Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

It is taught that $\mid z \mid ^{2} =x^{2}+y^{2}$ and $\bar{z}z=x^{2}+y^{2}$. Algebrically its fine to understand it. But what is the geometrical meaning of it? I tried multiplying couple of numbers but didn't find some good reason.

share|improve this question
Check this out: clarku.edu/~djoyce/complex/mult.html –  Git Gud Jan 16 '13 at 6:24
@JonasMeyer: ya I have corrected it... –  Rorschach Jan 16 '13 at 6:29

4 Answers 4

up vote 2 down vote accepted


when you multiply $\bar{z}$ by $z$ you are rotating $\bar{z}$ for a angle $\theta$ where $-\theta$ is the angle of $\bar{z}$(with respct to axis x). So $\bar{z}$ is "going" to axis x(with the same norm).

share|improve this answer

Consider the polar form of a complex number.

$$z = |z| e^{i \theta}$$

When you multiply two complex numbers, the magnitudes multiply and the angles add. Let $w = |w| e^{i \phi}$, so that

$$zw = |z||w| e^{i (\theta + \phi)}$$

Multiplying a complex number by another dilates and rotates according to these values.

share|improve this answer

Who knows if you will find this valuable, but this is a method I really like.

Identify each complex number $z = a + bi$ with the matrix $\left(\begin{array}{cc} a & -b \\ b & a \end{array}\right)$. If you multiply matrices for two complex numbers $z_1$ and $z_2$, you obtain the matrix of the product $z_1$ and $z_2$.

You can deduce that for any complex number $z \neq 0$ has a representation for $z/|z|$ which is a rotation matrix.

I'll leave it to you (or another answer) to spell it out exactly, but you can think of complex numbers as having two geometric parts - their magnitude and their rotation component.

share|improve this answer

It's easier to visualize the action in polar coordinates, (particularly exponential form using Euler's formula $z= r e^{i \phi}$.)

Standard ($x y$) and conjugate ($x \bar{y}$) multiplication both act in the same way with respect to the modulus: they multiply the moduli.

However, whereas standard multiplication adds the phases, conjugate multiplication subtracts the phases.

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.