Question

Simplify:

${(1+i)}^{(-1+i)}$

Answer 1

Let,

$A=(1+i{)}^{(-1+i)}$

$\log A=(-1+i)\log (1+i)$

$\Rightarrow \log A=(-1+i)\log (\sqrt{2}.e^i\frac{-\pi }{4})$

$=(-1+i)(\log \sqrt{2}+i\frac{-\pi }{4})$

$=(-1+i)(\frac{1}{2}\log 2+i\frac{-\pi }{4})$

$=\frac{-1}{2}\log 2-i\frac{\pi }{4}+\frac{i}{2}\log 2\frac{-\pi }{4}$

$\Rightarrow \log A=\frac{-1}{2}\log 2\frac{-\pi }{4}+i\left(\frac{1}{2}\log 2\frac{-\pi }{4}\right)$

$A=e^{\frac{-1}{2}\log 2\frac{-\pi }{4}+i\left(\frac{1}{2}\log 2\frac{-\pi }{4}\right)}$

$A=2^{\frac{i-1}{2}}.e^{\frac{-\pi }{4}-i\frac{-\pi }{4}}$