Help with Complex Numbers

**lidia** · September 22nd 2012, 01:36 AM

Hi,
I need help with this task:

Given that $z_1=1+2i$ and $z_2 =-3i$ find $(\frac{\bar {z_2}}{z_1})^6$ .

My solution so far

$(\frac{\bar {z_2}}{z_1})= \frac{3i}{1+2i}=\frac{3i(1-2i)}{(1+2i)(1-2i)}=\frac{3i+6}{1+4}=\frac{6+3i}{5}$

Now I need to find a trigonometric form

$|z|=\sqrt{(\frac{6}{5})^2+(\frac{3}{5})^2}=\frac{3 \sqrt{5}}{5}$

$\varphi=arctg\frac{\frac{3}{5}}{\frac{6}{5}}=arctg \frac{1}{2}$

**Prove It** · September 22nd 2012, 02:07 AM

Originally Posted by lidia

Hi,
I need help with this task:

Given that $z_1=1+2i$ and $z_2 =-3i$ find $(\frac{\bar {z_2}}{z_1})^6$ .

My solution so far

$(\frac{\bar {z_2}}{z_1})= \frac{3i}{1+2i}=\frac{3i(1-2i)}{(1+2i)(1-2i)}=\frac{3i+6}{1+4}=\frac{6+3i}{5}$

Now I need to find a trigonometric form

$|z|=\sqrt{(\frac{6}{5})^2+(\frac{3}{5})^2}=\frac{3 \sqrt{5}}{5}$

$\varphi=arctg\frac{\frac{3}{5}}{\frac{6}{5}}=arctg \frac{1}{2}$

Yeah, so $\displaystyle \begin{align*} \frac{\overline{z_2}}{z_1} = \frac{3\sqrt{5}}{5}\, e^{i\arctan{\frac{1}{2}}} \end{align*}$ , so

$\displaystyle \begin{align*} \left( \frac{\overline{z_2}}{z_1} \right)^6 &= \left( \frac{3\sqrt{5}}{5} \, e^{i\arctan{\frac{1}{2}}} \right)^6 \\ &= \frac{3^6}{5^3} \, e^{6i\arctan{\frac{1}{2}}} \\ &= \frac{729}{125} \left[ \cos{\left( 6\arctan{\frac{1}{2}} \right)} + i\sin{\left( 6\arctan{\frac{1}{2}} \right)} \right] \\ &= \frac{729}{125} \left( -\frac{117}{125} + \frac{44}{125}\,i \right) \end{align*}$

**HallsofIvy** · September 22nd 2012, 05:32 AM

Consider an equilateral triangle with sides of length 2. All angles have measure $\pi/3$ radians (60 degrees). If you draw a perpendicular from one vertex to the opposite side, it bisects the opposite side and bisects the angle, giving a right triangle with hypotenuse of length 2, one leg of length 1, and so, by the Pythagorean theorem, the other leg has length $\sqrt{3}$ . The half of the bisected angle, which is opposite the leg of length 1, has measure $\pi/6$ (30 degrees) so that $tan(\pi/6)= \frac{1}{2}$ and so $arctan(1/2)= \pi/6$ . That is a value a math student should have memorized.

**Prove It** · September 22nd 2012, 05:43 AM

Originally Posted by HallsofIvy

Consider an equilateral triangle with sides of length 2. All angles have measure $\pi/3$ radians (60 degrees). If you draw a perpendicular from one vertex to the opposite side, it bisects the opposite side and bisects the angle, giving a right triangle with hypotenuse of length 2, one leg of length 1, and so, by the Pythagorean theorem, the other leg has length $\sqrt{3}$ . The half of the bisected angle, which is opposite the leg of length 1, has measure $\pi/6$ (30 degrees) so that $tan(\pi/6)= \frac{1}{2}$ and so $arctan(1/2)= \pi/6$ . That is a value a math student should have memorized.

Last time I checked $\displaystyle \begin{align*} \tan{\frac{\pi}{6}} = \frac{1}{\sqrt{3}} \end{align*}$ , NOT $\displaystyle \begin{align*} \frac{1}{2} \end{align*}$ . That would be $\displaystyle \begin{align*} \sin{\frac{\pi}{6}} \end{align*}$ .

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