Q.4 (b) Show that log[sin(x+iy)]=1/2log(fraccosh2y-cos2x2)+itan⁻¹(cot xtanh y) .

sin(x+iy)=sin xcos iy+cos xsin iy=sin xcosh y+icos xsinh y . For log(a+ib)=1/2log(a²+b²)+itan⁻¹b/a : a²+b²=sin²xcosh²y+cos²xsinh²y=sin²x(1+sinh²y)+cos²xsinh²y=sin²x+sinh²y =frac1-cos2x2+fraccosh2y-12=fraccosh2y-cos2x2 b/a=cos xsinh y/sin xcosh y=cot xtanh y log[sin(x+iy)]=1/2log(fraccosh2y-cos2x2)+itan⁻¹(cot xtanh y) (proved).

Show that \(\log[\sin(x+iy)]=\frac12\log\left(\frac{\cosh2y-\cos2x}2\right)+i\tan^{-1}(\cot x\tanh y)\).

Q.4 (b) Show that \(\log[\sin(x+iy)]=\frac12\log\left(\frac{\cosh2y-\cos2x}2\right)+i\tan^{-1}(\cot x\tanh y)\).

Doubtly Asked 25 Oct 2026 mu engineering Save WhatsApp Report

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\(\sin(x+iy)=\sin x\cos iy+\cos x\sin iy=\sin x\cosh y+i\cos x\sinh y\).

For \(\log(a+ib)=\frac12\log(a^2+b^2)+i\tan^{-1}\frac ba\):

\(a^2+b^2=\sin^2x\cosh^2y+\cos^2x\sinh^2y=\sin^2x(1+\sinh^2y)+\cos^2x\sinh^2y=\sin^2x+\sinh^2y\)

\(=\frac{1-\cos2x}2+\frac{\cosh2y-1}2=\frac{\cosh2y-\cos2x}2\)

\(\frac ba=\frac{\cos x\sinh y}{\sin x\cosh y}=\cot x\tanh y\)

\(\log[\sin(x+iy)]=\frac12\log\left(\frac{\cosh2y-\cos2x}2\right)+i\tan^{-1}(\cot x\tanh y)\) (proved).

Doubtly AI · 25 Oct 2026 AI-verified · Oct 2026 Report

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Show that \(\log[\sin(x+iy)]=\frac12\log\left(\frac{\cosh2y-\cos2x}2\right)+i\tan^{-1}(\cot x\tanh y)\).

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