Suppose I have: $w=2+3i$ and $x=1+2i$. What does it really mean to divide $w$ by $x$?

EDIT: I am sorry that I did not tell my question precisely. (What you all told me turned out to be already known facts!) I was trying to ask the geometric intuition behind the division of complex numbers.

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    $\begingroup$ I'm reminded of this :) $\endgroup$ Jul 2, 2013 at 6:23
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    $\begingroup$ suitcaseofdreams.net/Geometric_division.htm $\endgroup$ Jul 2, 2013 at 6:25
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    $\begingroup$ Perhaps a polar form might be a little more 'intuitive'? $\endgroup$
    – copper.hat
    Jul 2, 2013 at 6:33
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    $\begingroup$ While multiplication/division of complex numbers can be interpreted geometrically, I don't think it is meant to be interpreted that way. $\endgroup$
    – user1551
    Jul 2, 2013 at 6:40
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    $\begingroup$ @user1551 au contraire it is meant to be interpreted geometrically. $\endgroup$
    – kumarharsh
    Jul 2, 2013 at 16:40

7 Answers 7


Complex numbers' multiplication is better understood if you forget the "cartesian vector in the complex plane" analogy: $$ z = a + b i \quad z \in \mathbb{C}, a, b \in \mathbb{R} $$ And in stead think in polar coordinates: $$z = r \angle \theta = r e^{i \theta} \quad z \in \mathbb{C}, r \in \mathbb{R}^+, \theta \in [0, 2 \pi) $$ Wherein $r$ is the magnitude, $\theta$ is the angle.

When multiplying it is easy: $$ z w = (r_z r_w) \angle (\theta _z + \theta _w) $$ You add the angles and multiply the magnitudes.

When dividing you do what comes naturally: $$ \frac z w = \left(\! \frac{r_z}{r_w} \!\right) \angle (\theta _z - \theta _w) $$ To divide means to find the difference in angles and the factor in magnitude.


It means to find another complex number $y$, such that $xy=w$. (Just as it does for real numbers.)


When you divide $a$ by $b$, you're asking "What do I multiply $b$ by in order to get $a$?".

Multiplying two complex numbers multiplies their magnitudes and adds their phases:

Complex multiplication

So when you divide a complex $a$ by a complex $b$ you are asking: "How much do I need to scale $b$ and rotate $b$ in order to get $a$? Please give me a complex number with a magnitude equivalent to how much I must scale and a phase equivalent to how much I must rotate.".


Consider $\frac{1 + i}{1 - i}$. How much do we need to scale and rotate $1-i$ in order to make it the same as $1+i$?

Well, when graphed on the complex plane you can see that $1-i$ has a 45 degree clockwise rotation and a magnitude of $\sqrt{2}$. $1+i$, on the other hand, has a 45 degree counter-clockwise rotation and the same magnitude of $\sqrt{2}$.

Since the magnitudes are the same, we don't need any scaling. Our result's magnitude will be 1.

To rotate from 45 degrees clockwise to 45 degrees counter-clockwise, we must rotate 90 degrees counter-clockwise. Thus our result will have a phase of 90 degrees counter-clockwise (which is upwards along the imaginary Y axis).

Move a distance of 1 up the imaginary Y axis and you get the answer... $\frac{1 + i}{1 - i} = i$. We can confirm this by doing the multiplication: $(1-i) \cdot i = i+1$.

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    $\begingroup$ wow for the gif $\endgroup$
    – kumarharsh
    Jul 2, 2013 at 16:39
  • $\begingroup$ @Harsh I'm glad you like it. I made it (and a bunch of others) myself as part of an explanation of Grover's Quantum Search. $\endgroup$ Jul 3, 2013 at 21:32
  • $\begingroup$ man... I was thinking you were very resourceful, and had taken it from somewhere... Now, to learn that you MADE IT! Salute $\endgroup$
    – kumarharsh
    Jul 3, 2013 at 22:01

Since multiplication can be nicely visualized as a rotation in the complex plane, you may find it helpful to think of division as a form of multiplication: $$\frac{2+3i}{1+2i} = \frac{2+3i}{1+2i}\cdot \frac{1-2i}{1-2i} = -\frac{1}{3}\cdot(2+3i)(1-2i)$$ So, instead of thinking about division, you can think of it as multiplying by the conjugate.


Geometrically, it means that the magnitudes get divided, and the angles get subtracted.

That is, imagine the complex numbers plotted in polar form.

ie: if

$w = r_1 e ^ {i \theta_1 } $

$x = r_2 e ^ {i \theta_2 } $


$w/x = (r_1 / r_2)e^{i(\theta_1 - \theta_2)}$



There are multiple ways by which you can describe this relation;

Since you have asked to give geometric representation so phasor form must be better to understand.

Dividing two complex numbers means to take the complex number that is of the magnitude equal to that of the division of amplitudes of the $X$ and $Y$ complex number and the phase of the new generated complex number is actually the difference of the phase between them.

Eg. $X = 2+3i \ , \ Y = 9+3i $

so the $|Z|= \frac{ |X|}{|Y|} = \dfrac{\sqrt{2^2+3^2}}{\sqrt{9^2+3^2}}$ and $\arg(Z)= \tan^{-1}\left(\frac32\right) - \tan^{-1}\left(\frac 39 \right)$

This $Z$ is also a complex number with amplitude and phase in geometric way.. but you to visualize the complex imaginary axis and real axis :D


Since $\frac wx = \frac{w\cdot x^*}{|x|^2}$, complex division is basically equivalent to multiplication of $w$ and the complex conjugate of $x$, but rescaled by the squared absolute value $|x|^2$.


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