Using De Moivre's theorem, prove that \((1+i)^8+(1-i)^8=32\).

Q.3 Using De Moivre's theorem, prove that \((1+i)^8+(1-i)^8=32\).

Doubtly AI Asked 27 Oct 2026 AI-generated · not yet verified Practice test mu engineering Save WhatsApp Report

1 answer

Best answer

Answer: \((1+i)^8+(1-i)^8=16+16=32\).

\(1+i=\sqrt2\left(\cos\frac\pi4+i\sin\frac\pi4\right)\) and \(1-i=\sqrt2\left(\cos\frac\pi4-i\sin\frac\pi4\right)\).

By De Moivre's theorem: \((1+i)^8=(\sqrt2)^8\left(\cos2\pi+i\sin2\pi\right)=16(1+0)=16\).

Similarly \((1-i)^8=(\sqrt2)^8\left(\cos2\pi-i\sin2\pi\right)=16\).

\(\therefore\ (1+i)^8+(1-i)^8=16+16=32\). Hence proved.

Doubtly AI · 27 Oct 2026 AI-verified · Oct 2026 Report

Was this answer helpful?