3

Calculate

i)$(1+i)^i$

ii)$(-1)^{\frac{1}{\pi}}$

I did

i)$(1+i)=\sqrt{2}e^{i\frac{\pi}{4}}$. Knowing that if $z$ and $c$ are complex numbers $z^c=e^{c\log z}$ $$(1+i)^i=e^{i\log(1+i)}=i\log(1+i)=i\log(\sqrt{2}e^{i\frac{\pi}{4}})=i\log\sqrt{2}-\frac{\pi}{4}$$

That's right?

What property should I use in item 2?

I again tried to question two and got it

$(-1)=e^{i\pi}$ $$(-1)^\frac{1}{\pi}=e^{\frac{1}{\pi}\log(e^{i\pi})}=\frac{1}{\pi}log(e^{i\pi)}=\frac{1}{\pi}i\pi=i$$

Roland
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  • should you not have $1 + i = \sqrt 2 e^{i(\pi/4 + 2k\pi)}$ instead with $k$ any integer. – abel Mar 27 '15 at 18:52
  • @abel I think this would not be necessary. How can I solve the second item? – Roland Mar 27 '15 at 18:57
  • Depends, @abel. Defining $x^y=e^{y\log x}$, some definitions use some principle value of $\log x$ and others consider a multi-valued definition. Multivalued is probably the more mathematically sound method, for a variety of reasons. – Thomas Andrews Mar 27 '15 at 18:57
  • @ThomasAndrews thanks . i suppose it an agreed upon convention. – abel Mar 27 '15 at 19:00
  • @askazy I really want to help here, so please let me know if there is any way I can improve my answer. I just want to provide the best answer I can give you. – Mark Viola Mar 27 '15 at 19:41
  • You wrote $$e^{i\log(1+i)} = i\log(1+i)$$ which is clearly wrong – Dylan Mar 28 '15 at 03:49

2 Answers2

2

For the first case, $(1+i)^i$ we may write $$\begin{align} (1+i)^{i}&= (\sqrt{2}e^{i(\pi /4 + 2n\pi)})^{i} \\ &=\left(\sqrt{2}\right)^{i}e^{-(\pi /4+2n\pi)}\\ &=e^{-(\pi /4+2n\pi)}\left(e^{\log(\sqrt{2})}\right)^{i}\\ &=e^{-(\pi /4+2n\pi)}e^{i\log(\sqrt{2})}\\ &=e^{-(\pi /4+2n\pi)}\left(\cos(\log(2)/2)+i\sin(\log(2)/2)\right) \end{align}$$

where $n$ is any integer ($n=0$ on the principal branch).

For the second case, $(-1)^{1/\pi}$ we may write $$\begin{align} (-1)^{1/\pi}&= (e^{i(2n+1)\pi})^{1/\pi}\\ &=e^{i(2n+1)}\\ &=\cos(2n+1)+i\sin(2n+1) \end{align}$$

where $n$ is any integer ($n=0$ on the principal branch).

Mark Viola
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1

let me see if $$z = (-1)^{1/\pi} \to \ln z = \frac 1{\pi} \ln(-1) = \frac 1{\pi} \ln (e^{i\pi})=\frac 1{\pi}(i\pi) = i $$ works. yes, it does. therefore, $$ z = e^{i}=\cos 1 + i \sin 1.$$

abel
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