# Absolute value of a complex number (can't uderstand notes)

I am taking a MOOC and in handouts there is the following expression:

$\left\lvert\alpha\right\rvert = \left\lvert\alpha_r + \alpha_c\right\rvert = \sqrt{\alpha^2_r+\alpha_c^2} = \sqrt{(\alpha_r+i\alpha_c)(\alpha_r-i\alpha_c)} = \sqrt{\alpha\bar\alpha} = \left\lvert\bar\alpha\right\rvert$

My question is how did they got from $\sqrt{\alpha^2_r+\alpha_c^2}$ to $\sqrt{(\alpha_r+i\alpha_c)(\alpha_r-i\alpha_c)}$? I am pretty sure that it should be $\sqrt{(\alpha_r\pm\alpha_c)^2 \pm 2\alpha_r\alpha_c}$.

Also in the handouts expression is missing some parentheses, so it looks like this: $\sqrt{\alpha_r+\alpha_c)(\alpha_r-\alpha_c}$

Can anyone clarify this for me?

• What is a MOOC? – Dietrich Burde Nov 5 '17 at 19:52
• @DietrichBurde Massive Open Online Course – Raskolnikov Nov 5 '17 at 19:56
• Massive open online course – Randall Nov 5 '17 at 19:57
• looks like there is a typo – mercio Nov 5 '17 at 19:57
• Thank you. I found it here. Hope it was not too massive, this course. – Dietrich Burde Nov 5 '17 at 19:58

When dealing with complex numbers $s=x+iy$ we can factor expressions such as $x^2+y^2=(x-iy)(x+iy)$ Due to $(i)(-i)=1$. It is essentially like a difference of 2 squares.
$\alpha$=$\alpha_r+i\alpha_c$ so the i was left out by mistake .Put it back in and you are fine.