= Solution
The equation requires $\sinh z=\pm i\cosh a$. Write $z=x+iy$, with real $x,y$. Comparing <real parts> and <imaginary parts> using part (b)(i) gives
$$
\sinh x\cos y=0,\qquad \cosh x\sin y=\pm\cosh a.
$$
If $x=0$, the second equality would require $|\sin y|=\cosh a>1$, impossible because $a>0$. Thus $\sinh x\ne0$, forcing $\cos y=0$ and
$$
y=\frac\pi2+k\pi,\qquad k\in\mathbb Z.
$$
Now $|\sin y|=1$, so $\cosh x=\cosh a$. The <hyperbolic cosine> is even and is a <strictly increasing function> on $[0,\infty)$, hence $x=\pm a$. Conversely, substituting these values gives $\sinh(x+iy)=i(-1)^k\cosh a$, so every value found does satisfy the original equation. Therefore the complete set is
$$
\boxed{z=\pm a+i\left(\frac\pi2+k\pi\right),\qquad k\in\mathbb Z.}
$$
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