Show that $|z| = 1$ if and only if $\bar{z} = \frac{1}{z}$. Maybe a very stupid question but I am stuck. Show that $|z| = 1$ if and only if $\bar{z} = \frac{1}{z}$.
Is it enough to simply multiply, i.e. $z\bar{z} = \frac{1\times z}{z} = 1$? Showhow I feel this is not correct. I know that if $z = \pm 1$ or $z \pm i$ then $|z| = 1$. Am I supposed to draw the circle $|z| = 1$? But what does $\frac{1}{z}$ represent?
If someone could give me a hint. 
 A: Write $z = x + i y $. Suppose $|z| = \sqrt{x^2+y^2} =1$. Then,
$$ \bar{z} = x - iy = \frac{ x^2 + y^2 }{x + iy} = \frac{1}{x+iy} = \frac{1}{z}$$
Now, suppose $ x - iy = \frac{1}{x + iy } $. Then,
$$ x^2 + y^2 = 1 \implies |z|=1$$
A: Take $z$ in the unit circle. $\bar{z}$ is the reflection of $z$ with respect to the real axis. Therefore, $\bar{z}$ has modulus $1$ and argument the negative of the argument of $z$. Since we multiply complex numbers by multiplying their modulus and adding their arguments, we have the $z\bar{z}$ has modulus $1$ and argument $0$ and so $z\bar{z}=1$.
A: The non-trivial direction: If $|z|=1$ you have $$z=|z|e^{i\arg z}=e^{i\arg z}$$
$$\bar{z}=e^{-i\arg z}$$
$$\frac{1}{z}=\frac{1}{e^{i\arg z}}=e^{-i\arg z}$$
A: $z = a + bi$
for $\frac{1}{z} = \bar{z}$
$\frac{1} {a + bi} = a - bi$
$\frac{1} {(a + bi)} \frac{a - bi}{a - bi} = a - bi$
$\frac{a - bi}{a^2 + b^2} = a - bi$
cancel and rearrange
$(a^2 + b^2) = 1$
recalling the formula for |z|
$|z| = \sqrt{a^2 + b^2} = \sqrt 1 =  1$
the other way
if |z| = 1 then $a^2 + b^2 = 1$
$z = a + bi$
again 
$\frac{1}{z} = \frac{a - bi} {a^2 + b^2} = \frac{a - bi}{|z|^2}$
$= \frac{a - bi}{1^2} = a - bi = \bar z$
A: When $\text{z}\in\mathbb{C}$:


*

*$$\overline{\text{z}}=\Re\left[\text{z}\right]-\Im\left[\text{z}\right]i$$

*$$\frac{1}{\text{z}}=\frac{\overline{\text{z}}}{\text{z}\overline{\text{z}}}=\frac{\overline{\text{z}}}{\left|\text{z}\right|^2}=\frac{\Re\left[\text{z}\right]-\Im\left[\text{z}\right]i}{\Re^2\left[\text{z}\right]+\Im^2\left[\text{z}\right]}$$


So, when we set 1. and 2. equal we get:
$$\overline{\text{z}}=\frac{1}{\text{z}}\Longleftrightarrow\Re\left[\text{z}\right]-\Im\left[\text{z}\right]i=\frac{\Re\left[\text{z}\right]-\Im\left[\text{z}\right]i}{\Re^2\left[\text{z}\right]+\Im^2\left[\text{z}\right]}$$
We get a system:
$$
\begin{cases}
\Re\left[\text{z}\right]=\frac{\Re\left[\text{z}\right]}{\Re^2\left[\text{z}\right]+\Im^2\left[\text{z}\right]}\\
\\
-\Im\left[\text{z}\right]=-\frac{\Im\left[\text{z}\right]}{\Re^2\left[\text{z}\right]+\Im^2\left[\text{z}\right]}
\end{cases}\space\space\space\Longleftrightarrow\space\space\space
\begin{cases}
\Re\left[\text{z}\right]=\frac{\Re\left[\text{z}\right]}{\Re^2\left[\text{z}\right]+\Im^2\left[\text{z}\right]}\\
\\
\Im\left[\text{z}\right]=\frac{\Im\left[\text{z}\right]}{\Re^2\left[\text{z}\right]+\Im^2\left[\text{z}\right]}
\end{cases}\space\space\space\Longleftrightarrow\space\space\space\Re^2\left[\text{z}\right]+\Im^2\left[\text{z}\right]=1
$$
A: This is a two part proof.
The equation $\bar{z}z = |z|^2$ is used for this proof.
This equation is true because $(a-bi)(a+bi)=a^2 -i^2b^2=a^2+b^2=|z|^2$
(=>)
Suppose |z|=1. Then, we must show $\bar{z} = \frac{1}{z}$.
Since |z|=1, $|z|^2=1$ and thus $\bar{z}z=1$. Therefore, after dividing by z,  $\bar{z}=\frac{1}{z}$
(<=) Suppose (1) $\bar{z}=\frac{1}{z}$. 
Then, we must show that |z| = 1
We can show that by multiplying each side of (1) by z to get
$\bar{z}z = \frac{z}{z}=1$. Hence, $|z|^2 = 1$ and thus $|z|=1$.
QED
