Prove that log(1 + cos 2θ + isin 2θ) = log(2 cosθ) + i θ | Complex Analysis

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log [log {cosθ+isinθ}] log [log {cosθ+isinθ}]
log(1 + e ^ (i*theta)) = log(2cos(theta/2)) + 1/2 * i*theta log(1 + e ^ (i*theta)) = log(2cos(theta/2)) + 1/2 * i*theta
cos within log within log -- an integration spectacular! cos within log within log -- an integration spectacular!
Show that 1+cosθ+isinθ=2cos(θ/2)(cos(θ/2)+isin(θ/2)) Show that 1+cosθ+isinθ=2cos(θ/2)(cos(θ/2)+isin(θ/2))
log(1+cos2θ+isin2θ)=log(2cosθ)+i(θ+2k.pi) log(1+cos2θ+isin2θ)=log(2cosθ)+i(θ+2k.pi)
Visualizing Euler Visualizing Euler's formula.
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If ρ(cosα+isinα)=sin(θ+iϕ),prove that ρ^2=1/2(cosh2ϕ-cos2θ) and tanα=cotθ.tanhϕ. If ρ(cosα+isinα)=sin(θ+iϕ),prove that ρ^2=1/2(cosh2ϕ-cos2θ) and tanα=cotθ.tanhϕ.