Convert the given complex number in polar form 1 - i and evaluate

Asked by Pragya Singh | 1 year ago |  58

##### Solution :-

The complex number is 1-i

Let $$rcos\theta=1$$ and $$rsin\theta=-1$$

$$r^2cos^2\theta+r^2sin^2\theta=1^2+(-1)^2$$

$$r^2(cos^2\theta+r^2sin^2\theta)=1+1$$

r= 2

$$r=\sqrt{2}$$

$$\sqrt{2}cos\theta=1$$ and  $$\sqrt{2}sin\theta=-1$$

$$cos\theta=\dfrac{1}{\sqrt{2}}$$ and $$sin\theta=-\dfrac{1}{\sqrt{2}}$$

$$\theta=-\dfrac{\pi}{4}$$ [ As θ liesin the IV quadrant ]

1- i = rcosθ + irsinθ

$$\sqrt{2}cos(-\dfrac{\pi}{4})+i\sqrt{2}sin(-\dfrac{\pi}{4})$$

$$\sqrt{2}[cos(-\dfrac{\pi}{4})+isin(-\dfrac{\pi}{4})]$$

Answered by Abhisek | 1 year ago

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