log((x+iy)/log(x-iy)=2i tan inverse y/x

Question

prove that

log((x+iy)/log(x-iy)=2i tan inverse y/x

Answer 1

If by "log" you mean the natural logarithm "ln" then, 

 

ln[(x+yi)/(x-yi)] = ln(x+yi) - ln(x-yi)

 

    = [(1/2)ln(x²+y²)+iTan^-1 (y/x)] - [(1/2)ln(x²+(-y)²)+iTan^-1(-y/x)]

 

    = (1/2)ln(x²+y²)+iTan^-1(y/x) - (1/2)ln(x²+y²) - iTan^-1(-y/x)

 

    = iTan^-1(y/x) + iTan^-1(y/x)   (since the inverse tangent is an odd                                                                               function)

 

    = 2iTan^-1(y/x)