1.4 Q-6

Exercise 1.4 — Question 6

Topic: Operations with Complex Numbers in Polar Form

This question applies polar (trigonometric) form of complex numbers and the rules for multiplication and division in polar form.

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Key Concepts

A complex number $z=a+ib$ can be written in polar form as:

$z=r(\cos \theta +i\sin \theta )$

where:

• $r=|z|=\sqrt{a^2+b^2}$ is the modulus
• $\theta =\arg (z)={\tan }^{-1}\left(\frac{b}{a}\right)$ is the argument

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Multiplication in Polar Form

If $z_1=r_1(\cos {\theta }_1+i\sin {\theta }_1)$ and $z_2=r_2(\cos {\theta }_2+i\sin {\theta }_2)$, then:

$z_1{z}_2=r_1{r}_2[\cos ({\theta }_1+{\theta }_2)+i\sin ({\theta }_1+{\theta }_2)]$

Rule: Multiply the moduli and add the arguments.

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Division in Polar Form

$\frac{z_1}{z_2}=\frac{r_1}{r_2}[\cos ({\theta }_1-{\theta }_2)+i\sin ({\theta }_1-{\theta }_2)],z_2\ne 0$

Rule: Divide the moduli and subtract the arguments.

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Worked Example (Typical Q6 Type)

Express $z_1=1+i$ and $z_2=\sqrt{3}-i$ in polar form, then find $z_1{z}_2$ and $\frac{z_1}{z_2}$.

Step 1 — Polar form of $z_1=1+i$:

$r_1=\sqrt{1^2+1^2}=\sqrt{2},{\theta }_1={\tan }^{-1}\left(\frac{1}{1}\right)=45°=\frac{\pi }{4}$

$z_1=\sqrt{2}\left(\cos 45°+i\sin 45°\right)$

Step 2 — Polar form of $z_2=\sqrt{3}-i$:

$r_2=\sqrt{(\sqrt{3}{)}^2+(-1{)}^2}=\sqrt{4}=2,{\theta }_2=-30°=-\frac{\pi }{6}$

$z_2=2\left(\cos (-30°)+i\sin (-30°)\right)$

Step 3 — Product $z_1{z}_2$:

$z_1{z}_2=\sqrt{2}\cdot 2[\cos (45°+(-30°))+i\sin (45°+(-30°))]=2\sqrt{2}\left(\cos 15°+i\sin 15°\right)$

Step 4 — Quotient $\frac{z_1}{z_2}$:

$\frac{z_1}{z_2}=\frac{\sqrt{2}}{2}[\cos (45°-(-30°))+i\sin (45°-(-30°))]=\frac{\sqrt{2}}{2}\left(\cos 75°+i\sin 75°\right)$

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Summary of Rules