Question

The imaginary part of log (1 + i) is​

Answer 1

Answer:

Let z = x + iy

According to Euler Theorem for Complex Numbers :

$z= re^{i\alpha }$

log z = log r + i$\alpha$

Clearly we can see that the imaginary part of log z is the argument $\alpha$

$\alpha$ = arctan(y/x) , where x = 1 = y

$\alpha$ = arctan(1) = $\frac{\pi }{4}$

Therefore the imaginary part of log(1+i) is $\frac{\pi }{4}$

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