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Thread: Find the domain of the complex square root funciton

  1. #1
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    Find the domain of the complex square root funciton

    Hello,

    The problem is: Define $\text{Sqrt}(z)=|z|^{1/2}e^{i \theta /2}$ where $z \in \mathbb{C} \backslash (-\infty,0]$ and $-\pi < \theta <\pi$.

    a) Find two numbers $z_1,z_2 \in \mathbb{C} \backslash (-\infty,0]$ such that $w=z_1z_2 \in \mathbb{C} \backslash (-\infty,0]$ but $\text{Sqrt}{(w)}\neq \text{Sqrt}{(z_1)}\text{Sqrt}{(z_2)}$.

    b) Find the subset $S \subseteq \mathbb{C} \backslash (-\infty,0]$ such that if $z \in S$ then $\text{Sqrt}{(iz)}=\text{Sqrt}{(i)}\text{Sqrt}{(z) }$.

    For part a) I think $2i$ and $-2i$ work (but I've screwed this up before). We'd have $\text{Sqrt}{(2i)}=\sqrt{2}e^{i\pi/4}$, $\text{Sqrt}{(-2i)}=\sqrt{2}e^{i \pi 3/4}$, and $\text{Sqrt}{(2i(-2i))}=\text{Sqrt}{(4)}=\sqrt{4}e^{i 0}$. Then $\sqrt{2}e^{i\pi/4}\sqrt{2}e^{i \pi 3/4} \neq \sqrt{4}e^{i 0}$.
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    Re: Find the domain of the complex square root funciton

    Quote Originally Posted by bkbowser View Post
    Hello,

    The problem is: Define $\text{Sqrt}(z)=|z|^{1/2}e^{i \theta /2}$ where $z \in \mathbb{C} \backslash (-\infty,0]$ and $-\pi < \theta <\pi$.

    a) Find two numbers $z_1,z_2 \in \mathbb{C} \backslash (-\infty,0]$ such that $w=z_1z_2 \in \mathbb{C} \backslash (-\infty,0]$ but $\text{Sqrt}{(w)}\neq \text{Sqrt}{(z_1)}\text{Sqrt}{(z_2)}$.

    b) Find the subset $S \subseteq \mathbb{C} \backslash (-\infty,0]$ such that if $z \in S$ then $\text{Sqrt}{(iz)}=\text{Sqrt}{(i)}\text{Sqrt}{(z) }$.

    For part a) I think $2i$ and $-2i$ work (but I've screwed this up before). We'd have $\text{Sqrt}{(2i)}=\sqrt{2}e^{i\pi/4}$, $\text{Sqrt}{(-2i)}=\sqrt{2}e^{i \pi 3/4}$, and $\text{Sqrt}{(2i(-2i))}=\text{Sqrt}{(4)}=\sqrt{4}e^{i 0}$. Then $\sqrt{2}e^{i\pi/4}\sqrt{2}e^{i \pi 3/4} \neq \sqrt{4}e^{i 0}$.
    According to the rule, $\text{Sqrt}{(-2i)}=\sqrt{2}\exp\left( {\frac{{ - i\pi }}{4}} \right)$ Is that correct?
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    Re: Find the domain of the complex square root funciton

    Quote Originally Posted by Plato View Post
    According to the rule, $\text{Sqrt}{(-2i)}=\sqrt{2}\exp\left( {\frac{{ - i\pi }}{4}} \right)$ Is that correct?
    I don't see why not.

    $-2i$ has polar form $|-2i|e^{i 3/2 \pi}=|-2i|e^{-i 1/2 \pi}$. Is that right?

    So that what? For some $z$ you get two outputs at least one of which falls on $(-\infty,0]$?

    I guess I can try that on part b.
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    Re: Find the domain of the complex square root funciton

    Quote Originally Posted by bkbowser View Post
    I don't see why not.
    $-2i$ has polar form $|-2i|e^{i 3/2 \pi}=|-2i|e^{-i 1/2 \pi}$. Is that right?
    So that what? For some $z$ you get two outputs at least one of which falls on $(-\infty,0]$?.
    $z_=-2i~\&~z_2=2i$ do not forget that $-\pi<\theta\le\pi$ is required.

    This $\text{Sqrt}(-2i)=2\exp\left( {\frac{{ - i\pi }}{4}} \right)~\&~\text{Sqrt}(2i)=2\exp\left( {\frac{{ i\pi }}{4}} \right)$

    So $\text{Sqrt}(-2i)\cdot\text{Sqrt}(2i)=4$
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    Re: Find the domain of the complex square root funciton

    We can see that magnitude of numbers we choose is going to have no effect on this problem.

    So suppose we have

    $z_1 = e^{i \theta_1},~z_2= e^{i \theta_2} \ni \theta_1 + \theta_2 > \pi$

    $w = z_1 z_2 = e^{i(\theta_1 + \theta_2)}$

    but as $-\pi < \theta < \pi$ this would have to be normalized as

    $w = e^{i(\theta_1 + \theta_2-2\pi)}$

    $\sqrt{w} = e^{i\left(\frac{\theta_1+\theta_2}{2}-\pi\right)}$

    $\sqrt{z_1}\sqrt{z_2} = e^{i\frac{\theta_1}{2}}e^{i\frac{\theta_2}{2}}= e^{i\left(\frac{\theta_1+\theta_2}{2}\right)}$

    The same phenomenon will occur when $\theta_1+\theta_2 < -\pi$
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    Re: Find the domain of the complex square root funciton

    Ah OK I think I get it.

    For part b then I would only need $z$s in $\mathbb{C}$ such that $-\pi< \pi/2+\text{Arg}z < \pi$. So complex numbers in Quadrant 4 are out of the set and I think that's all.
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    Re: Find the domain of the complex square root funciton

    Quote Originally Posted by romsek View Post
    So suppose we have
    $z_1 = e^{i \theta_1},~z_2= e^{i \theta_2} \ni \theta_1 + \theta_2 > \pi$
    $w = z_1 z_2 = e^{i(\theta_1 + \theta_2)}$
    but as $-\pi < \theta < \pi$ this would have to be normalized as
    $w = e^{i(\theta_1 + \theta_2-2\pi)}$
    $\sqrt{w} = e^{i\left(\frac{\theta_1+\theta_2}{2}-\pi\right)}$
    $\sqrt{z_1}\sqrt{z_2} = e^{i\frac{\theta_1}{2}}e^{i\frac{\theta_2}{2}}= e^{i\left(\frac{\theta_1+\theta_2}{2}\right)}$ The same phenomenon will occur when $\theta_1+\theta_2 < -\pi$
    I must confess that I do not understand the above. I have always assumed that a complex number has a standard form.
    That is $z=|z|\exp(\theta {\bf{i}})$ where $\pi<\theta\le\pi$. Now the number $z$ has two square roots one being $\sqrt{|z|}\exp \left( {\frac{{\theta i}}{2}} \right)$

    Given two complex numbers with arguments $\theta_1~\&\theta_2$ their principal arguments being in $(-\pi,\pi]$.
    Thus the arguments of the principal square roots are $\theta_1/2~\&~\theta_2/2$ both of which are in $\left( {\frac{{ - \pi }}{2},\frac{\pi }{2}} \right]$.
    So $\text{Sqrt}(z_1)\cdot\text{Sqrt}(z_2)$ will have a principal argument in $(-\pi,\pi]$.
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