Find limlimits_z→fracπ i2z^2coshdfrac4z3

Find $\ \lim\limits_{z\to\frac{\pi i}2}z^2\cosh\dfrac{4z}3$.

Step 1 · Compute $z_0^2$

$$ z_0^2=\left(\frac{\pi i}2\right)^2=\frac{\pi^2i^2}4=-\frac{\pi^2}4 $$

Step 2 · Reduce the cosh argument

$$ \frac{4z_0}3=\frac{4}{3}\cdot\frac{\pi i}2=\frac{2\pi i}3=i\cdot\frac{2\pi}3 $$

Step 3 · Evaluate $\cosh$ at a purely imaginary argument

Using $\cosh(iy)=\cos y$ for real $y$ (from $\cosh(iy)=\frac{e^{iy}+e^{-iy}}2=\cos y$):

$$ \cosh\frac{2\pi i}3=\cosh\left(i\cdot\frac{2\pi}3\right)=\cos\frac{2\pi}3=-\frac12 $$

Step 4 · Combine

$$ \boxed{z_0^2\cosh\frac{4z_0}3=\left(-\frac{\pi^2}4\right)\left(-\frac12\right)=\frac{\pi^2}8} $$

Verification

Direct expansion via Euler's formula: $\ \cosh\frac{2\pi i}3=\frac{e^{2\pi i/3}+e^{-2\pi i/3}}2=\frac{\left(-\frac12+i\frac{\sqrt3}2\right)+\left(-\frac12-i\frac{\sqrt3}2\right)}2=\frac{-1}2=-\frac12$, matching Step 3. Numerically, $z_0^2\cosh(4z_0/3)\approx1.23370$, and $\pi^2/8\approx1.23370$ ✓.