Loser66
  • Loser66
find real and imaginary parts of (1-i)^i Please, help
Mathematics
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
rvc
  • rvc
consider Z=(1-i)^i take log
Loser66
  • Loser66
@SithsAndGiggles if we do that way, we don't have imaginary part, right? @rvc I did and got stuck
anonymous
  • anonymous
Hold on, those rules don't hold up....

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More answers

Loser66
  • Loser66
we know that \(z^a = e^{alogz}\)
Loser66
  • Loser66
\((1-i)^i = e^{ilog(1-i)}\)
Loser66
  • Loser66
and log (1-i) = ln|1-i| + i arg (-pi/4)
rvc
  • rvc
\[\rm LogZ=i(\log(1-i))\\~~~~~=i(1/2\log2+i\pi/4)\]
myininaya
  • myininaya
have you tried writing 1-i in polar form ? r*e^(i theta) form is what I'm talking about
rvc
  • rvc
oops its negative pi/4
rvc
  • rvc
THen multiply the i
Loser66
  • Loser66
but ln|1-i| = ln (sqrt 2), right?
Loser66
  • Loser66
then, \(e^{ilog(1-i)}= e^{i (ln\sqrt 2+i(-\pi/4))}\) expand it, I got a complicated answer :(
rvc
  • rvc
\[\rm ~Z=(1-i)^i \\ logZ=i(\log[1-i]) \\ logZ=i(\log \sqrt2-i\pi/4) \\logZ=\pi/4+ilog \sqrt2\\Z=e^{\pi/4} \cdot e^{ilog \sqrt2} \]
rvc
  • rvc
now write e^ix = cos(x)+isin(x)
Loser66
  • Loser66
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 :(
rvc
  • rvc
\[\log(a+ib)=\log(r)+i(\theta) \\ where~r= \sqrt{a^2+b^2}~\ \theta= \tan^{-1}(b/a)\]
Kainui
  • Kainui
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. ---- 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 \[z = \Re\{z\} + i \Im\{z\}\]
rvc
  • rvc
whats the question?
Loser66
  • Loser66
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
rvc
  • rvc
hmmm i have no idea about that :/
rvc
  • rvc
@ganeshie8 please help :)
Loser66
  • Loser66
Thank you @rvc
rvc
  • rvc
just a sec please im asking others to help :)
ganeshie8
  • ganeshie8
\(c\) is correct but \(d\) may not hold always
Loser66
  • Loser66
c is correct iff the real of z >0 since log z is undefined on negative value of x axis.
Loser66
  • Loser66
c is false since \(z^a = e^{alogz}\\z^b = e^{blogz}\\z^az^b = e^{(a+b) logz} \neq e^{(a+b)}\)
rvc
  • rvc
z^a is the answer after separating the imaginary part n real?
rvc
  • rvc
@Loser66 ?
Loser66
  • Loser66
what do you mean?
Loser66
  • Loser66
Actually, I got most of them. Now I am working on radius of convergence of Taylor series now. :)
rvc
  • rvc
ohok

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