Different cases in polar form of complex number with illustrations

Polar Form of Complex Numbers

i) Find the polar form of \( z = 1 + i \)

Let \( x = 1 \), \( y = 1 \)
\( r = \sqrt{x^2 + y^2} = \sqrt{1^2 + 1^2} = \sqrt{2} \)
\( \cos \theta = \frac{x}{r} = \frac{1}{\sqrt{2}} \), \( \sin \theta = \frac{y}{r} = \frac{1}{\sqrt{2}} \)
\( \theta = \tan^{-1}\left(\frac{1}{1}\right) = \frac{\pi}{4} \)
Polar form:
\( z = \sqrt{2}(\cos \frac{\pi}{4} + i \sin \frac{\pi}{4}) \)

ii) Find the polar form of \( z = -1 + i \)

Let \( x = -1 \), \( y = 1 \); here, \( x < 0, y > 0 \) (2nd quadrant)
\( r = \sqrt{x^2 + y^2} = \sqrt{(-1)^2 + 1^2} = \sqrt{2} \)
\( \theta = \tan^{-1}\left(\frac{1}{-1}\right) = -\frac{\pi}{4} \), but since it’s in 2nd quadrant,
\( \theta = \pi - \frac{\pi}{4} = \frac{3\pi}{4} \)
Polar form:
\( z = \sqrt{2}(\cos \frac{3\pi}{4} + i \sin \frac{3\pi}{4}) \)

iii) Find the polar form of \( z = -1 - i \)

Let \( x = -1 \), \( y = -1 \); 3rd quadrant
\( r = \sqrt{x^2 + y^2} = \sqrt{2} \)
\( \theta = \tan^{-1}\left(\frac{-1}{-1}\right) = \tan^{-1}(1) = \frac{\pi}{4} \)
Since in 3rd quadrant: \( \theta = \pi + \frac{\pi}{4} = \frac{5\pi}{4} \)
Polar form:
\( z = \sqrt{2}(\cos \frac{5\pi}{4} + i \sin \frac{5\pi}{4}) \)

iv) Find the polar form of \( z = 1 - i \)

Let \( x = 1 \), \( y = -1 \); 4th quadrant
\( r = \sqrt{x^2 + y^2} = \sqrt{2} \)
\( \theta = \tan^{-1}\left(\frac{-1}{1}\right) = -\frac{\pi}{4} \)
Polar form:
\( z = \sqrt{2}(\cos(-\frac{\pi}{4}) + i \sin(-\frac{\pi}{4})) \)