# Help with solution to complex quadratic equation

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

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 $$|3+4i|^2$$?

• Multiplying complex numbers, absolute values are multiplied, from where $|w|^2=|w^2|.$ Feb 6, 2019 at 17:27
• This is misinterpreted $a^2+b^2=a^2+2abi-b^2.$ Here RHS is not $|w|^2.$ Feb 6, 2019 at 17:31

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 $$a+ib=\pm(2+i).$$

• Well sure but,,,I think I understand everything apart from that very first sentence which I feel like you'r rushing through. Feb 6, 2019 at 19:05
• Do you mean you don't understand ‘ the modulus is multiplicative ’? Feb 6, 2019 at 19:09
• Well, I'm guessing it has "something" to do with congruence (we rushed through that last chapter) but I have no idea how that is supposed to be applied to complex numbers, we haven't gone through that. Feb 6, 2019 at 19:13
• No, nothing to do with congruence. It corresponds to the well-known formula $\; |zz'|=|z|\,|z'|$. Feb 6, 2019 at 19:15
• It is just another complex number, possible equal to $z$. Feb 6, 2019 at 19:22

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

• Isn't the radicand $3+4i$? Feb 6, 2019 at 20:13
• Yes it is just corrected! Thank you for your hint! Feb 6, 2019 at 20:29

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

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