The complex number is 1-i

Let \( rcos\theta=1\) and \( rsin\theta=-1\)

squaring and adding

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

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

r^{2 }= 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 | 2 years agoShow that 1 + i^{10} + i^{20} + i^{30} is a real number?

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