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z^2=i 

solve z

Guest Aug 10, 2015

Best Answer 

 #1
avatar+18715 
+8

z^2=i  solve z

 

$$\small{\text{$
\begin{array}{rcl}
z^2&=&i \qquad | \qquad z^2= a+bi \quad a=0 $ and $ b = 1\\\\
r&=& \sqrt{a^2+b^2}=\sqrt{0^2+1^2}=\sqrt{1^2}=1\\\\
\varphi &=& \arctan{ \left(\dfrac{b}{a}\right) }=\dfrac{1}{0} \Rightarrow \varphi = \dfrac{\pi}{2}
\qquad
\varphi =
\begin{cases}
\arctan\left(\frac{b}{a}\right) & \text{for}\ a > 0,\ b \geq 0 \qquad \text{(1. Quadrant)}\\
\arctan\left(\frac{b}{a}\right) + 2\pi & \mathrm{f\ddot ur}\ a > 0,\ b < 0 \qquad \text{(4. Quadrant)}\\
\arctan\left(\frac{b}{a}\right) + \pi & \mathrm{f\ddot ur}\ a < 0 \qquad\qquad\quad \text{(2./3. Quadrant)}\\
\frac{\pi}{2} & \mathrm{f\ddot ur}\ a = 0,\ b > 0 \qquad \text{(positive Im-axis)}\\
\frac{3\pi}{2} & \mathrm{f\ddot ur}\ a = 0,\ b < 0 \qquad \text{(negative Im-axis)}\\
\text{indeterminate} & \mathrm{f\ddot ur}\ a = 0,\ b = 0 \qquad \text{(zero)}
\end{cases} \\\\
z^2&=& r\cdot e^{i\cdot \varphi}\\\\
z^2&=& 1\cdot e^{i\cdot \frac{\pi}{2}+2k\pi} \qquad | \qquad \sqrt{}\\\\
z_k &=& e^{i\cdot \frac{\pi}{2\cdot 2 }+\frac{2k\pi}{2} } \\\\
z_k &=& e^{i\cdot \frac{\pi}{4}+ k\pi } \\\\
k =0 \quad z_0 &=& e^{i\cdot \frac{\pi}{4}}\\\\
k =1 \quad z_1 &=& e^{i\cdot \frac{5\pi}{4}} \\\\
&& \boxed{~e^{i\varphi} =\cos{(\varphi)+i\sin{(\varphi)}} ~}\\\\
z_0 &=& e^{ i\frac{\pi}{4} } =\cos{(\frac{\pi}{4})+i\sin{(\frac{\pi}{4})}}\\\\
\mathbf{z_0} &\mathbf{=}& \mathbf{\dfrac{ \sqrt{2} }{2} +\dfrac{ \sqrt{2} }{2} i}\\\\
z_1 &=& e^{ i\frac{5\pi}{4} } =\cos{(\frac{5\pi}{4})+i\sin{(\frac{5\pi}{4})}}\\\\
\mathbf{z_1} &\mathbf{=}& \mathbf{-\dfrac{ \sqrt{2} }{2} -\dfrac{ \sqrt{2} }{2} i}
\end{array}
$}}$$

 

 

heureka  Aug 10, 2015
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1+0 Answers

 #1
avatar+18715 
+8
Best Answer

z^2=i  solve z

 

$$\small{\text{$
\begin{array}{rcl}
z^2&=&i \qquad | \qquad z^2= a+bi \quad a=0 $ and $ b = 1\\\\
r&=& \sqrt{a^2+b^2}=\sqrt{0^2+1^2}=\sqrt{1^2}=1\\\\
\varphi &=& \arctan{ \left(\dfrac{b}{a}\right) }=\dfrac{1}{0} \Rightarrow \varphi = \dfrac{\pi}{2}
\qquad
\varphi =
\begin{cases}
\arctan\left(\frac{b}{a}\right) & \text{for}\ a > 0,\ b \geq 0 \qquad \text{(1. Quadrant)}\\
\arctan\left(\frac{b}{a}\right) + 2\pi & \mathrm{f\ddot ur}\ a > 0,\ b < 0 \qquad \text{(4. Quadrant)}\\
\arctan\left(\frac{b}{a}\right) + \pi & \mathrm{f\ddot ur}\ a < 0 \qquad\qquad\quad \text{(2./3. Quadrant)}\\
\frac{\pi}{2} & \mathrm{f\ddot ur}\ a = 0,\ b > 0 \qquad \text{(positive Im-axis)}\\
\frac{3\pi}{2} & \mathrm{f\ddot ur}\ a = 0,\ b < 0 \qquad \text{(negative Im-axis)}\\
\text{indeterminate} & \mathrm{f\ddot ur}\ a = 0,\ b = 0 \qquad \text{(zero)}
\end{cases} \\\\
z^2&=& r\cdot e^{i\cdot \varphi}\\\\
z^2&=& 1\cdot e^{i\cdot \frac{\pi}{2}+2k\pi} \qquad | \qquad \sqrt{}\\\\
z_k &=& e^{i\cdot \frac{\pi}{2\cdot 2 }+\frac{2k\pi}{2} } \\\\
z_k &=& e^{i\cdot \frac{\pi}{4}+ k\pi } \\\\
k =0 \quad z_0 &=& e^{i\cdot \frac{\pi}{4}}\\\\
k =1 \quad z_1 &=& e^{i\cdot \frac{5\pi}{4}} \\\\
&& \boxed{~e^{i\varphi} =\cos{(\varphi)+i\sin{(\varphi)}} ~}\\\\
z_0 &=& e^{ i\frac{\pi}{4} } =\cos{(\frac{\pi}{4})+i\sin{(\frac{\pi}{4})}}\\\\
\mathbf{z_0} &\mathbf{=}& \mathbf{\dfrac{ \sqrt{2} }{2} +\dfrac{ \sqrt{2} }{2} i}\\\\
z_1 &=& e^{ i\frac{5\pi}{4} } =\cos{(\frac{5\pi}{4})+i\sin{(\frac{5\pi}{4})}}\\\\
\mathbf{z_1} &\mathbf{=}& \mathbf{-\dfrac{ \sqrt{2} }{2} -\dfrac{ \sqrt{2} }{2} i}
\end{array}
$}}$$

 

 

heureka  Aug 10, 2015

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