Complex logarithm question

**Glitch** · June 14th 2011, 09:54 PM

The question
Find the cartesian form of the principle value of:

$((\frac{1 + i\sqrt{3}}{2})^{-3})^{1 - i}$

My attempt
$e^{(1 - i)Log(e^{(\frac{\pi i}{3})}^{-3})$
$e^{(1 - i)Log(e^{-\pi i}})$
$e^{(1 - i)(ln|1| + i(-\pi))}$ //uh oh, this is on a branch cut!

How do I proceed, if the principle Log is not defined on the negative real axis?

Thank you.

**ojones** · June 15th 2011, 12:32 AM

Not sure there's a problem here. The only place Log isn't defined is at $z=0$ .

**Prove It** · June 15th 2011, 01:51 AM

Originally Posted by Glitch

The question
Find the cartesian form of the principle value of:

$((\frac{1 + i\sqrt{3}}{2})^{-3})^{1 - i}$

My attempt
$e^{(1 - i)Log(e^{(\frac{\pi i}{3})}^{-3})$
$e^{(1 - i)Log(e^{-\pi i}})$
$e^{(1 - i)(ln|1| + i(-\pi))}$ //uh oh, this is on a branch cut!

How do I proceed, if the principle Log is not defined on the negative real axis?

Thank you.

$\displaystyle \left[\left(\frac{1 + i\sqrt{3}}{2}\right)^{-3}\right]^{1 - i} = \left(\frac{1 + i\sqrt{3}}{2}\right)^{-3 + 3i}$

$\displaystyle \left|\frac{1 + i\sqrt{3}}{2}\right| = \sqrt{\left(\frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2} = \sqrt{\frac{1}{4} + \frac{3}{4}} = \sqrt{1} = 1$

$\displaystyle \arg{\left(\frac{1 + i\sqrt{3}}{2}\right)} = \arctan{\left(\frac{\frac{\sqrt{3}}{2}}{\frac{1}{2 }}\right)} = \arctan{\left(\sqrt{3}\right)} = \frac{\pi}{3}$

$\displaystyle \begin{align*} \left(\frac{1 + i\sqrt{3}}{2}\right)^{-3 + 3i} &= \left(e^{\frac{\pi i}{3}}\right)^{-3 + 3i} \\ &= e^{(-1 + i)\pi i} \\ &= e^{-\pi - \pi i} \\ &= e^{-\pi}\cos{(-\pi)} + i\,e^{-\pi}\sin{(-\pi)} \\ &= -e^{-\pi} \end{align*}$

**Glitch** · June 15th 2011, 02:06 AM

Thanks, that's the answer I eventually got. However the book answer is $-e^{\pi}$ . I think they just added $2\pi$ to the principle argument (not sure why).

**Prove It** · June 15th 2011, 02:16 AM

Originally Posted by Glitch

Thanks, that's the answer I eventually got. However the book answer is $-e^{\pi}$ . I think they just added $2\pi$ to the principle argument (not sure why).

It's because the principal argument is in the domain $\displaystyle (-\pi, \pi]$ .

**Opalg** · June 15th 2011, 03:58 AM

Originally Posted by Glitch

However the book answer is $-e^{\pi}$ . I think they just added $2\pi$ to the principle argument (not sure why).

I think the answer to that must be that they wanted you to evaluate the expression in the order indicated by the bracketing:

$\Bigl[\Bigl(\frac{1 + i\sqrt{3}}{2}\Bigr)^{-3}\Bigr]^{1 - i} = \left[(e^{i\pi/3})^{-3}\right]^{1-i} = (e^{-i\pi})^{1-i}.$

At that stage, they want you to replace $e^{-i\pi}$ by its principal value $e^{i\pi}$ before raising it to the power $1-i.$ Then you get $(e^{i\pi})^{1-i} = e^{(i+1)\pi} = -e^\pi$ .

**ojones** · June 15th 2011, 03:55 PM

Originally Posted by Prove It

$\displaystyle \left[\left(\frac{1 + i\sqrt{3}}{2}\right)^{-3}\right]^{1 - i} = \left(\frac{1 + i\sqrt{3}}{2}\right)^{-3 + 3i}$

$\displaystyle \left|\frac{1 + i\sqrt{3}}{2}\right| = \sqrt{\left(\frac{1}{2}\right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2} = \sqrt{\frac{1}{4} + \frac{3}{4}} = \sqrt{1} = 1$

$\displaystyle \arg{\left(\frac{1 + i\sqrt{3}}{2}\right)} = \arctan{\left(\frac{\frac{\sqrt{3}}{2}}{\frac{1}{2 }}\right)} = \arctan{\left(\sqrt{3}\right)} = \frac{\pi}{3}$

$\displaystyle \begin{align*} \left(\frac{1 + i\sqrt{3}}{2}\right)^{-3 + 3i} &= \left(e^{\frac{\pi i}{3}}\right)^{-3 + 3i} \\ &= e^{(-1 + i)\pi i} \\ &= e^{-\pi - \pi i} \\ &= e^{-\pi}\cos{(-\pi)} + i\,e^{-\pi}\sin{(-\pi)} \\ &= -e^{-\pi} \end{align*}$

You got to be careful here; $({z^\alpha})^\beta \ne z^{\alpha \beta}$ in general.

**ojones** · June 15th 2011, 03:57 PM

Originally Posted by Glitch

Thanks, that's the answer I eventually got. However the book answer is $-e^{\pi}$ . I think they just added $2\pi$ to the principle argument (not sure why).

You can't just add stuff. Check what principal argument range the book is using.

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