# Find principal value in a+ib form.

#### GaryBenton

##### New member
I need to find principal value of (1-i)^2i.

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.

#### Plato

##### Well-known member
MHB Math Helper
I need to find principal value of (1-i)^2i.
Let z = 1 - i
z^a = e^(a*lnz)
=e^2i*ln(1-i)
You are correct to this point.
$\exp \left[ {2i\log \left( {1 - i} \right)} \right] = \exp \left[ {2i\left( {\ln (\sqrt 2 ) - i\frac{\pi }{4}} \right)} \right]$

#### GaryBenton

##### New member
And then?

I know e^z = e^(x + iy) = e^x(cosy + isiny).

Is x simply 2i*ln(sqrt(2)) and y 2i*(i*pi/4) ?

Regards.

#### Plato

##### Well-known member
MHB Math Helper
I know e^z = e^(x + iy) = e^x(cosy + isiny).
Is x simply 2i*ln(sqrt(2)) and y 2i*(i*pi/4) ?
Write ${2i\log \left( {1 - i} \right)}$ in the $a+bi$ form.

If $z=\exp(a+bi)$ then $x=e^a\cos(b)~\&~y=e^x\sin(b)$.

• GaryBenton

#### Prove It

##### Well-known member
MHB Math Helper
I need to find principal value of (1-i)^2i.

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.
\displaystyle \begin{align*} (1 - i)^{2i} &= \left(2^{\frac{1}{2}}e^{-\frac{\pi}{4}i}\right)^{2i} \\ &= \left(2^{\frac{1}{2}}\right)^{2i}\left(e^{-\frac{\pi}{4}i}\right)^{2i} \\ &= 2^ie^{\frac{\pi}{2}} \\ &= e^{\log{\left(2^i\right)}}e^{\frac{\pi}{2}} \\ &= e^{i\log{2}}e^{\frac{\pi}{2}} \\ &= e^{\frac{\pi}{2}}\left[\cos{\left(\log{2}\right)} + i\sin{\left(\log{2}\right)}\right] \\ &= e^{\frac{\pi}{2}}\cos{\left(\log{2}\right)} + i\,e^{\frac{\pi}{2}}\sin{\left(\log{2}\right)} \end{align*}

• GaryBenton

#### GaryBenton

##### New member
Can you please explain where you got the 2^1/2 from?

My text book goes nowhere near explaining this material in depth. Frustrating...

Regards.

---------- Post added at 10:04 PM ---------- Previous post was at 08:42 PM ----------

All good. Understood (finally ).

Thanks and regards.

#### Prove It

##### Well-known member
MHB Math Helper
Can you please explain where you got the 2^1/2 from?

My text book goes nowhere near explaining this material in depth. Frustrating...

Regards.

---------- Post added at 10:04 PM ---------- Previous post was at 08:42 PM ----------

All good. Understood (finally ).

Thanks and regards.
It's the magnitude of 1 - i 