إذا كان $Z=\cos \theta +i\sin \theta$ أثبت أن $\frac{Z^n}{1+Z^{2n}}=\frac{1}{2\cos n\theta }$

$\begin{array}{rl}\mathbf{LHS}& :\frac{Z^n}{1+Z^{2n}}=\frac{(\cos \theta +i\sin \theta {)}^n}{1+(\cos \theta +i\sin \theta {)}^{2n}}\\ & \frac{(\cos \theta +i\sin \theta {)}^n}{1+(\cos \theta +i\sin \theta {)}^{2n}}=\frac{\cos n\theta +i\sin n\theta }{1+\cos 2n\theta +i\sin 2n\theta }\\ & =\frac{\cos n\theta +i\sin n\theta }{1+2{\cos }^{2}n\theta -1+i(2\sin n\theta \cos n\theta )}=\frac{\cos n\theta +i\sin n\theta }{2{\cos }^{2}n\theta +i(2\sin n\theta \cos n\theta )}\\ & =\frac{\cos n\theta +i\sin n\theta }{2\cos n\theta (\cos n\theta +i\sin n\theta )}=\frac{1}{2\cos n\theta }:\mathbf{RHS}\end{array}$

التمارين العامة الخاصة بالفصل الأول

(1)- جد قيم $x,y\in R$ والتي تحقق $\frac{y}{1+i}=\frac{x^2+4}{x+2i}$

بما أن $4=-4i^2$

$\begin{array}{rl}& \frac{y}{1+i}=\frac{x^2-4i^2}{x+2i}\Rightarrow \frac{y}{1+i}=\frac{(x+2i)(x-2i)}{x+2i}\Rightarrow \frac{y}{1+i}=(x-2i)\\ & y=(1+i)(x-2i)\\ & y=x-2i+xi-2i^2\\ & y+0i=x+2-2i+xi\\ & y+0i=(x+2)+(-2+x)i\\ & y=x+2\dots \left(1\right)\\ & 0=-2+x\dots \left(2\right)\Rightarrow x=2\end{array}$

نعوض في معادلة (1)

$\therefore y=2+2=4$

(2)- إذا كان $z=\frac{1-\sqrt{3}i}{1+\sqrt{-3}}$ عدداً مركباً جد باستخدام مبرهنة ديموافر $z^{\frac{1}{2}}$

$\begin{array}{rl}& \because \sqrt{-3}=\sqrt{3}i\\ & z=\frac{1-\sqrt{3}i}{1+\sqrt{3}i}\cdot \frac{1-\sqrt{3}i}{1-\sqrt{3}i}=\frac{1-\sqrt{3}i-\sqrt{3}i-3i^2}{1+3}=\frac{-2-2\sqrt{3}i}{4}\\ & \therefore z=\frac{-1}{2}-\frac{\sqrt{3}i}{2}\end{array}$

المقياس:

$$\begin{gathered} r=\sqrt{x^2+y^2}=\sqrt{\frac{1}{4}+\frac{3}{4}}=\sqrt{1}=1 \\ \cos \theta =\frac{x}{r}=\frac{-\frac{1}{2}}{1}=-\frac{1}{2} , \sin \theta =\frac{y}{r}=\frac{\frac{-\sqrt{3}}{2}}{1}=\frac{-\sqrt{3}}{2} \end{gathered}$$

نستنتج أن الزاوية $\theta$ تقع في الربع الثالث.

$\therefore \arg \left(z\right)=\pi +\frac{\pi }{4}=\frac{4\pi }{3}$

الصيغة القطبية للعدد المركب:

$\begin{array}{rl}& z=r(\cos \theta +i\sin \theta )\\ & z=1(\cos \frac{4\pi }{3}+i\sin \frac{4\pi }{3})\\ & \sqrt[n]{Z}=r^{\frac{1}{n}}[\cos \left(\frac{\theta +2k\pi }{n}\right)+i\sin \left(\frac{\theta +2k\pi }{n}\right)]\\ & z^{\frac{1}{2}}=1^{\frac{1}{2}}{(\cos \frac{4\pi }{3}+i\sin \frac{4\pi }{3})}^{\frac{1}{2}}\\ & z^{\frac{1}{2}}=[\cos \left(\frac{\frac{4\pi }{3}+2k\pi }{2}\right)+i\sin \left(\frac{\frac{4\pi }{3}+2k\pi }{2}\right)]k=0,1\end{array}$

$\frac{2\pi }{3}$ تقع في الربع الثاني.

$\begin{array}{c}k=0\Rightarrow Z_1=\cos \frac{4\pi }{6}+i\sin \frac{4\pi }{6}=\cos \frac{2\pi }{3}+i\sin \frac{2\pi }{3}\\ Z_1=-\cos \frac{\pi }{3}+i\sin \frac{\pi }{3}=(-\frac{1}{2}+\frac{\sqrt{3}}{2}i)\\ k=1\Rightarrow Z_2=[\cos \left(\frac{\frac{4\pi }{3}+2\pi }{2}\right)+i\sin \left(\frac{\frac{4\pi }{3}+2\pi }{2}\right)]\end{array}$

$\frac{10\pi }{6}$ تقع في الربع الرابع.

$\begin{array}{rl}& Z_2=\cos \left(\frac{\frac{4\pi +6\pi }{3}}{2}\right)+i\sin \left(\frac{\frac{4\pi +6\pi }{3}}{2}\right)=\cos \left(\frac{10\pi }{6}\right)+i\sin \left(\frac{10\pi }{6}\right)\\ & Z_2=\cos \left(\frac{5\pi }{3}\right)+i\sin \left(\frac{5\pi }{3}\right)=\cos \left(\frac{\pi }{3}\right)-i\sin \left(\frac{\pi }{3}\right)=(\frac{1}{2}-\frac{\sqrt{3}}{2}i)\end{array}$

(3)- إذا كان $Z=\cos 2x+i\sin 2x$ فأثبت أن $\frac{2}{1+z}=1-i\tan x$

ط1:

$\begin{array}{rl}\mathbf{LHS}& :\frac{2}{1+z}=\frac{2}{1+\cos 2x+i\sin 2x}\\ & =\frac{2}{1+2{\cos }^{2}x-1+2i\sin x\cos x}\\ & =\frac{2}{2\cos x(\cos x+i\sin x)}=\frac{{\cos }^{2}x+{\sin }^{2}x}{\cos x(\cos x+i\sin x)}=\frac{{\cos }^{2}x-i^2{\sin }^{2}x}{\cos x(\cos x+i\sin x)}\\ & =\frac{(\cos x-i\sin x)(\cos x+i\sin x)}{\cos x(\cos x+i\sin x)}=\frac{(\cos x-i\sin x)}{\cos x}=\frac{\cos x}{\cos x}-\frac{i\sin x}{\cos x}=1-i\tan x:\mathbf{RHS}\end{array}$

ط2:

$\begin{array}{rl}\mathbf{LHS}& :\frac{2}{1+z}=\frac{2}{1+\cos 2x+i\sin 2x}\\ & =\frac{2}{1+2{\cos }^{2}x-1+2i\sin x\cos x}\\ & =\frac{2}{2\cos x(\cos x+i\sin x)}=\frac{(\cos x+i\sin x{)}^{-1}}{\cos x}=\frac{(\cos x-i\sin x)}{\cos x}\\ & =\frac{\cos x}{\cos x}-\frac{i\sin x}{\cos x}=1-i\tan x:\mathbf{RHS}\end{array}$

(4)- إذا كان $Z=\cos \theta +i\sin \theta$ أثبت أن $\frac{Z^n}{1+Z^{2n}}=\frac{1}{2\cos n\theta }$

$\begin{array}{rl}\mathbf{LHS}& :\frac{Z^n}{1+Z^{2n}}=\frac{(\cos \theta +i\sin \theta {)}^n}{1+(\cos \theta +i\sin \theta {)}^{2n}}\\ & \frac{(\cos \theta +i\sin \theta {)}^n}{1+(\cos \theta +i\sin \theta {)}^{2n}}=\frac{\cos n\theta +i\sin n\theta }{1+\cos 2n\theta +i\sin 2n\theta }\\ & =\frac{\cos n\theta +i\sin n\theta }{1+2{\cos }^{2}n\theta -1+i(2\sin n\theta \cos n\theta )}=\frac{\cos n\theta +i\sin n\theta }{2{\cos }^{2}n\theta +i(2\sin n\theta \cos n\theta )}\\ & =\frac{\cos n\theta +i\sin n\theta }{2\cos n\theta (\cos n\theta +i\sin n\theta )}=\frac{1}{2\cos n\theta }:\mathbf{RHS}\end{array}$