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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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    There's no need to resort to taking real and imaginary parts. Multiply the numerator and denominator by $\overline{z}$.2012-08-29
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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.2012-08-29
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    It was one of the very first things I was ever taught about complex numbers. The problem is that people needing to ask the question are most probably going about solving the problem in the wrong way. Continuing along their path of reasoning may encourage them to continue in an inefficient manner. It's better to give them a little nudge in the right direction.2012-08-29
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    @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.2012-08-29
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    I have already voiced my opinions regarding your approach, please read one comment up. For your own peace of mind, let me assure you that I am neither worried nor in a state of anxiety. We have both agreed that your solution is "a little inefficient", and that was the sole purpose of my post.2012-08-29
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    @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.2012-08-29
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    The OP used real & imaginary parts, seems reasonable to use them.2012-08-29
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    @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.2012-08-29
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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 negative vote?2013-05-11
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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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    This answers seems more complicated than necessary: multiply the numerator and denominator by $\overline{z}$. Done!2012-08-29
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    @FlybyNight I don't understand your solution. All I did was divide both side by $|z|^2$. That is pretty easy already.2012-08-29
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    You don't understand that $1/z = (1/z) \times (\overline{z}/\overline{z}) = \overline{z}/|z|^2$?2012-08-29
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    @FlybyNight You didn't say "multiply the numerator and denominator by $\bar{z}$" to what. But it is essentially the same thing.2012-08-29
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    The problem is "Show that $1/z$ is equal to such and such." What else could I have reasonably meant?2012-08-29
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    @FlybyNight Relax! There are many ways to prove this; for instance, I think copper.hat's answer is what you recommended. My answer emphasized proving the inverse formula by finding a number $a$ such that $az = 1$.2012-08-29
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    Yes, there are many ways of proving many things. But some answers are better than others. That is why we can vote-up and vote-down answers, as well as comment upon them. Your answer was overly complicated and I brought that to your attention. Perhaps it might be better to stay away from public fora if you are not willing to accept public criticism.2012-08-29
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    @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.2012-08-29
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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.