1. ## Complex Analysis Confusion!

Hi,

I'm doing revision for an exam in complex analysis at the mo. I've been going through examples and I've found a couple that I've never seen before and have no idea how to do-any help would be much appreciated!

a)
Find all values of the following:
$\log\imath$ and log (i+1)

(Can't figure it out in Latex, I'm only learning!)

b) Let D= C\(-oo,0], and suppose log 1=0
Find (1-i)^4i and [e(-1-ROOT(3)i)/2]^(3PIi)
(Again if someone could show me this in latex it'd be appreciated, I kept getting 1-i^4 when I tried)

c)Given D= C\(-oo,0] and ${\imath}^i$=exp(PI/2), its principle value, show that
(-1+(ROOT(3))i)^(3/2)= -2ROOT2

Please help, I'm so confused. I'm thinking it has to do with the section we did on simply connected domains and isolated singularities of an analytic domain
but I can't find any examples to follow and I'm lost!

Thanks
PS I'm trying with the latex but its slow so sorry about that!

2. Originally Posted by musicmental85
a)
Find all values of the following:
$\log\imath$ and log (i+1)
$\log i = \log |i| + i \arg (i)$ and $\log (i+1) = \log |1+i|+i\arg (1+i)$
Can you finish?

3. ## Complex Analysis

A. For all values of $\log(\imath)$, what we want is, from the definition of the logarithm, all z such that $e^z=\imath$.
This is, of course, $z=\imath(\tfrac{\pi}{2}+2n\pi)$, where n is any integer.
Similarly, note that $|1+\imath|=\sqrt{2}$ and $\arg (1+\imath) \equiv \dfrac{\pi}{4} \pmod{2\pi}$, and use these to find all z such that $e^z=1+\imath=|1+\imath|e^{\imath\arg (1+\imath)}$ .

B. Our branch cut (and log(1)=0) gives us $-\pi < \arg z < \pi$ for any z not on the cut. Thus we use the fact that $z = |z|e^{\imath\arg z}$ for any complex z to get
$z^{a+b\imath} = \left ( |z|e^{\imath\arg z} \right )^{a+b\imath}$
$= |z|^{a+b\imath}e^{(\imath\arg z)(a+b\imath)}$
$= |z|^a e^{\imath b \log |z|} e^{-b \arg z} e^{\imath a \arg z}$

So you can use this (along with basic exponentiation rules like $(ab)^c=a^c b^c$ and $r^{a+b} = r^a r^b$) for $(1 - \imath)^{a\imath}$ and for $\left [ e \left ( -1-\dfrac{\imath\sqrt{3}}{2} \right ) \right ]^{3\pi\imath}$

C. Are you sure that it's $\imath^\imath = e^{\tfrac{\pi}{2}}$ and not $\imath^\imath = e^{-\tfrac{\pi}{2}}$? I dont think the former is possible:
$|\imath| = 1$, so $\imath = e^{\imath \arg \imath}$ and $\imath^\imath = e^{-\arg \imath}$.
Thus $\imath^\imath = e^{\tfrac{\pi}{2}}$ gives $\arg \imath = -\dfrac{\pi}{2}$, but $\arg \imath \equiv \dfrac{\pi}{2} \pmod{2\pi}$.

I hope this helps point you in the right direction.

--Kevin C.

4. Ok, just working on the first time (I'm slow at these)
Do I need the perfect hackers comment?

log i= log|i|+iarg(i)
=log 1 +i(PI/2 +2nPI)
=0 + i(PI/2 +2nPI)

Can you just use the fact that z=|z|e^(iargz) and state answer as z=e^(i)(PI/2+2nPI) for example in the first case?

For the second does the final answer become z=[ROOT(2)][e^(i)(7PI/4 + 2nPI)] ?

In part b)
Does the answer come out something like (1-i)^4i= e^(-4logROOT(2)-4i(7PI/4 +2nPI)) ?

and the second comes out as e^(3PI(-1/ROOT3 +2nPI +ilog e^2)?

In part c) you are right but it is supposed to be (-i)^i=exp(PI/2)
Thanks so much for the help!