Modulus of a complex number

Question

I am now encountering a problem regarding on complex analysis

Lets say we have $w=u+iv$

What would it be for $$|w|^{2}$$

I check a lot of videos and lecture notes, and realize the answer is $$u^{2}+v^{2}$$

Can someone explain to me why is it like that instead of doing $$u^{2}+2iuv+v^{2}$$ Where did the $2iuv$ go? How to deduce $u^{2}+v^{2}$?

Thank you.

For a complex number $z=x+iy,~x,y\in\mathbb{R}$, $|z|$ is the distance of the point $(x,y)$ from the origin in the Cartesian plane. — Prasun Biswas, Mar 22 '15 at 16:00
Even if you had in mind $w^2$ instead of $|w|^2$, you have $u^2+2iuv-v^2$, not $u^2+2iuv+v^2$. — user26486, Mar 22 '15 at 18:58

score 11 · Accepted Answer · answered Mar 22 '15 at 15:59

11

The definition of $|w|^2$ is $$|w|^2=w\bar{w},$$ where $\bar{w}$ is the conjugate of $w$. Hence, $$|w|^2=(u+iv)(u-iv)=u^2+v^2.$$

answered Mar 22 '15 at 15:59

Spenser

19,469

score 4 · Answer 2 · answered Mar 22 '15 at 16:00

The modulus of a complex number is defined as $\lvert z\rvert = \sqrt{x^2+y^2}$ where $z=x+iy$. That is, the modulus is the distance. Another way $z$ can be written is in polar form which lends itself to this form $z = re^{i\theta} = \lvert z\rvert e^{i\theta}$. What you are confusing is the square of $z$ with the modulus squared. $$ z^2 = (x+iy)^2 = x^2 - y^2 + 2iyx\neq x^2 + y^2 = z\bar{z} = \lvert z\rvert^2 $$

score 1 · Answer 3 · answered Mar 22 '15 at 16:00

1

$|w|$ represents the distance of $w$ from origin in complex plane, now you can use distance formula to calculate the value of $|w|$ i.e distance form $(u,v)$ to $(0,0)$ is the value if $|w|.$

answered Mar 22 '15 at 16:00

K_user

2,438

Modulus of a complex number

3 Answers3