How to find the real and imaginary parts of the complex number z=ln(1+i)

We have Ln w = Ln | w | + i Arg w (I use w instead of z since you have already used z.)

Where w = a + b i is any complex number with a, b the real and imaginary parts respectively.

Here w = 1 + i, so a = 1, b = 1, and the modulus is | w | = sqrt(1^2 + 1^2) = sqrt(2)

Also Arg w = Arctan(b/a) = Arctan(1) = Pi/4 is the phase or argument

Hence z = Ln(1 + i) = Ln(sqrt(2)) + i Pi/4

It follows the real part is Ln(sqrt(2)) and the imaginary part is Pi/4

Remark: The phase is periodic; you can add multiples of 2Pi to it and land back on the original spot after a 360 degrees or 2Pi radian rotation. For example tan Pi/4 = tan 9Pi/4 etc.

So the imaginary part in general should be 2kPi + Pi/4 with k an integer. (Pi/4 is the simplest of them.)

Lastly, note that these are the EXACT values I have obtained. The first responder above has given you approximate solutions, most probably using a calculating utility. My method works by hand.

ln(1+i) = ln(sqrt2) +i (pi/4+2kpi)
In general ln ( x+i y )= ln (sqrt(x^2+y^2) +i (arctan y/x+2k pi)
ln(sqrt(x^2+y^2) is the real numbered ln
A complex number has infinite ln.

z = ln(1 + i)
e^z = 1 + i
e^z = √2 (cos(π/4) + i sin(π/4))
e^z = √2 e^((π/4) i)

z = ln(√2 e^((π/4) i))
z = ln(2^(1/2)) + ln((π/4) i)
z = 1/2 ln(2) + (π/4) i

Check: http://www.wolframalpha.com/input/?i=ln%…

Mαthmφm

real 0.346574

imaginary 0.785398