rectangle where $\cos{z} =iz$ has exactly one solution Determine a rectangle inside which there is exactly one solution of the equation $\cos{z} = iz$.
I know the following result:
Let $f$ be holomorphic in $\Omega$ with $a \in \Omega$. Let $f(a)= b$ is a value of multiplicity $n$. Then $\exists \epsilon ,\delta > 0$ such that  $\forall \omega \in \dot{D}(b,\epsilon), \exists  z_1, z_2, \ldots, z_n$ with $z_k \neq z_l$ for $k \neq l$ in $\dot{D}(a,\delta)$, where $f(z_1) = f(z_2)= \ldots = f(z_n) = \omega$. (Here the dot signifies a punctured neighborhood).
In the context of the problem this means that there is a neighborhood around every simple zero of $\cos{z}$ in which all the values are assumed only once. This proves the existence of such a rectangle. But in order to determine a particular example, I believe probably we need to use Rouche's theorem or something similar and find estimates along the boundary. I am getting stuck here.It is worth noting that the equation does not have a purely real or purely imaginary solution. Can someone help please. 
 A: Following the suggestion by LutzL, you can try to "box in" the root near $1-i$. A natural idea is to look at the rectangle $0\le x\le \pi$, $-2\le y\le 0$, and apply the argument principle to $f(z)=\cos z-iz$. In real terms, the function is 
$$(x+iy)\mapsto (\cos x \cosh y +y, -\sin x \sinh y-x) \tag{1}$$
You need to show that $\arg f(z)$ increases by $2\pi$ as $z$ traverses the boundary of the rectangle, moving counterclockwise. For this, it suffices to determine how the image of the boundary crosses some particular half-line emanating from $0$. Maybe the positive imaginary semiaxis is convenient for this problem: see below.
When you plug in one of four boundary pieces, the equation becomes a simple parametric curve. The vertical boundaries $x=0$ and $x=\pi$   are mapped to horizontal line segments: the imaginary part of $f$ is constant on each. The horizontal boundary $y=0$ is mapped to a rotated piece of cosine wave. None of these cross the positive imaginary semiaxis. 
When $y=-2$,  the real part of $f$ is a decreasing function of $x$. It turns into zero when  $\cos x=2/\cosh 2$, which is within $x\in (0,\pi/2)$. Since
$$\sin x \sinh 2-x > \frac{2}{\pi} (\sinh 2) x - x >0$$  the image indeed crosses the  positive imaginary semiaxis, moving from right to left.  
(source of the picture below)

A: Use $\cos z\approx 1-\tfrac12 z^2$ for a first estimation of the root location,
$$0=z^2+2iz-2=(z+i)^2-1=(z-1+i)(z+1+i)$$
Try then to use error estimates and, as you said, Rouche's theorem.
