1.4 Q-8

Exercise 1.4 — Question 8

Topic: Operations with Complex Numbers in Polar Form

This question involves applying operations (multiplication and/or division) to complex numbers expressed in polar (trigonometric) form.

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Key Concepts

Polar Form of a Complex Number:

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 (argument magnitude)
• $\theta =\arg (z)={\tan }^{-1}\left(\frac{b}{a}\right)$ is the argument (angle with positive real axis)

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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:

$z_1\cdot z_2=r_1{r}_2[\cos ({\theta }_1+{\theta }_2)+i\sin ({\theta }_1+{\theta }_2)]$

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Division in Polar Form

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

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De Moivre's Theorem (Power in Polar Form)

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

This is used when a complex number in polar form is raised to a power $n$.

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Worked Example

Let $z_1=2\left(\cos 60^{\circ }+i\sin 60^{\circ }\right)$ and $z_2=3\left(\cos 30^{\circ }+i\sin 30^{\circ }\right)$.

Find $z_1\cdot z_2$:

$z_1\cdot z_2=(2)(3)[\cos (60^{\circ }+30^{\circ })+i\sin (60^{\circ }+30^{\circ })]$ $=6(\cos 90^{\circ }+i\sin 90^{\circ })=6(0+i)=6i$

Find $\frac{z_1}{z_2}$:

$\frac{z_1}{z_2}=\frac{2}{3}[\cos (60^{\circ }-30^{\circ })+i\sin (60^{\circ }-30^{\circ })]$ $=\frac{2}{3}(\cos 30^{\circ }+i\sin 30^{\circ })=\frac{2}{3}\left(\frac{\sqrt{3}}{2}+\frac{i}{2}\right)=\frac{\sqrt{3}}{3}+\frac{i}{3}$

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Steps to Solve Exercise 1.4 Q8

1. Convert each complex number to polar form: find $r$ and $\theta$.
2. Apply the appropriate polar operation rule (multiply/divide/power).
3. Simplify the resulting modulus and argument.
4. Convert back to rectangular form $a+ib$ if required.