**Question**: Using argument principle along line on the picture, calculate the number of zeroes of the following function in the left half-plane:

$z+a=e^z, (a>0)$

(The graph will be attached.)

(Plus: I think Professor Victor originally wants to give us $a>1$ rather than $a>0$?)

**Solution**:

$\begin{align}

h(iy)&=iy+a-e^{iy}\\

\nonumber &=iy+a-cos(y)-isin(y)\\

\nonumber &=(a-cos(y))+i(y-sin(y))

\end{align}

$

If $a>0$, we cannot conclude anything for $\text{Re} h(iy)$, but if $a>1$, then $\text{Re} h(iy)$ is always positive because the range of $cos(y)$ is consistently from $-1$ to $1$.

$\text{Im}h(iy)$ will increase when $y$ goes from $-R$ to $R$.

$\begin{align}

h(\text{Re}^{it})=\text{Re}^{it}+a=e^{\text{Re}^{it}}

\end{align}

$

Where t is from $\frac{\pi}{2}$ to $\frac{3\pi}{2}$, and $z$ goes from $iR$ to $-iR$. $h(z)$ in this circumstance has been travelled a counterclockwise circuit.

Therefore, the argument for $h(z)$ should be $2\pi$.

By The Argument Principle, $\frac{1}{2\pi}\cdot 2\pi=1$.

Hence, $h(z)$ has a total of one zero in this plane.