Complex Multiplication and Division as Rotations and Dilations

Hello! Last lesson, you wrote nonzero complex numbers in modulus–argument form: a distance from the origin and a direction. Today, those two pieces of information will tell you what multiplication and division do on an Argand diagram. This is useful when an exam question asks for an image point or an unknown multiplier without explicitly suggesting polar form.

---

Multiplication moves a point in two ways

Start with a fixed complex number $w$. Multiplying another number $z$ by $w$ changes both its distance from the origin and its direction. Watch this graphical example first:

To establish the rule, write the two nonzero numbers in the form from last lesson:

$$z=r\mathrm{cis}\alpha ,w=s\mathrm{cis}\beta ,r,s>0.$$

The trigonometric addition identities give

$$\mathrm{cis}\alpha \mathrm{cis}\beta =\mathrm{cis}(\alpha +\beta ).$$

Consequently,

$$\boxed{zw=rs\mathrm{cis}(\alpha +\beta ).}$$

There are two simultaneous geometric effects. The distance of $z$ from the origin is multiplied by $s=|w|$, and its direction changes by $\beta$, an argument of $w$. A positive $\beta$ means an anticlockwise rotation; a negative $\beta$ means a clockwise rotation. If $s>1$, the point moves farther from the origin; if $0<s<1$, it moves closer. Dilation always has a positive scale factor—a negative real multiplier achieves its half-turn through its angle, not through a negative scale factor.

For the algebra behind this interpretation, read the product rule and its short derivation:

This picture summarises the effect. Its multiplier is labelled $w=r{e}^{i\theta }$; for this lesson, read that as $w=r\mathrm{cis}\theta$.

The order in which you imagine the rotation and dilation does not affect the result: both are centred at the origin. The formulas tell you the final distance and direction directly.

---

Division reverses the transformation

Division by $w\ne 0$ must undo multiplication by $w$. If multiplying by $w=s\mathrm{cis}\beta$ scales distances by $s$ and rotates through $\beta$, dividing by it scales distances by $1/s$ and rotates through $-\beta$. Thus, for nonzero $z$,

$$\boxed{\frac{z}{w}=\frac{r}{s}\mathrm{cis}(\alpha -\beta ).}$$

You can justify this without memorising a separate geometric rule. Let $q=z/w$, so $qw=z$. The product rule requires $|q|s=r$, hence $|q|=r/s$. It also requires the angle of $q$, followed by the angle $\beta$ of $w$, to give the direction $\alpha$ of $z$. An angle for $q$ is therefore $\alpha -\beta$.

Be careful with the words clockwise and shrink. They describe division only when $\beta >0$ and $s>1$, respectively. Dividing by a multiplier with a negative argument rotates anticlockwise; dividing by one with modulus less than one enlarges distances.

For example, let

$$z=2\mathrm{cis}\left(\frac{\pi }{6}\right),w=\frac{1}{2}\mathrm{cis}\left(-\frac{\pi }{3}\right).$$

Then

$$\frac{z}{w}=4\mathrm{cis}\left(\frac{\pi }{6}-\left(-\frac{\pi }{3}\right)\right)=4\mathrm{cis}\left(\frac{\pi }{2}\right)=4i.$$

Here division doubles the distance of $z$ from the origin and rotates it anticlockwise through $\pi /3$. That is the opposite of the shrink-and-clockwise example in the lower diagram, but it follows exactly the same rule.

---

Angles may cross the principal-argument boundary

The angle $\alpha +\beta$ or $\alpha -\beta$ gives the correct direction, but it need not be the principal argument. As in the previous lesson, angles differing by a full revolution describe the same point.

Suppose

$$z=2\mathrm{cis}\left(\frac{5\pi }{6}\right),w=3\mathrm{cis}\left(\frac{2\pi }{3}\right).$$

Multiplication gives

$$zw=6\mathrm{cis}\left(\frac{3\pi }{2}\right)=6\mathrm{cis}\left(-\frac{\pi }{2}\right)=-6i.$$

The first angle records the total turn accurately. The second is the principal argument, because $-\pi /2$ lies in $(-\pi ,\pi ]$. In contrast, division gives

$$\frac{z}{w}=\frac{2}{3}\mathrm{cis}\left(\frac{\pi }{6}\right);$$

this time the angle difference already lies in the principal interval.

So avoid writing $\mathrm{Arg}(zw)=\mathrm{Arg}(z)+\mathrm{Arg}(w)$ without qualification. Arguments add or subtract up to a full revolution; principal arguments must then be adjusted into $(-\pi ,\pi ]$. At the negative real axis, that interval chooses $\pi$, not $-\pi$.

There are two boundary cases worth keeping distinct. Multiplying $z=0$ by any $w$ still gives $0$, and dividing $0$ by nonzero $w$ still gives $0$; neither result has an argument. Division by $w=0$ is undefined.

---

When the multiplier is hidden

In an unfamiliar question, you may be told where a point starts and ends rather than being given the multiplier. Suppose multiplication by a fixed complex number takes $1+2i$ to $-4+2i$. What geometric action occurred?

The starting distance is $\sqrt{5}$; the final distance is $2\sqrt{5}$. The scale factor is therefore $2$. On an Argand diagram, a quarter-turn anticlockwise takes the direction of $(1,2)$ to the direction of $(-2,1)$. Scaling by $2$ then gives $(-4,2)$. The multiplier should have modulus $2$ and argument $\pi /2$, namely $2i$.

Cartesian arithmetic verifies the deduction:

$$(1+2i)(2i)=-4+2i.$$

If the rotation is not obvious from a sketch, you can recover the multiplier by division. Calling it $q$,

$$q=\frac{-4+2i}{1+2i}=\frac{(-4+2i)(1-2i)}{5}=2i.$$

This suggests a useful exam strategy: compare distances and directions first when a diagram makes the transformation clear; use the quotient to confirm it or to determine a less familiar angle. For a nonzero starting point, the quotient also shows that the multiplier is unique.

---

Key takeaways

Multiplying by $w=s\mathrm{cis}\beta$ dilates distances from the origin by $s$ and rotates through $\beta$. Dividing by nonzero $w$ dilates by $1/s$ and rotates through $-\beta$. Neither operation translates the origin, and any calculated angle must be adjusted if a principal argument is requested.

The next lesson uses repeated multiplication—and therefore repeated scaling and turning—to evaluate integer powers with De Moivre’s theorem.