Show that $z^{-1} = \frac{\bar z}{|z|^2}$ I'm stuck on this question, I have a feeling the answer is very straightforward but I just can't figure it out. 
Question: Considering $z= x + iy$, show that: $$z^{-1} = \frac{\bar z}{|z|^2}$$
So far this is what I have: $\bar z=x-iy$ and $|z|^2= x^2 + y^2$
Therefore: $$\frac1{x+iy}=\frac{x-iy}{x^2 + y^2}$$
Where do I go from here? Thanks!
 A: $\frac{1}{z} = \frac{\overline{z}}{\overline{z}} \frac{1}{z} = \frac{\overline{z}}{|z|^2}$ (since $z \overline{z} = |z|^2$).
A: A multiplicative inverse $z^{-1}$ of a number $z$ is defined as any number with the property that $z\cdot z^{-1}=1$. This is easy to verify in this instance since $z\cdot z^*=|z|^2$:
$$z\cdot\frac{z*}{|z|^2}=\frac{zz^*}{zz*}=1$$
So indeed, your proposed multiplicative inverse is, in fact, a multiplicative inverse, and since it is unique we can say that
$$z^{-1}=\frac{z^*}{|z|^2}$$
A: HINT: $$\frac1z=\frac{1}{x+iy}=\frac{1}{x+iy}\cdot\frac{x-iy}{x-iy}=\dots ?$$
A: The step you are missing is to multiply the left hand side by $\dfrac{x-iy}{x-iy}$ and then see that it equals the right hand side.
A: Recall that by definition, $z^{-1}$ is the number such that $z^{-1}z = 1$. 
Suppose $z \neq 0$. You have that 
$\bar{z}z = |z|^2$
Dividing both side by $|z|^2$ (which does not equal $0$ since $z \neq 0$)
$\left(\frac{\bar{z}}{|z|^2}\right) z = 1$
Hence $\frac{\bar{z}}{|z|^2}$ is a such a number whose product with $z$ is $1$. It satisfies the definition of being the inverse of $z$.
A: This is analogous to rationalizing denominators, except here we are realizing them. Suppose that we are given a ring $\rm\:R\:$ of "real" numbers contained in a ring $\rm\,C\,$ of "complex" numbers, such that every complex number $\rm\,z\ne 0\,$ has a nonzero real multiple $\rm\, z\,\hat z\, =\, r\in R.\:$ Then 
$$\rm z\,\hat z\, =\, r\ne 0\,\ \Rightarrow\,\ \frac{y}{z}\ =\ \frac{y}z\,\frac{\hat z}{\hat z}\ =\ \frac{y\hat z}{r}$$ 
Thus we've reduced division by a "complex" number $\rm\,z\,$ to "simpler" division by a "real" number $\rm\,r.\:$ An analgous technique works for any algebraic extension. For much further discussion see my posts on rationalizing denominators. 
A: You can use uniqueness of the inverse: $\frac{\overline z}{|z|^2}z = {(x-iy)(x+iy)\over |z|^2} = {x^2+y^2\over x^2+y^2}=1.$
