Complex Variables Inequality

Question

If $\mathrm{Im}(z) > 0$ then $\mathrm{Im}(1/z) < 0$. I know that $z=x+yi$, so then the $\mathrm{Im}(z)$ must equal $y$, which should be greater than zero to exist correct? I'm not really sure if there is a way so show this with the equalities that might make it easier? And then $1/z$ is equal to the conjugate of $z$, so then it must be $-y$, which is less than zero.

Answer 1

If $z=x+yi$, with $x,y\in\mathbb R$, then $\operatorname{Im}z>0$ means that $y>0$. And$$\frac1z=\frac x{x^2+y^2}-\frac y{x^2+y^2}i.$$Besides,$$y>0\implies-\frac y{x^2+y^2}<0.$$

Answer 2

I'm typing this answer since none of the existing ones are built up based on OP's observations.

Let $z = x+yi, x,y\in\Bbb{R}$. OP states that "then $1/z = \bar{z}$, so then it must be $−y$, which is less than zero". In fact, we need $|z| = 1$ to apply $1/z = \bar{z}$. Therefore, we need to show that it suffices to show that it is sufficient to prove only the case $|z| = 1$.

However, $\mathrm{Im}(z) > 0 \iff \mathrm{Im}(z/|z|) > 0$ is obvious, since we are just multiplying a complex number by a positive number. Similarly, $\mathrm{Im}(1/z) < 0 \iff \mathrm{Im}(|z|/z) < 0$. Q.E.D.

Answer 3

If you consider the polar form representation of $z$ then $Im(z)= \left| z \right|\sin \theta$ whereas $Im(\frac{1}{z})=\sin (-\theta)\cdot \frac{1}{\left|z\right|}$ The result follows immediately since sinus is an odd function.