The real and imaginary parts of this complex number

Question

The number is $(i+1)^{(i-1)}$

I tried but couldn't solve it.

Answer 1

\begin{align*} \mathrm{i}+1 &= \sqrt{2} \mathrm{e}^{\mathrm{i}\pi/4} \\ &= \mathrm{e}^{\log \sqrt{2}} \mathrm{e}^{\mathrm{i}\pi/4} \\ &= \mathrm{e}^{\frac{1}{2}\log 2} \mathrm{e}^{\mathrm{i}\pi/4} \\ (\mathrm{i}+1)^{1-\mathrm{i}} &= \left( \mathrm{e}^{\frac{1}{2}\log 2} \mathrm{e}^{\mathrm{i}\pi/4} \right)^{1-\mathrm{i}} \\ &= \mathrm{e}^{ \left( \frac{1}{2}\log 2 \right)(1-\mathrm{i})}\mathrm{e}^{(\mathrm{i}\pi/4)(1-\mathrm{i})} \\ &= \mathrm{e}^{\left( \frac{1}{2}\log 2 \right)}\mathrm{e}^{-\mathrm{i} \left( \frac{1}{2}\log 2 \right)}\mathrm{e}^{\mathrm{i}\pi/4}\mathrm{e}^{-\mathrm{i}^2\pi/4} \\ &= \sqrt{2}\mathrm{e}^{\mathrm{i} \left( -\frac{\log 2}{2} \right)}\frac{1+\mathrm{i}}{\sqrt{2}}\mathrm{e}^{\pi/4} \\ &= \mathrm{e}^{\pi/4} (1+\mathrm{i})\left(\cos \left( -\frac{\log 2}{2} \right) + \mathrm{i} \sin \left( -\frac{\log 2}{2} \right)\right)\\ &= \mathrm{e}^{\pi/4} (1+\mathrm{i})\left(\cos \log \sqrt{2} - \mathrm{i} \sin \log \sqrt{2} \right)\\ &= \mathrm{e}^{\pi/4} \left(\cos \log \sqrt{2} - \mathrm{i} \sin \log \sqrt{2} + \mathrm{i} \cos \log \sqrt{2} - \mathrm{i}^2 \sin \log \sqrt{2} \right)\\ &= \mathrm{e}^{\pi/4} \left(\cos \log \sqrt{2} + \sin \log \sqrt{2} + \mathrm{i} \cos \log \sqrt{2} - \mathrm{i} \sin \log \sqrt{2} \right) \end{align*} from which we directly read off the real part $$ \mathrm{e}^{\pi/4} \left(\cos \log \sqrt{2} + \sin \log \sqrt{2}\right) $$ and imaginary part $$ \mathrm{e}^{\pi/4} \left(\cos \log \sqrt{2} - \sin \log \sqrt{2}\right) $$

Answer 2

$$(1+i)^{-1+i}=\left(e^{\text{Log}(1+i)}\right)^{-1+i}=e^{(-1+i)\text{Log}(1+i)}$$

Now let $\displaystyle1=r\cos\phi, 1=r\sin\phi$ where $r>0$

Squaring & adding we get $\displaystyle(r\cos\phi)^2+(r\sin\phi)^2=2\implies r^2=2\implies r=\sqrt2$

$\displaystyle\implies\sin\phi=\cos\phi=\frac1{\sqrt2}\implies\phi=2n\pi+\frac\pi4 $ where $n$ is any integer

$\displaystyle\implies1+i=\sqrt2\left[\cos\left(2n\pi+\frac\pi4\right)+i\sin\left(2n\pi+\frac\pi4\right)\right]=\sqrt2e^{i\left(2n\pi+\dfrac\pi4\right)}$ (Using Euler Formula)

$\displaystyle\implies\text{Log}(1+i)=\log\sqrt2+i\left(2n\pi+\frac\pi4\right)$

$\displaystyle\implies(-1+i)\text{Log}(1+i)=(-1+i)\left[\log\sqrt2+i\left(2n\pi+\frac\pi4\right)\right]$

$\displaystyle=-\frac12\log2-2n\pi+\frac\pi4+i\left(\frac12\log2-2n\pi-\frac\pi4\right)$

$\displaystyle\implies e^{(-1+i)\text{Log}(1+i)}=e^{-\dfrac12\log2-2n\pi+\frac\pi4+i\left(\dfrac12\log2-2n\pi-\dfrac\pi4\right)}$

$\displaystyle=e^{-\dfrac12\log2-2n\pi+\frac\pi4}\text{cis}\left(\dfrac12\log2-2n\pi-\dfrac\pi4\right)$

$\displaystyle=e^{-\dfrac12\log2-2n\pi+\frac\pi4}\text{cis}\left(\dfrac12\log2-\dfrac\pi4\right)$ where $\text{cis}(x)=\cos x+i\sin x$

For the principal value set $n=0$