MHB Effie's question via email about Complex Numbers

[Prove It]

If $\displaystyle \begin{align*} z = -2 + 2\,\mathbf{i} \end{align*}$ what is $\displaystyle \begin{align*} \textrm{arg}\,\left( z^5 \right) \end{align*}$?

First let's write this number in its polar form.

$\displaystyle \begin{align*} \left| z \right| &= \sqrt{\left( -2 \right) ^2 + 2^2} \\ &= \sqrt{4 + 4} \\ &= \sqrt{8} \\ &= 2\,\sqrt{2} \end{align*}$

and as the number is in Quadrant 2

$\displaystyle \begin{align*} \textrm{arg}\,\left( z \right) &= \pi - \arctan{ \left| \frac{2}{-2} \right| } \\ &= \pi - \arctan{ \left( 1 \right) } \\ &= \pi - \frac{\pi}{4} \\ &= \frac{3\,\pi}{4} \end{align*}$

thus we can say

$\displaystyle \begin{align*} z &= -2 + 2\,\mathrm{i} \\ &= 2\,\sqrt{2}\,\mathrm{e}^{ \frac{3\,\pi}{4}\,\mathrm{i} } \\ z^5 &= \left( 2\,\sqrt{2}\,\mathrm{e}^{\frac{3\,\pi}{4}\,\mathrm{i}} \right) ^5 \\ &= 128\,\sqrt{2}\,\mathrm{e}^{ \frac{15\,\pi}{4}\,\mathrm{i} } \end{align*}$

so that means $\displaystyle \begin{align*} z^5 \end{align*}$ makes an angle of $\displaystyle \begin{align*} \frac{15\,\pi}{4} \end{align*}$ with the positive real axis, but as we define $\displaystyle \begin{align*} \textrm{arg}\,\left( Z \right) \in \left( -\pi , \pi \right] \end{align*}$, that means we keep adding or subtracting integer multiples of $\displaystyle \begin{align*} 2\,\pi \end{align*}$ until we have an angle in this region.

Thus $\displaystyle \begin{align*} \textrm{arg}\,\left( z^5 \right) = -\frac{\pi}{4} \end{align*}$.

 

[SammyS]

First let's write this number in its polar form.

$\displaystyle \begin{align*} \left| z \right| &= \sqrt{\left( -2 \right) ^2 + 2^2} \\ &= \sqrt{4 + 4} \\ &= \sqrt{8} \\ &= 2\,\sqrt{2} \end{align*}$

and as the number is in Quadrant 2

$\displaystyle \begin{align*} \textrm{arg}\,\left( z \right) &= \pi - \arctan{ \left| \frac{2}{-2} \right| } \\ &= \pi - \arctan{ \left( 1 \right) } \\ &= \pi - \frac{\pi}{4} \\ &= \frac{3\,\pi}{4} \end{align*}$

thus we can say

$\displaystyle \begin{align*} z &= -2 + 2\,\mathrm{i} \\ &= 2\,\sqrt{2}\,\mathrm{e}^{ \frac{3\,\pi}{4}\,\mathrm{i} } \\ z^5 &= \left( 2\,\sqrt{2}\,\mathrm{e}^{\frac{3\,\pi}{4}\,\mathrm{i}} \right) ^5 \\ &= 128\,\sqrt{2}\,\mathrm{e}^{ \frac{15\,\pi}{4}\,\mathrm{i} } \end{align*}$

so that means $\displaystyle \begin{align*} z^5 \end{align*}$ makes an angle of $\displaystyle \begin{align*} \frac{15\,\pi}{4} \end{align*}$ with the positive real axis, but as we define $\displaystyle \begin{align*} \textrm{arg}\,\left( Z \right) \in \left( -\pi , \pi \right] \end{align*}$, that means we keep adding or subtracting integer multiples of $\displaystyle \begin{align*} 2\,\pi \end{align*}$ until we have an angle in this region.

Thus $\displaystyle \begin{align*} \textrm{arg}\,\left( z^5 \right) = -\frac{\pi}{4} \end{align*}$.

This problem is solved correctly.

 

[PeroK]

Well, we can see that $\arg (z) = \frac{3\pi}{4}$ by considering its position in the plane. Then $arg (z^5) = \frac{15\pi}{4} = -\frac {\pi}{4}$.