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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 $(x+iy)(x-iy) = x^2+y^2$.2012-08-29

7 Answers 7

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

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    @FlybyNight: "The whole idea of an "add comment" button is to suggest how an answer could be improved."$-$This is true, but the answer writer is free to disagree with your opinion. You voiced your concerns, voted accordingly, and added to the discussion with your first comment. There is no need to continue the argument, and there is certainly no need to be condescending. (I quote "Maybe you should write to the web designers and tell them a nasty man was mean to you because he down-voted your answer.") I have deleted your last two comments as they do not contribute constructively.2012-08-29
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$\frac{1}{z} = \frac{\overline{z}}{\overline{z}} \frac{1}{z} = \frac{\overline{z}}{|z|^2}$ (since $z \overline{z} = |z|^2$).

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    Why the extra downvote?2017-07-16
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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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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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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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    The only lesson that I seem able to draw is that people become very bitter when people criticise their replies. The fact that the OP accepted an answer different to both yours and Brian's just adds weight to my initial comments.2012-08-29
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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.