# Convert The Given Complex Number In Polar Form: 1 - i.

Sol:

Given, z = 1− i.

Let r cos θ = 1 and r sin θ = −1

On squaring and adding, we obtain:

$$\Rightarrow r^{2}\;cos^{2} \Theta + r^{2}\;sin^{2} \Theta \\\Rightarrow = 1^{2} + (-1)^{2}\\\Rightarrow r^{2}\;(cos^{2} \Theta + \;sin^{2}\Theta) = 2\\\Rightarrow r^{2} = 2\\\Rightarrow r = \sqrt{2}\;( since, r > 0)\\\Rightarrow \sqrt{2} cos \Theta = 1 \;and\; \sqrt{2}sin \Theta = -1.$$

Therefore, $$\Theta = \frac{\Pi}{4}$$

As θ lies in the fourth quadrant, the polar form is

$$1 – i = r \;cos\Theta + ir\;sin\Theta = \sqrt{2} cos (\frac{-\Pi }{4}) + i\sqrt{2}sin (\frac{-\Pi }{4}) \\\Rightarrow \sqrt{2}[cos (\frac{-\Pi}{4}) + i\;sin (\frac{-\Pi}{4})]$$

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