Exam question: Showing $z = \left(\frac{1 +z}{\left|1 + z\right|} \right)^2$ for $z\in\mathbb{C}$ and $\left| z \right| = 1$

Question

The following was a midterm exam question in an abstract algebra course I was not able to solve:

Let $z \in \mathbb{C}$ with $\left| z \right| = 1$ and $z \neq -1$. Show that the following holds true:

$z = \left(\frac{1 +z}{\left|1 + z\right|} \right)^2$

Hint: This is a possible definition for a "principal root" of $z$.

In the exam, I tried to attack this using the fact that if $\left| z \right| = 1$ holds true, I can write $z$ as $\exp(it)$ for some $t\in \left[0, 2\pi \right[$. I tried to work on the left side of the equation using the properties of $\exp$, but didn't know how to deal with the norm in the denominator.

In hindsight, how can this be solved? Is the hint actually helpful?

Answer 1

How about this:

$$\frac{(1+z)^2}{|1+z|^2}=\frac{(1+z)^2}{(1+z)\overline{(1+z)}}=\frac{1+z}{\overline{(1+z)}}=\frac{1+z}{1+\overline z},$$ where we used the fact that $z\neq -1$.

Replace $z=e^{i\theta}$ as you suggest, and obtain

$$\frac{1+ e^{i\theta}}{1+ e^{-i\theta}}=\frac{e^{i\theta}\left(e^{-i\theta}+ 1\right)}{1+ e^{-i\theta}}= e^{i\theta} = z.$$

Thanks to Peter for the further simplification.

Answer 2

I assume you know the polar notation for complex numbers.

Try drawing a rhombus in the complex plane to determine the angle of $\frac{1+z}{|1+z|}$ in relation to the angle of $z$ itself (the diagonal of a rhombus is a bissector).

Why does $\frac{1+z}{|1+z|}$ lie on the goniometric circle? So knowing its angle, what is its polar notation?

Answer 3

We have that

$$ \left(\frac{1 +z}{\left|1 + z\right|} \right)^2= \frac{(1 +z)^2}{(1+z)(\overline{1+ z})}= \frac{1 +z}{1+\bar z}=\frac{1 +z}{1+\frac1z}=z\frac{1 +z}{1 +z}=z$$

Answer 4

Your approach failed probably because you're trying to simplify and deal directly with the quantity $|\exp(it)+1|$, which is however very difficult, as you have the sum of $1$ and a complex number in polar notation. That's in general not going to give you something helpful, and polar coordinates will usually only help you when you're multiplying two such numbers, since the indices can then add up. Of course, there are tricks you can employ to make this work in this case (see the other answers), but I think in general it's not going to be your first approach.

Here there's a very simple observation to make: that $|1+z|$, the denominator, is just the magnitude (i.e. modulus) of $1+z$, the numerator. Hence, it has unit length, the same as $z$ by assumption. The other thing to do now is to verify that they have the same direction, which is equivalent to the argument of $(1+z)$ (the angle between it and the real axis on the complex plane) being half of that of $z$.

Draw this out on the complex plane. You will see that the points $0,z,(1+z)$ form an isosceles triangle. After that, you will be able to drive that the angle formed by the line joining $0$ and $(1+z)$ with the positive real axis is exactly half of the angle formed by the line joining $0$ and $z$ with the positive real axis.

Of course, the simpler way to do this would be to just substitute $z=x+iy$ assuming $x^2+y^2=1$, then do a lot of simplifications, or equivalently substitute $z=\cos\theta+i\sin\theta$, and indeed that is probably the expected answer. But I think this approach gives a nice geometric intuition.