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I am asked to show that the only solution of the equation $ 2- \frac{1}{z} = z^* $ is $z=1$.

I have tried this a few ways the two where i get closest are as follows.

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

$0=z^* z -2z +1$

$\left|z^2 \right| -2z +1=0 = \left|z \right|^2 -2z +1 $

id really like to say $(\left| z \right| -1)^2=0 $ or even better $(z -1)^2=0 $

ive tried a few other things but i always end up with $ \left| z \right| $ instead of $z$

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    Note that $|z|$ is a rel number ,but $z$ may not be just real $z=x+iy$ .I think its better to put $z$ in form of $x+iy$ – 2017-01-20
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    Is $z^c $ supposed to be the complex conjugate that is usually notated as $\overline z $? I've never seen it expressed as $z^c$. – 2017-01-20
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    As @Khosrotash has said, usually these type of questions usually require substituting in $z=x+iy$. However, there is an alternate way. Instead you can notice that by conjugating the equation we get a second equation $2 -\frac{1}{\overline{z}} = z$ where we can now substitute the first equation into. – 2017-01-20
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    If $|z|^2-2z+1=0$ then z is a real positive number. You are done. – 2017-01-20
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    No need to substitute x +y until you need to. And in this question you never need to. But if to do you have $(x^2+y^2)-2 (x+yi)+1 = [x^2+y^2-2x+1]-2yi =0=0+0i$. So $2y=0$ and $x^2+y^2-2x+1=0$. So $y=0$ and $x^2-2x+1=0$ so $(x-1)^2=0$ so $x=1$. So $z=1+0i=1$. But why go through the trouble when you know $2z=|z|^2+1$ is a positive real number so $z^2-2z+1=0$ and that only has 1 double root solution. – 2017-01-20

4 Answers 4

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Let's start from

$$z\overline{z}=2z-1$$

This means with $x=x+iy$

$$x^2+y^2=2x-1+2iy$$

This means identifying the real parts and the imaginary parts of each side

$$\begin{align} y=&0\\ (x-1)^2+y^2=&0 \end{align}$$

and we have proven $z=1$

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    I think you made typo (in line 4) $2x-1+i(2y)$!? – 2017-01-20
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... so $2z=|z|^2+1$ which is a positive real number so $|z|=z $ and....

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We have $|z|^2=2z-1$, hence $2z-1 \in \mathbb R$, therefore $z \in \mathbb R$.

It follows that $|z|^2=z^2$, thus we have

$$z^2-2z+1=0,$$

which gives $z=1$.

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$$\left|z^2 \right| -2z +1=0 \\ \to \sqrt{x^2+y^2} ^2 -2(x+iy) +1=0\\ x^2+y^2-2x+1-i(2y)=0 $$ I t is better to say $$x^2+y^2-2x+1-i(2y)=0 +i0 \to \\ \begin{cases}x^2+y^2-2x+1=0 & (x-1)^2+y^2=0\\-2y=0 & \to y=0 \end{cases}\\ \begin{cases}x^2+y^2-2x+1=0 & (x-1)^2+0^2=0 \to\\-2y=0 & \to y=0 \end{cases}\\(x-1)^2=0 \to x=1$$ so $$z=x+iy=1+0y $$

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    Correct. But a heck of a lot of unnecessary work when you can deduce from the very first line that z is real. – 2017-01-20