Prove that |tanhdfracπ(1+i)4|=1

Prove that $\ \left|\tanh\dfrac{\pi(1+i)}4\right|=1$.

Step 1 · Set up the two moduli

With $x=y=\frac\pi4$:

$$ |\sinh z|^2=\sinh^2\frac\pi4+\sin^2\frac\pi4, \qquad |\cosh z|^2=\cosh^2\frac\pi4-\sin^2\frac\pi4 $$

Step 2 · Show the two are equal

Equality $|\sinh z|^2=|\cosh z|^2$ requires:

$$ \sinh^2\frac\pi4+\sin^2\frac\pi4=\cosh^2\frac\pi4-\sin^2\frac\pi4 \ \Longleftrightarrow\ 2\sin^2\frac\pi4=\cosh^2\frac\pi4-\sinh^2\frac\pi4=1 $$

using $\cosh^2-\sinh^2=1$. Since $\sin\frac\pi4=\frac{\sqrt2}2$: $\ 2\left(\frac{\sqrt2}2\right)^2=2\cdot\frac12=1$ ✓ — exactly true.

Step 3 · Conclude

Since $|\sinh z|^2=|\cosh z|^2$ and both are positive:

$$ \boxed{\left|\tanh\frac{\pi(1+i)}4\right|=\frac{|\sinh z|}{|\cosh z|}=1} $$

Verification · direct exponential computation

With $e^{i\pi/4}=\frac{\sqrt2}2(1+i)$: $\ e^z=e^{\pi/4}\cdot\frac{\sqrt2}2(1+i),\ e^{-z}=e^{-\pi/4}\cdot\frac{\sqrt2}2(1-i)$. Combining, $\ e^z-e^{-z}=\sqrt2(\sinh\frac\pi4+i\cosh\frac\pi4)$ and $\ e^z+e^{-z}=\sqrt2(\cosh\frac\pi4+i\sinh\frac\pi4)$, so:

$$ \tanh z=\frac{\sinh\frac\pi4+i\cosh\frac\pi4}{\cosh\frac\pi4+i\sinh\frac\pi4} \quad\Longrightarrow\quad |\tanh z|=\frac{\sqrt{\sinh^2\frac\pi4+\cosh^2\frac\pi4}}{\sqrt{\cosh^2\frac\pi4+\sinh^2\frac\pi4}}=1 $$

Numerically: $\tanh\!\left(\frac{\pi(1+i)}4\right)\approx0.91715+0.39854i$, whose modulus computes to exactly $1.00000$.