# Finding roots of complex quadratic equation

I'm trying to solve for the following equation:

$$|(1+50*i*x)^2|$$ I keep getting the form $$-2500x^2 + 100ix + 1$$ when the problem needs to have the following form:

$$2500x^2 + 1$$

What steps are necessary for me to take to reach the latter response?

• 1) What you wrote is not an equation but an expression. 2) Is $x$ real or complex?
– 5xum
May 12, 2014 at 10:07

$|1+50ix|^2=(1+50ix)(1-50ix)=2500x^2+1$,

you just needed to take the conjugate, but this is assuming that $x$ is real.

• Interesting, but how can (1+50ix)(1−50ix) = (1+50ix)(1+50ix)? May 12, 2014 at 10:18
• $(1+50ix)(1-50ix)=1+50ix-50ix-50^2i^2x^2=1+2500x^2$ :) May 12, 2014 at 10:22
• @hacke: You are missing the absolute value signs. Then you have $|z|=|\bar z|$ which completely justifies the equality. Otherwise, $|z^2|=|z|^2=z\bar z$. May 12, 2014 at 10:23
• $(a+b)(a-b) = a^2 - b^2$
– mvw
May 12, 2014 at 10:23

recall that $|a+bi|^2=|(a+bi)^2|$, because $$|a+bi|^2=(\sqrt{a^2+b^2})^2=a^2+b^2$$ and $$|(a+bi)^2|=|a^2-b^2+2abi|=\sqrt{(a^2-b^2)^2+4a^2b^2}=a^2+b^2$$ So $$|(1+50xi)^2|=|1+50xi|^2=(\sqrt{1+(50x)^2})^2=2500x^2+1$$

For a complex number $z=a+bi$, where $a,b$ are real numbers, the value $|z|$ is defined as $\sqrt{a^2+b^2}$. The value $|z^2|$ is equal to $|z|^2$, does this answer your question?