Find all complex numbers \(z = x + iy\), \((x, y \in \mathbb{R})\), such that \(e^z = 2i\).
<1>2. If \(e^z=2i\), then \(e^x=2\).
<1>3. The argument condition is \[ y=\frac\pi2+2\pi k, \qquad k\in\mathbb Z. \]
<1>4. Therefore all solutions are \[ \boxed{z=\ln2+i\left(\frac\pi2+2\pi k\right),\qquad k\in\mathbb Z.} \]
Solution
1. Write \(z=x+iy\) with \(x,y\in\mathbb R\). Then\[ e^z=e^x(\cos y+i\sin y). \]
Taking absolute values gives \[ e^x=|e^z|=|2i|=2. \] Hence \(x=\ln2\).
After 2, the equation becomes \[ \cos y+i\sin y=i. \] Thus \(\cos y=0\) and \(\sin y=1\), which occurs exactly when \(y=\pi/2+2\pi k\).
Every such \(z\) satisfies \[ e^z=2e^{i(\pi/2+2\pi k)}=2i, \] and <1>2–<1>3 show that no other solutions are possible.