \cdot e^{ilog \sqrt2} \]

Answer 14

now write e^ix = cos(x)+isin(x)

Answer 15

I want the form z = 1-i because I still have part b, c, d, e to work on that form. If I do your way , z = (1-i)^i, I can't work further :(

Answer 16

\[\log(a+ib)=\log(r)+i(\theta) \\ where~r= \sqrt{a^2+b^2}~\ \theta= \tan^{-1}(b/a)\]

Answer 17

I would just plug the number into this and simplify:
\[\Re\{z\} = \frac{z+z^*}{2}\]\[\Im\{z\} = \frac{z-z^*}{2i}\]


In case you think I'm lying to you, here are some things you can do to prove it to yourself now that you wouldn't need to do ever again once you know that above.

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You can tell that the real part is real because:

\[\Re\{z\}^* = \frac{(z+z^*)^*}{2} = \frac{z+z^*}{2} = \Re\{z\}\]
Similarly purely imaginary numbers obey this, \[\Im\{z\}^* = - \Im\{z\}\]

And check to make sure that \[z = \Re\{z\} + i \Im\{z\}\]

Answer 18

whats the question?

Answer 19

This is the whole question
b) find the domain of analicity of z^n and show that (z^a)'= az^(a-1) where z^(a-1) is the principal branch of z^(a-1)
c) z^az^b = z^(a+b) ?
d) \(z_1^az_2^a = (z_1z_2)^a?
e) find the Taylor series for f(z) = z ^a at z =1. Determine the radius of convergence

Answer 20

hmmm
i have no idea about that :/

Answer 21

@ganeshie8 please help :)

Answer 22

Thank you @rvc

Answer 23

just a sec please
im asking others to help :)

Answer 24

\(c\) is correct

but \(d\) may not hold always

Answer 25

c is correct iff the real of z >0 since log z is undefined on negative value of x axis.

Answer 26

c is false since
\(z^a = e^{alogz}\\z^b = e^{blogz}\\z^az^b = e^{(a+b) logz} \neq e^{(a+b)}\)

Answer 27

z^a is the answer after separating the imaginary part n real?

Answer 28

@Loser66 ?

Answer 29

what do you mean?

Answer 30

Actually, I got most of them. Now I am working on radius of convergence of Taylor series now. :)

Answer 31

ohok