complex numbers inequality proof

Question

Show that if $z$ is real, then |$\frac{e^{iz}}{z^2 + 1}$| $\leq$ $\frac{1}{|z|^2 + 1}$.

This was my working out: let $z$ be real i.e. Im($z$) = 0. Let $z = x +iy$, $x,y$ ∈ $\mathbb{R}$.

So $z = x$ ==> |$z$| = |$x$| = $x$.

LHS = |$\frac{e^{iz}}{z^2 + 1}$|= |$\frac{e^{ix}}{x^2 + 1}$| = $\frac{|e^{iz}|}{|x^2 + 1|}$ = $\frac{1}{|x^2 + 1|}$ = $\frac{1}{x^2 + 1}$ (since $x$ is real) = $\frac{1}{|z|^2 + 1}$ = RHS.

Is this correct because I never had an inequality in my working is it valid to say LHS $\leq$ RHS?

Also, how do I figure out if this still holds when $z$ is not real?

Answer 1

What you did is correct.

This does not hold in general. Take $z=-2i$, for instance.

Answer 2

This is almost correct, except your (unnecessary) claim that $|x|=x$ which only holds for $x\ge 0$.

If $z$ is not real, then you can try to find $z$ such that $|z^2+1|$ is close to zero. Then $\frac1{|z^2+1|}$ will be much larger that $\frac1{|z|^2+1}$, a number that is always at most $1$. Consider what $e^{iz}$ does for such numbers and you should finish finding a counterexample.