The complex number is -1+i
Let rcosθ = -1 and rsinθ = 1
Squaring and adding
\( r^2cos^2θ + r^2sin^2θ = (-1)^2+1^2\)
\( r^2(cos^2θ + sin^2θ) =1+1\)
\( r^2=2\)
\( r=\sqrt{2}\)
\(\sqrt{2} cosθ = -1 \;and\; \sqrt{2}sinθ = 1\)
\( cosθ =- \dfrac{1}{\sqrt{2}}and \sqrt{2}sinθ =1\)
\( θ ...(\pi-\dfrac{\pi}{4})...-\dfrac{3\pi}{4}L\) [As θ lies in the II quadrant]
It can be written,
-1+i = rcosθ + irsinθ
=\(\sqrt{2}cos \dfrac{3\pi}{4}+i\sqrt{2}sin \dfrac{3\pi}{4}\)
= \( \sqrt{2}(cos \dfrac{3\pi}{4}+isin \dfrac{3\pi}{4})\)
Answered by Pragya Singh | 1 year agoShow that 1 + i10 + i20 + i30 is a real number?
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