# 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!

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Note that $(x+iy)(x-iy) = x^2+y^2$. – copper.hat Aug 29 '12 at 17:37

$\frac{1}{z} = \frac{\overline{z}}{\overline{z}} \frac{1}{z} = \frac{\overline{z}}{|z|^2}$ (since $z \overline{z} = |z|^2$).

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Why the negative vote? – copper.hat May 11 '13 at 7:11

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.

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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 $z$.

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This answers seems more complicated than necessary: multiply the numerator and denominator by $\overline{z}$. Done! – Fly by Night Aug 29 '12 at 18:27
@FlybyNight I don't understand your solution. All I did was divide both side by $|z|^2$. That is pretty easy already. – William Aug 29 '12 at 18:43
You don't understand that $1/z = (1/z) \times (\overline{z}/\overline{z}) = \overline{z}/|z|^2$? – Fly by Night Aug 29 '12 at 19:03
@FlybyNight You didn't say "multiply the numerator and denominator by $\bar{z}$" to what. But it is essentially the same thing. – William Aug 29 '12 at 19:08
@FlybyNight I am sorry if this is not what you or the OP is looking for; however, a different perspective can not hurt. However, you should probably review your comments on Brian's and my post and reflex on who appears to be more bellicose. Enough said on these unmathematical matters; we should all return to enjoying some math. – William Aug 29 '12 at 20:17

You can use uniqueness of the inverse: $\frac{\overline z}{|z|^2}z = {(x-iy)(x+iy)\over |z|^2} = {x^2+y^2\over x^2+y^2}=1.$

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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.

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HINT: $$\frac1z=\frac{1}{x+iy}=\frac{1}{x+iy}\cdot\frac{x-iy}{x-iy}=\dots ?$$

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@FlybyNight: Of course $-$ provided that the OP knows that $z\bar z=|z|^2$. Since it’s not clear that that’s the case, and since the OP was already looking at real and imaginary parts, I chose the more elementary approach. – Brian M. Scott Aug 29 '12 at 18:30
@FlybyNight: If there were anything actually wrong with writing $z$ in rectangular form to solve the problem, I’d have chosen a different approach. However, there isn’t. Yes, it’s just a little inefficient, but it uses a manipulation that one should know in any case. You’re worrying about a non-problem. – Brian M. Scott Aug 29 '12 at 19:22
@FlybyNight: I read it, and as I said, I disagree with your notion of what constitutes good pædagogy in this instance. (By the way, you might want to improve your understanding of colloquial usage of worrying about.) An even better answer would have gone on to point out that the rectangular coordinates were unnecessary, but copper.hat’s answer already covered that ground, so I didn’t bother with the addition. And I’ve no more to say on the subject. – Brian M. Scott Aug 29 '12 at 19:41
The OP used real & imaginary parts, seems reasonable to use them. – copper.hat Aug 29 '12 at 20:39
@FlybyNight: Best for whom? Different people will have different notions of what is the best answer. Is it the one that the OP finds most useful? Is it the one that gets the most upvotes? Is it the one that is helpful to the widest range of users of the site? Is the most efficient one? Is it the most elegant one? In some cases these might be five different answers, and they are frequently more than one different answer. Unlike comments suggesting possible improvements, downvotes of a correct answer, especially one that others have found helpful, are not really helpful to the site’s audience. – Brian M. Scott Aug 29 '12 at 21:55

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

$$z^{-1}=\frac{z^*}{|z|^2}$$

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