+0  
 
0
589
1
avatar

z^2=i 

solve z

Guest Aug 10, 2015

Best Answer 

 #1
avatar+20025 
+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
 #1
avatar+20025 
+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

18 Online Users

avatar
avatar

New Privacy Policy

We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive information about your use of our website.
For more information: our cookie policy and privacy policy.