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I have to find all the solutions of this complex equation.

I am trying to do: $z = r^{1/4} e^{i(\theta +2\pi k)/4}$ but I don't know how to find the angle because is 2/0 so any hints are welcome .

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Write $2i = 2exp(i\pi/2)$ and proceed from there. –  James S. Cook Mar 17 '13 at 23:39
1  
What do you mean, "because is 2/0"? –  Gerry Myerson Mar 17 '13 at 23:41
    
i mean whats the angle if theta when arctan(2/0) ? (2 is the coefficient of 2i ) –  nicolas Mar 17 '13 at 23:42
    
Try $\frac{\pi}{2}$ –  amWhy Mar 17 '13 at 23:44
    
$z^4=2i=2(0+1i)=2($cos$(\Pi/2) +i$sin$(\Pi/2))$ –  jim Mar 17 '13 at 23:44

4 Answers 4

up vote 1 down vote accepted

for $z^4 = 2i$ rewrite as $z^4 = 2 \cdot e^{(\pi\cdot \tfrac{i}{2})}$ then extract 4'th root on appropriate branch and get $2^{\tfrac{1}{4}}\cdot e^{\tfrac{\pi}{8}}$,$2^{\tfrac{1}{4}}\cdot e^{\tfrac{5\pi}{8}}$, $2^{\tfrac{1}{4}}\cdot e^{\tfrac{9\pi}{8}}$ and $2^{(1/4)}\cdot e^{\tfrac{13\pi}{8}}$

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Try to format answers using LaTeX. –  nbubis Mar 18 '13 at 0:54

$$z^4 = 2i = 2 e^{i\pi/2+2\pi i n}$$

where $n \in \mathbb{Z}$

Then with principal branch, as there are 4 unique solutions:

$$z_1^4 =2 e^{i\pi/2} \implies z_1 = \sqrt[4]{2} e^{i\pi/8} = I$$ $$z_2^4 =2 e^{i\pi/2+2\pi i} \implies z_1 = \sqrt[4]{2} e^{i\pi/8+\pi i /2} = iI$$ $$z_3^4 =2 e^{i\pi/2+4\pi i} \implies z_1 = \sqrt[4]{2} e^{i\pi/8+\pi i} = -I$$ $$z_4^4 =2 e^{i\pi/2+6\pi i} \implies z_1 = \sqrt[4]{2} e^{i\pi/8+3i \pi/2} = -iI$$

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When solving equations of the form $z^n=z_0$ for some $z_0\in \Bbb C$, first you write $z_0$ in polar form $z_0=re^{i\theta+2\pi k i}$ then we have$$ z=r^\frac{1}{n}e^\frac{i\theta+2\pi k i}{n},$$$k=0,1,2,...,n-1$. if $z_0=x+iy$. we know that $r=\sqrt{x^2+y^2}$ and $\theta=arctan(\frac{y}{x})$, in your case $r=2$ and $\theta=\frac{\pi}{2}$, the angle is $\frac{\pi}{2}$ because we want to find $arctan(\frac{2}{0})$ in some scenes, we know that$$\lim_{x\to\infty}arctan(x)=\frac{\pi}{2}$$so you can look at $$arctan\left(\frac{2}{0}\right)=\lim_{x\to 0^+}arctan\left(\frac{2}{x}\right)=\lim_{y\to\infty}arctan\left(2y\right)=\frac{\pi}{2}.$$once you found the angle you can find the solutions.

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$z^4= 2 e^{\frac{\pi}{2}}$

$e^{\theta}=1$

$\theta= 2n\pi$.

Therefore, general solution has to be:

$z= 2^{\frac{1}{4}}. e^{\frac{2n\pi+\frac{\pi}{2}}{4}}$

Where $k$ $\epsilon(1,2,3,4)$

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correct but its for k ϵ(0,1,2,3). thanks –  nicolas Mar 18 '13 at 16:21
    
Yeah! I have mentioned that. –  Inceptio Mar 19 '13 at 7:51

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