Matematické Fórum

stuart clark · 25. 02. 2013 10:03

If $z = \cos \theta + i.\sin \theta$ . Then How can i prove

(1) $\displaystyle \arg(z^2+\bar{z}) = \frac{\arg(z)}{2}\;\;,$ If $\displaystyle 0 \leq \theta <\frac{\pi}{3}\;\;,\pi < \theta <\frac{5\pi}{3}$

(2) $\displaystyle \arg(z^2+\bar{z}) = \pi+\frac{\arg(z)}{2}\;\;,$ If $\displaystyle \frac{\pi}{3} < \theta < \pi \;\;,\frac{5\pi}{3} < \theta < 2\pi$

(3) $\displaystyle \arg(z^2+\bar{z}) = 0\;\;,$ If $\displaystyle \theta = \frac{\pi}{3} \;,\pi\;,\frac{5\pi}{3}$

Brano · 25. 02. 2013 12:46

For $0<\theta<\pi/3$ draw $\overline{z},z,z^2$ on unit circle. They have angles (arguments) $-\theta,\theta,2\theta$ respectively. Now complete $0,\overline{z},z^2$ into parallelogram and observe that line from $0$ to $z^2+\overline{z}$ is its diagonal and since the legths of $z^2$ and $\overline{z}$ are the same it splits its angle into even parts so the argument of $z^2+\overline{z}$ is $\theta/2$ .

Matematické Fórum

#1 25. 02. 2013 10:03

Complex number

#2 25. 02. 2013 12:46

Re: Complex number

Zápatí