oblicz podane pierwiastki

[wnoros89]

oblicz podane pierwiastki
1)\(\displaystyle{ \sqrt{j}}\)
2)\(\displaystyle{ \sqrt{4j-3}}\)
3)\(\displaystyle{ \sqrt[3]{27}}\)
4)\(\displaystyle{ \sqrt[4]{( \sqrt{3} -j) ^{12} }}\)

a)z definicji
b)z postaci trygonometrycznej

[raphel]

1. \(\displaystyle{ \sqrt{j} =|j| = 1}\)
2. \(\displaystyle{ ... = |4j-3| = \sqrt{16+9} =5}\)

[soku11]

raphel, czego utozsamiasz pierwiastek z modulem ?? Przeciez twoje rozwiazania sa zle...



1.
a)
\(\displaystyle{ \sqrt{j} =z\\
j=z^2\\
j=(x+jy)^2\\
j=x^2+2xyj-y^2\\
\begin{cases}
x^2-y^2=0\\
2xy=1\end{cases}\\
y=\frac{1}{2x}\\
x^2-\left(\frac{1}{2x}\right)^2=0\\
x^2-\frac{1}{4x^2}=0\\
4x^4-1=0\\
(2x^2-1)(2x^2+1)=0\\
2x^2-1=0\\
(\sqrt{2}x-1)(\sqrt{2}x+1)=0\\
x=\pm\frac{\sqrt{2}}{2}\\
\begin{cases}
x=\frac{\sqrt{2}}{2}\\
y=\frac{\sqrt{2}}{2}\end{cases}\;\;\vee\;\;
\begin{cases}
x=-\frac{\sqrt{2}}{2}\\
y=-\frac{\sqrt{2}}{2}\end{cases}\\
z=\frac{\sqrt{2}}{2}+j\frac{\sqrt{2}}{2}\;\;\vee\;\;z=-\frac{\sqrt{2}}{2}-j\frac{\sqrt{2}}{2}\\}\)
b)
\(\displaystyle{ z^2=j=\cos \frac{\pi}{2}+j\sin\frac{\pi}{2}\\
z_k=\cos \frac{\frac{\pi}{2}+2k\pi}{2}+j\sin\frac{\frac{\pi}{2}+2k\pi}{2}\;\;k\in\{0,1\}\\
z_1=\cos \frac{\pi}{4}+j\sin\frac{\pi}{4}=\frac{\sqrt{2}}{2}+j \frac{\sqrt{2}}{2}\\
z_2=\cos \frac{5\pi}{4}+j\sin\frac{5\pi}{4}=-\frac{\sqrt{2}}{2}-j\frac{\sqrt{2}}{2}}\)

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