Question #1907

Could someone shine some light on how to find the inverse,the domain and the rule of the
following function?
f(z)= ln [(z+1+i)/3]

Expert's answer

Similar to the functions of real variable. The inverse function to logariphm is exponenta:

ln [(z+1+i)/3] = ln (z+a) - ln3, where a = 1+i.

ln(f(z)+a) - ln3 = z

f(z)+a = e(z+ln3)

f(z) = e^{(z+ln3)} - a = 3e^{z} -a

Thus inverse is 3e^{z}-1-i.

The domain is z+a > 0, z > -a, z > (1+i).

The domain is

ln [(z+1+i)/3] = ln (z+a) - ln3, where a = 1+i.

ln(f(z)+a) - ln3 = z

f(z)+a = e(z+ln3)

f(z) = e

Thus inverse is 3e

The domain is z+a > 0, z > -a, z > (1+i).

The domain is

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