Solve for z : sqrt3 e^(iπ/4)· z=2sqrt3 e^(i7π/12)

Solve for $z$: $\ \sqrt3\,e^{i\pi/4}\cdot z=2\sqrt3\,e^{i7\pi/12}$.

Step 1 · Isolate $z$

$$ z=\frac{2\sqrt3\,e^{i7\pi/12}}{\sqrt3\,e^{i\pi/4}} $$

Step 2 · Divide moduli, subtract arguments

$$ z=\frac{2\sqrt3}{\sqrt3}\cdot e^{i\left(\frac{7\pi}{12}-\frac\pi4\right)}=2\,e^{i\left(\frac{7\pi}{12}-\frac{3\pi}{12}\right)}=2\,e^{i\frac{4\pi}{12}}=2\,e^{i\pi/3} $$

Step 3 · Convert to rectangular form

$$ z=2\left(\cos\frac\pi3+i\sin\frac\pi3\right)=2\left(\frac12+i\frac{\sqrt3}2\right)=1+i\sqrt3 $$

Step 4 · Verify

Substituting back: $\sqrt3\,e^{i\pi/4}\cdot2e^{i\pi/3}=2\sqrt3\,e^{i(\pi/4+\pi/3)}=2\sqrt3\,e^{i7\pi/12}$, matching the right-hand side exactly (checked numerically to full precision as well).