Adding to what Bipinji is saying

 

ln(a+ib) = 2ni + (1/2)ln(a² + b²) + i.tan^-1(b/a) is the more general expression.

It may puzzle you that if we put b = 0, i.e. for a real number, you get infinite values for its logarithm of a real number and that too in complex numbers. The explanation is quite simple.

 

Remember the polynomial expansion or more appropriately, the Taylor expansion of ln(1+x) = x - x²/2 + x³/3 -.....

 

That's a polynomial with infinte degree. So infinite solutions in complex numbers.