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Thread: Is the square root of z multiplied by the square root of z* equal to mod(z) squared?

  1. #1
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    Is the square root of z multiplied by the square root of z* equal to mod(z) squared?

    I am a bit confused about the rule about "passing" the root:

    \sqrt { z } \sqrt { z } = \sqrt { z*z }


    \sqrt { z } \sqrt { z } = \sqrt { z*z }
    Is this true?
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  2. #2
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    Re: Is the square root of z multiplied by the square root of z* equal to mod(z) squar

    in all cases you can combine arguments to the product of square roots under a single radical

    $\sqrt{a}\sqrt{b}=\sqrt{ab}$

    so

    $\sqrt{z}\sqrt{z^*}=\sqrt{zz^*}=\sqrt{|z|^2}=|z|$
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    Re: Is the square root of z multiplied by the square root of z* equal to mod(z) squar

    Quote Originally Posted by romsek View Post
    in all cases you can combine arguments to the product of square roots under a single radical

    $\sqrt{a}\sqrt{b}=\sqrt{ab}$
    but what if a = b = -1?

    then  $\sqrt{a}\sqrt{b}=i x i = -1$

    $\sqrt{a}\sqrt{b}=i x i = -1$

    but  $\sqrt{ab}=\sqrt{-1 x -1}=\sqrt{1}= 1$

    $\sqrt{ab}=\sqrt{-1 x -1}=\sqrt{1}= 1$


    ?

    So does the rule not work for pure imaginary numbers?
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    Re: Is the square root of z multiplied by the square root of z* equal to mod(z) squar

    Quote Originally Posted by Applestrudle View Post
    but what if a = b = -1?

    then  $\sqrt{a}\sqrt{b}=i x i = -1$

    $\sqrt{a}\sqrt{b}=i x i = -1$

    but  $\sqrt{ab}=\sqrt{-1 x -1}=\sqrt{1}= 1$

    $\sqrt{ab}=\sqrt{-1 x -1}=\sqrt{1}= 1$


    ?

    So does the rule not work for pure imaginary numbers?
    I'm wrong. I'll have to take a further look later but your counter-example is clear enough.

    Technically....

    if $z=-1 = e^{\imath \pi}$

    then yes $z^* = -1$ but it more accurately is $z^*=e^{-\imath \pi}$

    and using this you get the correct answer.
    Last edited by romsek; Dec 2nd 2014 at 01:25 PM.
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    Re: Is the square root of z multiplied by the square root of z* equal to mod(z) squar

    Quote Originally Posted by Applestrudle View Post
    but what if a = b = -1?

    then  $\sqrt{a}\sqrt{b}=i x i = -1$

    $\sqrt{a}\sqrt{b}=i x i = -1$

    but  $\sqrt{ab}=\sqrt{-1 x -1}=\sqrt{1}= 1$

    $\sqrt{ab}=\sqrt{-1 x -1}=\sqrt{1}= 1$


    ?

    So does the rule not work for pure imaginary numbers?
    I think you confused romsek with your notation. "x" is a clumsy way to write the "times" operator. In the future please use (...) or \cdot In LaTeX code use \cdot to indicate "times."

    So to supplement romsek's comments, yes the square root operation does not always work smoothly with negative numbers under the radical. The problem occurs because i^2 = -1 is the definition whereas i = \sqrt{-1} is not.

    -Dan
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    Re: Is the square root of z multiplied by the square root of z* equal to mod(z) squar

    Quote Originally Posted by romsek View Post
    I'm wrong. I'll have to take a further look later but your counter-example is clear enough.
    Don't feel bad about it, you are not alone. Many well-known authors have made the same error .

    The problem is very simple: does $\sqrt z$ exist? Now that question does NOT mean "do the square roots exist?" Absolutely yes For any complex number $z$ there are two square roots of $z=2-2i$, but is $\sqrt z$ a proper representation of one of them? I for one say no. I say that $\sqrt{2-2i}$ has absolutely no meaning in standard material on complex numbers. NOTE that of course I did not say that $z=2-2i$ does not have square roots. Of course it does. The error is 'applying' $\sqrt{~}$ to $z$. What I do say is that the radical sign should not be applied to complex numbers nor negative real numbers. No $\sqrt{i}$ nor $\sqrt{-1}$. Then you ask how do we define $\bf{i}~?$

    We posit a number $\bf{i}$ that is a root of the real equation $x^2+1=0$. From that one enlargement complex numbers are developed.
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