Question

“not right but left half plane.” complex analysis

CC be defined by or each real number a, let fo : Prove that if a > 1, then fa has exactly one zero in the right half-plane E C:Ra)o and that this zero is a real number

Answer 1

Let $z=x+iy$ be a zero of the function on left half-plane; we want to show that $y=0$.

Suppose that $y\neq 0$; we will derive a contradiction. Since

$0=(x+iy)+\alpha-e^xe^{iy}=(x+\alpha-e^x\cos y)+i(y-e^x\sin y)$

which implies $x+\alpha=e^x\cos y$ , and

$e^{-x}={\frac{\sin y}y}$

Since $z=x&plus;iy$ is on left half-plane, we know $x<0$, so that $e^{-x}>1$. Thus, the second equality above implies

${\frac{\sin y}y}>1\hspace{2cm}(\ast)$

Now, because $|\sin y|\leq1$, the inequality above is possible only if $|y|\leq1$.

Consider the function $g(y)=\sin y-y$ where $y\in (0,1)$. We have $g'(y)=\cos y-1$ . Thus, for $y\in (0,1)$, we have $g'(y)=\cos y-1<0$ , which implies that $g(y)=\sin y-y$ is decreasing for $y\in (0,1)$. Since

$\lim_{y\rightarrow0}(\sin y-y)=0$

this means that $\sin y-y\leq 0$ for $y\in (0,1)$, implying

${\frac{\sin y}y}\not>1$

when $y\in (0,1)$.

Consider the function $g(y)=\sin y-y$ where $y\in (-1,0)$. We have $g'(y)=\cos y-1$ . Thus, for $y\in (-1,0)$, we have $g'(y)=\cos y-1<0$ , which implies that $g(y)=\sin y-y$ is decreasing for $y\in (-1,0)$. Since

$\lim_{y\rightarrow0}(\sin y-y)=0$

this means that $\sin y-y\geq 0$ for $y\in (-1,0)$. Thus, $-\sin(-y)&plus;(-y)\geq 0$ for $-y\in (0,1)$ , implying

$0<{\frac{\sin (-y)}{-y}}\leq1$

Thus,

${\frac{\sin y}{y}}\leq1$

Therefore, we have shown that if $y\neq 0$ then

${\frac{\sin y}{y}}\not>1$

when $y\in (0,1)$​​​​​​​. This contradicts $(\ast)$, as desired.

Thus, we have proved that if $y\neq 0$ then $z=x&plus;iy$ in left half-plane can not be a zero of

$f_\alpha(z)=z&plus;\alpha-e^z$

for any real $\alpha$. Thus, the only root of this function (if any) must be real $z=x&plus;i0=x$. Hence, it suffices to show that the real variable function $f_\alpha(x)=x&plus;\alpha-e^x$ has a unique solution in $(-\infty,0)$ if $\alpha>1$.

We have $f'_\alpha(x)=1-e^x>0$ in $(-\infty,0)$, showing that $f_\alpha(x)=x&plus;\alpha-e^x$ is increasing; this ensures that in $(-\infty,0)$ the function $f_\alpha(x)=x&plus;\alpha-e^x$ can have at most one root.

Note that

$\lim_{x\rightarrow-\infty}f_{\alpha}(x)=\lim_{x\rightarrow-\infty}x&plus;\alpha-\lim_{x\rightarrow-\infty}e^x=-\infty&plus;\alpha-0=-\infty$

and

$\lim_{x\rightarrow0}f_{\alpha}(x)=0&plus;\alpha-1=\alpha-1>0$

In particular, the function $f_\alpha(x)=x&plus;\alpha-e^x$ takes both positive and negative values in $(-\infty,0)$. Since $f_\alpha(x)=x&plus;\alpha-e^x$ is continuous, and takes both positive and negative values in $(-\infty,0)$, it has to take the value $0$ at some point in $(-\infty,0)$. Thus, the function must have a root in $(-\infty,0)$.

This completes the proof.