Difference between $z^2$ and $|z|^2$

Question

Are $z^2$ and $|z|^2$ same? Where $z$ is a complex number. If imaginary part of $z$ is zero, then surely we can say they are both are same. What about if imaginary part of $z$ is non-zero?

Answer 1

Suppose $z=re^{i\theta}$, $\theta\neq 0,\pi$. We see that $|z|=r$, and $|z|^2=r^2$ but $z^2=r^2e^{2i\theta}$.

Then $z^2\neq|z|^2$ if the imaginary part of $z$ is non-zero.

Answer 2

It turns out that $z^2=|z|^2$ if and only if $z$ is real. In fact, it is clear that $0^2=|0|^2$. On the other hand, if $z\neq0$, then$$z^2=|z|^2\iff z^2=z.\overline z\iff z=\overline z\iff z\in\mathbb R.$$

Answer 3

If $z=0$ then evidently $z^2=0=|z|^2$.

If $z\neq0$ then we can write $z=|z|e^{i\phi}$ and we have $z^2=|z|^2e^{2i\phi}$ where in general $e^{2i\phi}\neq1$ and consequently $z^2\neq|z|^2$.

The exceptions are achieved for $e^{i2\phi}=1$.

Summarized this leads to: $$z^2=0=|z|^2\iff z\in\mathbb R$$

Answer 4

Not at all: $i^2=-1$, and $|i|^2=1$.

More generally, you may ask when $\;z^n=|z|^n$.

Writing $z$ in exponential form: $\;z=|z|\,\mathrm e^{\arg z}$, and using the properties of module and argument, one obtains that $\; z^n=|z|^n\,\mathrm e^{n\arg z}$, so that $$n\arg z\equiv 0\mod 2\pi\iff \arg z\equiv 0\mod \frac{2\pi}n.$$

In the case $n=2$, this means $z=\pm|z|$, i.e. $z$ is real.

Answer 5

$|z|^2$ is always a real number (positive, at that), while $z^2$ can be any complex number.

Answer 6

You can calculate the square of $z=x+iy$ for $x,y \in \mathbb{R}$ by using the first binomial formula. Furthermore notice that $|z|$ is given by Pythagoras in the complex plane. Putting $y=1$, for instance, yields your answer that I'll not directly give here. Try it by yourself first and consult Wikipedia.

Answer 7

If you want a geometric interpretation as well, here it is.

Think of $z$ as an operator on a real plane through which any vector (based at the origin) is multiplied by a given positive (real) number and rotated by a given angle. They are the modulus of $z$ and its phase.

Now $z^2$ being $z\cdot z$ correspond to an operator given by two consecutive application of the operator $z$. So any vector will be multiplied by $\vert z^2\vert=\vert z\vert^2$ and rotated by an angle that is the double of the phase of $z$.

On the contrary $\vert z\vert^2$ is the square of a real. In the considered plane a real can be interpreted as an operator of the kind I've been talking about whose phase is zero. So $\vert z\vert$ multiplies any vector of the plane by $\vert z\vert$ with no rotation. It comes that $\vert z\vert^2$, representing two consecutive applications of $\vert z\vert$, multiplies any vector by $\vert z\vert^2$ (as it happens with the operator $z^2$) but does not rotate it (as opposed to the operator $z^2$).