Help with solution to complex quadratic equation

Question

I'm learning about solving complex quadratic equations, in the examle I'm following, we start with:

$$z^2+4iz-(7+4i)=0$$

This can be rewritten as:

$$(z+2i)^2-(3+4i)=0$$

The square can be rewritten as $w^2$ so:

$$w^2=3+4i$$

Any complex number can be written on the form $a+bi $ so:

$$w^2=(a+bi)^2=a^2+2abi-b^2=3+4i$$

From this we learn that $2ab=4$ and that $a^2-b^2=3$ provided that $a$ and $b$ are real numbers. I'm prepared to accept all of this, but the last part of the solution really bothers me.

According to my book the following applies:

$$x^2+y^2=|w|^2=|w^2|=|3+4i|=5$$

I can accept the first part, but why on earth would $|w|^2=|w^2|$?

This would mean that $a^2+b^2=a^2+2abi-b^2$, and why on earth would this be the case?

Also, why would $|w^2|=|3+4i|$?

If $|w^2|=|w|^2$ then by all rights, $|w^2|$ should equal $|3+4i|^2$?

Answer 1

Since the modulus a complex numbers is multiplicative, if $w^2=z$, then $\;|z|=|w^2|=|w|^2$, so here $$|z|=\sqrt{9+16\mathstrut}=5=a^2+b^2.$$ On the other hand, identifying real parts and imaginary parts, we obtain \begin{cases} a^2-b^2=3,\\ 2ab=4. \end{cases} We'll determine $a^2$ and $b^2$, and deduce $a$ and $b$, taking into account that $a$ and $b$ have the same sign, by the last equation. Adding and subtracting the first and second equation results in $$2a^2=8\iff a=\pm 4,\qquad 2b^2=2\iff b=\pm 1.$$ Therefore, as $a$ and $b$ have the same sign by the last equation, the square roots of $3+4i$ are $$a+ib=\pm(2+i).$$

Answer 2

By the quadratic formula we get $$z_{1,2}=-2i\pm\sqrt{-4+7+4i}$$ Note that $$\sqrt{3+4i}=2+i$$

Answer 3

$$|(a+ib)|^2=a^2+b^2.$$

$$|(a+ib)^2|=|a^2-b^2+2iab|=\sqrt{(a^2-b^2)+(2ab)^2}=\sqrt{(a^2+b^2)^2}.$$