Question

Write the polar form of 2+i/2-i

Answer 1

Let x + iy = (2 + i)/(2 - i) where 'x' represents the real component & 'y' represents the imaginary component of the given equation.

x + iy = (2 + i)/(2 - i)

Multiplying the numerator and the denominator of RHS by (2+i), we get :

x + iy = (2+i)(2+i)/(2-i)(2+i)

x + iy = (2² + i² + 2(2)(i))/(2×2 + 2×i - (2×i) - (i×i))

We know that i² = -1.

Therefore,

x + iy = (4 + (-1) + 4i)/(4 + 2i - 2i - (-1))

x + iy = (4 - 1 + 4i)/(4 + 1)

x + iy = (3 + 4i)/5

i.e., x + iy = (3/5) + i(4/5)

Polar form of a complex number is written as :

x + iy = r(cos$\theta$ + isin$\theta$)

Where, r = $\sqrt{\smash[b]{x² + y²}}$ & $\theta$ = arctan(y/x)

r = $\sqrt{\smash[b]{(3/5)² + (4/5)²}}$ & $\theta$ = arctan((4/5)/(3/5))

r = $\sqrt{\smash[b]{(9/25) + (16/25)}}$ & $\theta$ = arctan(4/3)

r = $\sqrt{\smash[b]{(25/25}}$ & $\theta$ = arctan(4/3)

r = $\sqrt{\smash[b]{1}}$ & $\theta$ = arctan(4/3)

$\therefore$ r = 1 & $\theta$ = 53.13$\degree$

Hence, the polar form of the given equation is :

(2 + i)/(2 - i) = (3/5) + i(4/5) = 1(cos(53.13$\degree$) + isin(53.13$\degree$))

$\therefore$ (2 + i)/(2 - i) = cos(53.13$\degree$) + isin(53.13$\degree$)