Which of these numbers [$1, 4i, 3+3i, 3-3i, 1 - 2i, 2+ 3i, 6$] satisfy $Re\left( \frac{1}{z} \right) = \frac{1}{6}$?

Question

Consider the curve with equation $$\operatorname{Re}{\left( \dfrac{1}{z} \right)} = \frac{1}{6}.$$ For each complex number in the following list,
$$1, 4i, 3+3i, 3-3i, 1 - 2i, 2+ 3i, 6$$

How would I test these values? I am not sure where to start.I know the $\operatorname{Re}()$ means the real part. Does $\operatorname{Re}{(4i)}$ mean undefined?

Answer 1

Let's think about what $\operatorname{Re}\bigg(\frac1z\bigg)$ means. We know that $$\frac1z = \frac{\bar{z}}{z\cdot\bar{z}}$$where $\operatorname{Re}$ is the Real part of the complex number, $\operatorname{Im}$ is the imaginary part, and $\bar{z}$ is the conjugate of $z$. So $$\operatorname{Re}\bigg(\frac1z\bigg) = \frac{\operatorname{Re}(z)}{\operatorname{Re}(z)^2+\operatorname{Im}(z)^2}$$

So, using this formula, we find:

$\operatorname{Re}(\frac11)=\frac1{1^2+0^2}=1\neq\frac16$

$\operatorname{Re}(\frac1{4i})=\frac0{0^2+4^2}=0\neq\frac16$

$\operatorname{Re}(\frac1{3+3i})=\frac3{3^2+3^2}=\frac16$

$\operatorname{Re}(\frac1{3-3i})=\frac3{3^2+(-3)^2}=\frac16$

$\operatorname{Re}(\frac1{1-2i})=\frac1{1^2+(-2)^2}=\frac15\neq\frac16$

$\operatorname{Re}(\frac1{2+3i})=\frac2{2^2+3^2}=\frac2{13}\neq\frac16$

$\operatorname{Re}(\frac16)=\frac6{6^2+0^2}=\frac16$

Answer 2

Let $z=a+bi.$

Then $$\text{Re}\Big(\frac{1}{z}\Big)=\text{Re}\Big(\frac{1}{a+bi}\Big)=\text{Re}\Big(\frac{1}{a+bi}\cdot\frac{a-bi}{a-bi}\Big)=\text{Re}\Big(\frac{a-bi}{a^2+b^2}\Big)=\frac{a}{a^2+b^2}.$$

Now you just need to substitute in the formula.