Unit-I: Sets and Functions
Unit-II: Algebra
Unit-III: Coordinate Geometry
Unit-IV: Calculus
Unit-V: Mathematical Reasoning
Unit-VI: Statistics and Probability

Polar representation of Complex Numbers

Argand Plane / Complex Plane:

The plane that has a complex number assigned to each of its points is called the Complex plane or Argand Plane.

The x-axis and the y-axis in the Argand plane are called,respectively the real axis and the imaginary axis.

The identification of a complex number on a plane was proposed by Jean-Robert Argand (1768-1822), hence the complex plane is known as the Argand plane.

Polar Representation of a Complex Number

Let the point  P have a polar coordinates (r, \theta) .

Since  P has a cartesian  (x, y) we have:

 x=r \cos \theta, ~ y=r\sin \theta –(i)

Clearly, r=\sqrt{x^2+y^2} \text { and } \tan \theta= \dfrac{y}{x}, x \neq 0 —(ii)

Now,  z=x+iy=r \left( \cos \theta+i \sin \theta \right)—(iii)

For any value of  \theta (i), (ii) and (iii) holds is called an Argument of  z denoted by \theta=\arg z , when  z=0=0+i0 , \arg z is not defined.

A unique value to \arg z by restricting  \theta to the interval (-\pi, \pi] i.e., -\pi < \theta \leq \pi , then it is called the Principal Argument.

 -\pi < \arg z \leq \pi

 \arg z is determined as follows:

\arg z=\begin{Bmatrix}\tan^{-1} & \text { if} & x >0, y > 0 \\\pi+\tan ^{-1}\dfrac{y}{x} &\text {if} & x<0, y > 0 \\-\pi +\tan^{-1}\dfrac{y}{x} &\text {if} &x<0,y <0 \\\dfrac{\pi}{2} &\text {if} & x=0, y >0 \\-\dfrac{\pi}{2} &\text {if} & x=0, y <0\end{Bmatrix}

where -\dfrac{\pi}{2} < \tan ^{-1} \dfrac{y}{x} < \dfrac{\pi}{2}

Example:

Find  \arg \left( \dfrac{-1+\sqrt{3}i}{2} \right) \text { and } \arg \left( \dfrac{-1-\sqrt{3}i}{2} \right)

Solution:

  (i)\begin{aligned}\arg\left( \dfrac{-1+\sqrt{3}i}{2} \right) &=\tan^{-1}\dfrac{\dfrac{\sqrt{3}}{2}}{\dfrac{-1}{2}}\\&=\tan^{-1}-\sqrt{3}=\pi-\tan {-1}\sqrt{3}\\&=\pi-\dfrac{\pi}{3}=\dfrac{2\pi}{3} \end{aligned}

  (ii)\begin{aligned} \arg \left( \dfrac{-1-\sqrt{3}i}{2} \right)&=\tan^{-1}\left( \dfrac{-\dfrac{\sqrt{3}}{2}}{-\dfrac{1}{2}} \right)=\tan^{-1} \sqrt{3}\\&=\dfrac{\pi}{3} \end{aligned}

Some Properties in Polar Form:

 \begin{aligned} \text {If }x&=r\left( \cos \theta+i \sin \theta \right)\\\implies \overline{z}&=r \left( \cos-i\sin \theta \right)\\z\cdot\overline{z}&= \left\{ r\left( \cos \theta+i \sin \theta \right) \right\}\left\{ r\left( \cos \theta-i \sin \theta \right) \right\}\\&=r^2\left( \cos^2\theta+\sin^2\theta \right)\\&=r^2\\\implies |z|^2&=r^2 \implies |z|=r\end{aligned}

(i)  \left( \cos \alpha+i \sin \alpha \right)\left( \cos \beta+i \sin \beta \right)=\cos (\alpha+\beta)+i \sin (\alpha\beta)

(ii) (\cos \alpha+i \sin \alpha)(\cos \beta+i \sin \beta)=\cos (\alpha-\beta)+i \sin (\alpha-\beta)

If z_1 and  z_2 are the two complex numbers expressed by:

 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_1z_2=r_1r_2\left[ \cos (\theta_1+\theta_2)+i\sin(\theta_1+\theta_2) \right]

and  \dfrac{z_1}{z_2}=\dfrac{r_1}{r_2}\left[ \cos (\theta_1-\theta_2)+ i \sin (\theta_1-\theta_2) \right], z_2 \neq 0

If  |\cos \theta+i \sin \theta|=1, we get:

(i)  |z_1z_2|=r_1r_2=|z_1||z_2|

(ii) \left| \dfrac{z_1}{z_2} \right|=\dfrac{r_1}{r_2}=\dfrac{|z_1|}{|z_2|}

(iii)  \arg z_1z_2=\theta_1+\theta_2=\arg z_1+\arg z_2

(iv) \arg \left( \dfrac{z_1}{z_2} \right)=\theta_1-\theta_2=\arg z_1-\arg z_2

NOTE:

If z \neq 0, ~ \arg z+\arg \overline{z}=2n \pi , where  n is an integer or zero.

If z_1=a_1+ib_1, \text { and } z_2=a_2+ib_2 , then graphical representation of  z_1+z_2=\left\{ (a_1+a_2)+i(b_1+b_2) \right\}

Example 1:

Express  \dfrac{2+3i}{5-2i} in the form of a+ib .

Solution:

\begin{aligned} \dfrac{2+3i}{5-2i}&=\dfrac{(2+3i)(5+2i)}{(5-2i)(5+2i)}=\dfrac{10-6+i(15+4)}{5^2+2^2}\\&=\dfrac{4+19i}{29}=\dfrac{4}{29}+i\dfrac{19}{29} \end{aligned}

Example 2:

If z_1=2\left[ \cos \dfrac{\pi}{3}+i \sin \dfrac{\pi}{3} \right] and z_2=3\left[ \cos \dfrac{\pi}{6}+i \sin \dfrac{\pi}{6} \right]

\begin{aligned}\text { then ( i ) } z_1z_2&=2\times 3\left[ \cos \left( \dfrac{\pi}{3}+\dfrac{\pi}{6} \right)+i \sin \left( \dfrac{\pi}{3}+\dfrac{\pi}{6} \right) \right] \\&=6\left[ \cos\left( \dfrac{2\pi+\pi}{6} \right)+i \sin \left( \dfrac{2\pi+\pi}{6} \right) \right]\\&=6\left[ \cos \dfrac{\pi}{2}+i \sin \dfrac{\pi}{2} \right] \\&=6 \times [0+i \times 1]=6i \end{aligned}

\begin{aligned}(ii)\dfrac{z_1}{z_2}&=\dfrac{2}{3} \left[ \cos \left( \dfrac{\pi}{3}-\dfrac{\pi}{6} \right)+i \sin \left( \dfrac{\pi}{3}-\dfrac{\pi}{6} \right) \right]\\&=\dfrac{2}{2}\left[ \cos \dfrac{\pi}{6}+i \sin \dfrac{\pi}{6} \right] \\&=\dfrac{2}{3}\left[ \dfrac{\sqrt{3}}{2}+i \times \dfrac{1}{2} \right]=\dfrac{\sqrt{3}+i}{3} \end{aligned}

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