Why does \(e^z=0\) have no complex solution?

Give the general solution of \(e^z=1\).

Solve \(e^z=-1\).

Solve \(e^z=e^3\).

Solve \(e^z=i\).

Solve \(e^z=1+i\).

Solve \(e^{2z}=1\).

Solve \(e^{z+1}=-1\).

Solve \(\sinh z=0\).

Solve \(\cos z=0\).

Solve \(\sin z=0\) using exponentials.

Verify that every \(z=2\pi ik\) solves \(e^z=1\).

Solve \(e^{3z}=8\).

Solve \(e^{z}=2e^{i\pi/3}\).

For which integers \(k\) does \(z=\pi ik\) solve both \(e^{2z}=1\) and \(e^z=1\)?

Find all \(z\) satisfying \(e^z=\overline{e^z}\).

Solve \(\cosh z=0\).

A learner solves \(e^z=1\) and gives only \(z=0\). Explain the error.

Explain why the fundamental theorem of algebra gives no count for the solutions of \(e^z=1\).

Show that \(e^{az}=1\), with non-zero complex \(a\), has infinitely many solutions and give them.