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if $$|z| = 1$$

prove that $$\frac{1 + z}{1 - z}$$ is purely imaginary

I have tried multiply by the conjugate like this

$$\frac{(x+1)+iy}{(1-x)-iy} * \frac{(x+1)-iy}{(x+1)-iy}$$

giving me $$\frac{x^2+2x+1 + y^2}{1-x^2-y^2 - 2iy}$$

But I have no idea how to prove that this is in the form of ix where x is a real number.

I know the question insinuates that $$z = -1$$ $$x^2 + y^2 = -1$$However if I substitute that into the equation I just get $$\frac{2x}{2iy}$$

Po Chen Liu
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  • prove that...what? – user May 13 '18 at 06:24
  • How does one prove ‘that’ – Prince M May 13 '18 at 06:24
  • From the title, I guess that the expression is purely imaginary. – badjohn May 13 '18 at 06:37
  • You have your conjugate wrong. Only negate the imaginary component. Also, remember that $x$ and $y$ are real. – badjohn May 13 '18 at 06:43
  • You've multiplied numerator and denominator by the conjugate of the numerator: OK. The numerator of the result is real. OK. But you've miscalculated the denominator - it is $(1-x^2-y^2)-2 iy=2iy$ – ancient mathematician May 13 '18 at 06:45
  • "I know the question insinuates that z=−1" No, it doesn't. $z$ can be anything with absolute value of $1$ (except 1). So $z$ could be $i$ or $z$ could be $-1$ (remember $0 = 0i$ so $0$ is purely imaginary-- $0$ is the only number that is both real and purely imaginary) and $z$ could be $-i$ and $z$ could be $\frac 1{\sqrt 2} + \frac 1{\sqrt 2}i$ or $z$ could be $\frac 2{\sqrt 5} + \frac 3{\sqrt 5} i$ or $\cos \theta + i\sin \theta$ for any $\theta \ne 2k\pi$. – fleablood May 13 '18 at 07:28
  • The conjugate of $(1-x) - yi$ is $(1-x) + yi$; not $(x+1) - yi$. You are supposed to multiply by the conjugate of the denominator; not the numerator. – fleablood May 13 '18 at 07:31
  • $x^2 + y^2 \ne -1$ (that's impossible) but $x^2 + y^2 = 1$. so $\frac{x^2+2x+1 + y^2}{1-x^2-y^2 - 2iy}= \frac{x^2+2x+1 + y^2}{1-1 - 2iy}=\frac{x^2+2x+1 + y^2}{ - 2iy}$. The numerator is purely real and the denominator is purely imaginary. As $\frac 1i = -i$ the number is purely imaginary and you are done. – fleablood May 13 '18 at 07:36
  • See also : https://math.stackexchange.com/questions/118868/for-complex-z-z-1-implies-textre-left-frac1-z1z-right-0 – Arnaud D. Aug 29 '18 at 13:19

5 Answers5

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You want to prove that $w=\dfrac{1+z}{1-z}$ is purely imaginary, that is $w+\overline w=0$. But $$w+\overline w=\frac{1+z}{1-z}+\frac{1+\overline z}{1-\overline z}.\tag{*}$$ We are given that $|z|=1$. This means that $1=|z|^2=z\overline z$, so $\overline z=z^{-1}$. Try substituting this into (*).

Angina Seng
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To finish your work:

$\frac{x^2+2x+1 + y^2}{1-x^2-y^2 - 2iy}$.

But $|z| = x^2 + y^2 = 1$

So $\frac{x^2+2x+1 + y^2}{1-x^2-y^2 - 2iy}= \frac{x^2+2x+1 + y^2}{1-1 - 2iy}=\frac{x^2+2x+1 + y^2}{ - 2iy}$

$x^2+2x+1 + y^2$ is reall. Let $x^2+2x+1 + y^2= r$ so

$\frac{x^2+2x+1 + y^2}{ - 2iy} = \frac {r}{-2yi} = \frac {r}{-2yi}\frac ii=\frac {ri}{2y}$.

Which is purely imaginary.

I know the question insinuates that z=−1

No, it doesn't.

$z$ can be any complex number number with absolute value of $1$ (except $z \ne 1$).

$z$ could be $\frac {12}{\sqrt{13}} + \frac {5}{\sqrt {13}}i$ for all we know.

So long as $x^2 + y^2 = 1$ and $(x,y) \ne (1,0)$ then $z = x + yi$ is possible.

====

The point of multiplying by the conjugate is to get the imaginary terms out of the denominator not the numerator.

$\frac {1+z}{1-z} = \frac {(1+a) + bi}{(1-a) -bi} = \frac {(1+a) + bi)((1-a)+bi)}{((1-a) -bi)((1-a) + bi)}=$

$\frac {(1+a)(1-a)-b^2 + bi(1+a + 1-a)}{(1-a)^2 + b^2}$

$=\frac {1- a^2 -b^2 + 2bi}{(1-a)^2 + b^2}$

$= \frac {1- (a^2 +b^2) + 2bi}{(1-a)^2 + b^2}$

$= \frac {1- |z| + 2bi}{(1-a)^2 + b^2}$

And we were told that $|z| = a^2 + b^2 = 1$ so

$\frac {1+z}{1-z}=\frac {1- |z| + 2bi}{(1-a)^2 + b^2}=$

$\frac {1- 1 + 2bi}{(1-a)^2 + b^2}=$

$\frac {2b}{(1-a)^2 + b^2)}i$

===

But we should be at the point where we don't need to write out $z = x + yi$ all the time. So long as we know that for real $k$ that $\overline {k \pm z}=k \pm \overline z$, $z + \overline z$ is purely real and $z -\overline z$ is purely imaginary, we are good. (Because $\overline z = Re(z) - Im(z)$ and $Re(k\pm z) = k \pm Re(z)$ and $Im(k \pm z) = \pm Im(z)$. And $z + \overline z = 2Re(z)$ etc.)

Then $\frac {1+z}{1-z} = \frac {(1+z)(1-\overline z)}{(1-z)\overline {(1-z)}}$ and.... well, now I'm just doing someone else's fairly excellent answer.

But really, the math and the notation is much simpler this way.

fleablood
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1

If $z=x+iy$, I get $$\frac{1+z}{1-z} = \frac{1+z \cdot \overline{1-z}}{(1-z)\cdot \overline{1-z}} = \frac{((1+z)(1-\overline{z})}{|1-z|^2}= \frac{1-z\overline{z}+z -\overline{z}}{|1-z|^2}$$

Now, $z\overline{z} = |z|^2= 1$ and $z-\overline{z} = 2iy$, so that the above fraction becomes $$i\frac{2y}{|1-z|^2}$$

which is purely imaginary.

Henno Brandsma
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Consider $w=\frac{1 + z}{1 - z}$ then

$$w+\bar w=\frac{1 + z}{1 - z}+\frac{1 + \bar z}{1 - \bar z}=\frac{1 + z-\bar z-z \bar z+1-z+\bar z-z\bar z}{1 - z-\bar z}=0$$

then

$$w+\bar w=0\implies w=-\bar w\implies Re(w)=0$$

user
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The other answers have already helped you along, so I'm just providing an alternate method.

In polar form, $z$ can be expressed as $$ z = \cos\theta + i\sin\theta, \ \theta \in (-\pi,\pi] $$

So \begin{align} \frac{1+z}{1-z} &= \frac{(1+\cos\theta)+i\sin\theta}{(1-\cos\theta)-i\sin\theta} \\ &= \frac{2\cos^2 \frac{\theta}{2} + 2i\sin\frac{\theta}{2}\cos\frac{\theta}{2}}{2\sin^2 \frac{\theta}{2} - 2i\sin\frac{\theta}{2}\cos\frac{\theta}{2}} \\ &= \frac{\cos\frac{\theta}{2}\left[\cos\frac{\theta}{2} + i\sin\frac{\theta}{2}\right]}{-i\sin\frac{\theta}{2}\left[i\sin\frac{\theta}{2}+\cos\frac{\theta}{2}\right]} \\ &= i\cot \frac{\theta}{2} \end{align}

which is imaginary indeed

Dylan
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