Purely imaginary

Question

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 $$\frac{2x}{2iy}$$

Answer 1

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.

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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

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

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

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

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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.

Answer 2

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 (*).

Answer 3

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.

Answer 4

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

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

Answer 5

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