- Thread starter
- #1

#### GaryBenton

##### New member

- Jan 31, 2012

- 3

Let z = 1 - i

z^a = e^(a*lnz)

=e^2i*ln(1-i)

r = z = sqrt(1^2 + (-1)^2 = sqrt(2)

tan theta = y/x = -1/1 = -1 => theta = argz = -45 degrees = -pi/4 rads.

ln z= loge(z) = i(theta + 2*pi*n)

= loge(sqrt(2)) + i(-pi/4 + 2*pi)

= loge(sqrt(2)) - i*pi/4

Could someone please help me out here finishing this one off - into a + ib form? Struggling a little with completion.

Much appreciated if you could.

Regards.