Write cot((π)/(4)-iln2) in the form u+iv
Write $\cot\left(\dfrac{\pi}{4}-i\ln2\right)$ in the form $u+iv$.
Explanation of Method
Use $\cot(x+iy)=\dfrac{\sin(2x)-i\sinh(2y)}{\cosh(2y)-\cos(2x)}$, and evaluate the hyperbolic terms exactly using $e^{\ln2}=2$.
Step 1 · Apply with $x=\frac{\pi}4,\ y=-\ln2$
$$ \cot\left(\frac\pi4-i\ln2\right)=\frac{\sin\left(\frac\pi2\right)-i\sinh(-2\ln2)}{\cosh(-2\ln2)-\cos\left(\frac\pi2\right)}=\frac{1+i\sinh(2\ln2)}{\cosh(2\ln2)} $$
(using $\sin\frac\pi2=1,\ \cos\frac\pi2=0$, and $\sinh,\cosh$ odd/even.)
Step 2 · Evaluate $\sinh(2\ln2)$ and $\cosh(2\ln2)$ exactly
Since $e^{2\ln2}=2^2=4$ and $e^{-2\ln2}=2^{-2}=\frac14$:
$$ \sinh(2\ln2)=\frac{4-\frac14}2=\frac{15}8, \qquad \cosh(2\ln2)=\frac{4+\frac14}2=\frac{17}8 $$
Step 3 · Substitute and simplify
$$ \cot\left(\frac\pi4-i\ln2\right)=\frac{1+i\frac{15}8}{\frac{17}8}=\frac{8+15i}{17}=\frac8{17}+\frac{15}{17}i $$
Step 4 · Result
$$ \boxed{u=\frac8{17}, \qquad v=\frac{15}{17}} $$
Verification
Direct numerical evaluation of $\cot(0.785398-0.693147i)$ gives $0.470588\ldots+0.882353\ldots i$, which matches $\frac8{17}\approx0.470588$ and $\frac{15}{17}\approx0.882353$ exactly.
Result
An unusually clean result — the use of $\ln2$ as the imaginary part makes $e^{\pm2\ln2}$ rational, so the $8$–$15$–$17$ Pythagorean triple falls straight out of the hyperbolic evaluation.