Let ln(i)=p+iq, then i=ep+iq=epeiq=ep(cos(q)+isin(q)).
i=epcos(q)+iepsin(q)⇒cos(q)=0⇒q=(2n+1)π/2 and epsin(q)=1. But sin((2n+1)π/2)=±1. sin(π/2)=1, sin(3π/2)=-1, sin(5π/2)=1, etc., that is, sin((4n+1)π/2)=1, cos((4n+1)π/2)=0, and epsin((4n+1)π/2)=1, so ep=1⇒p=0, q=(4n+1)π/2.
ln(i)=p+iq=0+i(4n+1)π/2=i(4n+1)π/2.
If ia+ib=a+ib, then:
(a+ib)ln(i)=ln(a+ib), so i(a+ib)(4n+1)π/2=(ai-b)(4n+1)π/2=ln(a+ib).
(ai-b)(4n+1)π/2=ai(4n+1)π/2-b(4n+1)π/2,
a+ib=e-½b(4n+1)πe½ai(4n+1)π=e-½b(4n+1)π(cos(½a(4n+1)π)+isin(½a(4n+1)π)).
If a+ib=rcosθ+irsinθ, then a=rcosθ, b=rsinθ, so r2=a2+b2, tanθ=b/a.
Therefore, a=rcosθ=e-½b(4n+1)πcos(½a(4n+1)π), b=rsinθ=e-½b(4n+1)πsin(½a(4n+1)π).
r=e-½b(4n+1)π, θ=½a(4n+1)π.
r2=a2+b2=e-b(4n+1)π (QED), [also: b=atan(½a(4n+1)π)].