The complex number is 1-i
Let \( rcos\theta=1\) and \( rsin\theta=-1\)
squaring and adding
\( r^2cos^2\theta+r^2sin^2\theta=1^2+(-1)^2\)
\( r^2(cos^2\theta+r^2sin^2\theta)=1+1\)
r2 = 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 agoShow that 1 + i10 + i20 + i30 is a real number?
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