= Solution
Using the complex-angle identity,
$$
\cos(x+iy)=\cos x\cosh y-i\sin x\sinh y.
$$
Since $\cosh y>0$, the real part can vanish only when $x=\pi/2+k\pi$. Then the imaginary part requires $\sinh y=(-1)^{k+1}$. Therefore all roots are
$$
\boxed{z=\frac\pi2+k\pi+i(-1)^{k+1}\ln(1+\sqrt2),\qquad k\in\mathbb Z.}
$$
Substitution gives $\cos z=i$ for every integer $k$, and the positivity of $\cosh y$ shows that no other real parts were missed. The imaginary offsets are the two values $\pm\operatorname{arsinh}1$.
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