2
$\begingroup$

Given that $z=\cos\theta+i\sin\theta$, show that $Re\left(\frac{z-1}{z+1}\right)=0, \quad z\ne-1$

For this question I had to show that the real part of $\frac{z-1}{z+1}=0$

To find that I first substituted $z$ with $\cos\theta+i\sin\theta$ to get

$$\frac{\cos\theta+i\sin\theta-1}{\cos\theta+i\sin\theta+1}$$

I then multiplied by the conjugate of the denominator

$$\frac{\cos\theta+i\sin\theta-1}{\cos\theta+i\sin\theta+1}\cdot\frac{\cos\theta-i\sin\theta+1}{\cos\theta-i\sin\theta+1}$$

Which, when expanded, gives me

$$\frac{\cos^2\theta+\sin^2\theta+2i\sin\theta-1}{\cos^2\theta+\sin^2\theta+2i\cos\theta+1}$$

After that,

$$\frac{2i\sin\theta}{2\cos\theta+2}=\frac{i\sin\theta}{\cos\theta+1}$$

How do I proceed?

Edit: Just realized (after being told by a commenter) that $\frac{i\sin\theta}{\cos\theta+1}$ is imaginary. I won't delete the question. Ínstead I'll leave it here, as a testament to my stupidity, for everyone to see. May God have mercy on my soul during my exam.

$\endgroup$
  • 5
    $\begingroup$ You have already finished. Your number is imaginary. $\endgroup$ – Dog_69 Aug 29 '18 at 20:58
1
$\begingroup$

You are done indeed

$$\frac{i\sin\theta}{\cos\theta+1}=i\cdot \frac{\sin\theta}{\cos\theta+1}$$

is purely imaginary.

As an alternative since $z\bar z=|z|^2=\cos^2 \theta+\sin^2 \theta=1$ we have that

$$\frac{z-1}{z+1}=\frac{z-1}{z+1}\frac{\bar z+1}{\bar z+1}=\frac{z\bar z+z-\bar z-1}{z\bar z+z+\bar z+1}=i\cdot \frac{\,\Im(z)}{\Re(z)+1}$$

which is again purely imaginary.

$\endgroup$
3
$\begingroup$

A complex number $w$ is purely imaginary (that is, its real part is $0$) if and only if $w+\bar{w}=0$.

You're given that $|z|=1$; then the number plus its conjugate is $$ \frac{z-1}{z+1}+\frac{\bar{z}-1}{\bar{z}+1}= \frac{z\bar{z}-\bar{z}+z-1+z\bar{z}+\bar{z}-z-1}{(z+1)(\bar{z}+1)}=0 $$ because $z\bar{z}=1$.

$\endgroup$
0
$\begingroup$

Yet another way:   $|z| = 1 \iff z\bar z=1 \iff \bar z = \dfrac{1}{z}\,$, then using that $\,|\operatorname{Re}(z)| \le |z| \le 1\,$:

$$ \left(\frac{z-1}{z+1}\right)^2=\frac{z^2-2z+1}{z^2+2z+1}=\frac{z-2+\dfrac{1}{z}}{z+2+\dfrac{1}{z}} = \frac{z+\bar z - 2}{z+\bar z +2} = \frac{2\operatorname{Re}(z)-2}{2\operatorname{Re}(z)+2} = \frac{\operatorname{Re}(z)-1}{\operatorname{Re}(z)+1} \;\le\; 0 $$

Since the square of $\,\dfrac{z-1}{z+1}\,$ is a non-positive real, $\,\dfrac{z-1}{z+1}\,$ is either $\,0\,$, or purely imaginary.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.