Multiplication and Division in Polar Form

This lesson explores how multiplication and division of complex numbers work geometrically in polar form. The key insight is that multiplying complex numbers multiplies their moduli and adds their arguments, and dividing subtracts arguments. This geometric perspective is beautiful and powerful.

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Multiplication in Polar Form

Let $z_1 = r_1(\cos\theta_1 + i\sin\theta_1)$z1​=r1​(cosθ1​+isinθ1​) and $z_2 = r_2(\cos\theta_2 + i\sin\theta_2)$z2​=r2​(cosθ2​+isinθ2​).

Their product is:

$z_1 z_2 = r_1 r_2 (\cos(\theta_1 + \theta_2) + i\sin(\theta_1 + \theta_2))$z1​z2​=r1​r2​(cos(θ1​+θ2​)+isin(θ1​+θ2​))

Proof: Expand using the angle addition formulae:

$z_1 z_2 = r_1 r_2 [(\cos\theta_1 \cos\theta_2 - \sin\theta_1 \sin\theta_2) + i(\sin\theta_1 \cos\theta_2 + \cos\theta_1 \sin\theta_2)]$z1​z2​=r1​r2​[(cosθ1​cosθ2​−sinθ1​sinθ2​)+i(sinθ1​cosθ2​+cosθ1​sinθ2​)]

$= r_1 r_2 [\cos(\theta_1 + \theta_2) + i\sin(\theta_1 + \theta_2)]$=r1​r2​[cos(θ1​+θ2​)+isin(θ1​+θ2​)]

Geometric Interpretation

Multiplying by a complex number $w = r(\cos\theta + i\sin\theta)$w=r(cosθ+isinθ):

• Scales the distance from the origin by $r$r
• Rotates anticlockwise by angle $\theta$θ

This means complex multiplication is a spiral similarity (a combination of scaling and rotation).

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Division in Polar Form

$\frac{z_1}{z_2} = \frac{r_1}{r_2}(\cos(\theta_1 - \theta_2) + i\sin(\theta_1 - \theta_2))$z2​z1​​=r2​r1​​(cos(θ1​−θ2​)+isin(θ1​−θ2​))

Division divides moduli and subtracts arguments.

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Worked Examples

Example 1: Multiply $z_1 = 2(\cos\frac{\pi}{3} + i\sin\frac{\pi}{3})$z1​=2(cos3π​+isin3π​) and $z_2 = 3(\cos\frac{\pi}{6} + i\sin\frac{\pi}{6})$z2​=3(cos6π​+isin6π​).

$z_1 z_2 = 6\left(\cos\frac{\pi}{2} + i\sin\frac{\pi}{2}\right) = 6(0 + i) = 6i$z1​z2​=6(cos2π​+isin2π​)=6(0+i)=6i

Example 2: Compute $\frac{z_1}{z_2}$z2​z1​​ for the same $z_1, z_2$z1​,z2​.

$\frac{z_1}{z_2} = \frac{2}{3}\left(\cos\frac{\pi}{6} + i\sin\frac{\pi}{6}\right) = \frac{2}{3}\left(\frac{\sqrt{3}}{2} + \frac{1}{2}i\right) = \frac{\sqrt{3}}{3} + \frac{1}{3}i$z2​z1​​=32​(cos6π​+isin6π​)=32​(23​​+21​i)=33​​+31​i

Example 3: Find the modulus and argument of $(1 + i)^2$(1+i)2 using polar form.

First convert: $1 + i = \sqrt{2}(\cos\frac{\pi}{4} + i\sin\frac{\pi}{4})$1+i=2​(cos4π​+isin4π​).

$(1 + i)^2 = (\sqrt{2})^2 \left(\cos\frac{\pi}{2} + i\sin\frac{\pi}{2}\right) = 2\left(\cos\frac{\pi}{2} + i\sin\frac{\pi}{2}\right) = 2i$(1+i)2=(2​)2(cos2π​+isin2π​)=2(cos2π​+isin2π​)=2i

Modulus: 2. Argument: $\frac{\pi}{2}$2π​.

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Repeated Multiplication and Powers

By repeatedly applying the multiplication rule:

$z^n = r^n(\cos(n\theta) + i\sin(n\theta))$zn=rn(cos(nθ)+isin(nθ))

This is De Moivre's Theorem (which we will prove formally in a later lesson). For now, we can use this result to compute powers efficiently.

Example 4: Compute $(1 + \sqrt{3}\, i)^3$(1+3​i)3.

Convert to polar: $|1 + \sqrt{3}\, i| = 2$∣1+3​i∣=2, $\arg = \frac{\pi}{3}$arg=3π​.

$(1 + \sqrt{3}\, i)^3 = 2^3 \left(\cos\pi + i\sin\pi\right) = 8(-1 + 0) = -8$(1+3​i)3=23(cosπ+isinπ)=8(−1+0)=−8

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Rotation and Scaling Transformations

Multiplying every point $z$z in the Argand diagram by a fixed complex number $w$w transforms the entire plane.