chapter complex number....
Question: Find log(1+i)
Suppose z = x + i.y
Then, e^z = e^x.e^(iy) = e^x(cosy - i.siny)
e^z = e^x(cosy - i.siny)
Now, let w = e^z, then log(w) = z
setting w = 1 + i, then 1 + i = e^z = e^x.cosy - i.e^x.siny, i.e.
1 = e^x.cosy
1 = -e^x.siny
Division of the above eqns gives tan(y) = -1 => y = -π/4.
y = -π/4 gives cosy = -siny = 1/√2, therefore e^x = √2 and x = log(√2)
log(1 + i) = log(w) = z = x + iy = log(√2) - i.π/4
Answer: log(1 + i) = log(√2) - i.π/4 Read More: ...