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# solve z

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z^2=i

solve z

Guest Aug 10, 2015

#1
+18829
+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
+18829
+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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