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Complex numbers problem - lots of Algebra

  1. May 24, 2010 #1
    1. The problem statement, all variables and given/known data
    Show that [tex]\sqrt{\frac{1} {2} (a + \sqrt {a^2+b^2})} + i \sqrt{\frac{1} {2} (-a + \sqrt {a^2+b^2})}= a+ib[/tex]

    2. Relevant equations

    3. The attempt at a solution
    Distributed the i and then the 1/2's in each term which gave:
    [tex] \sqrt{\frac{a} {2} + \frac{ \sqrt {a^2+b^2}}{2}}} -({-\frac{a} {2} + \frac {\sqrt {a^2+b^2}}{2}}) = a+ib [/tex].

    Next I squared both sides to eliminate the roots, which gives:
    [tex]\frac{a}{2}+\frac{\sqrt {a^2+b^2}}{2} - \frac {a^2}{4} - \frac {a^2+b^2}{4}+\frac {2a\sqrt{a^2+b^2}}{4}= a^2+2abi-b^2[/tex].

    From that point it doesn't seem to work out. Was I going in the right direction?

    Thanks for your help.
  2. jcsd
  3. May 25, 2010 #2


    Staff: Mentor

    You lost the i on the left side.

    As stated, you are supposed to show this is an identity, but it isn't. As a simple counterexample, let a = 3 and b = 4. Then

    [tex]\sqrt{\frac{1} {2} (a + \sqrt {a^2+b^2})} + i \sqrt{\frac{1} {2} (-a + \sqrt {a^2+b^2})}= \sqrt{\frac{1} {2} (3 + \sqrt {3^2+4^2})} + i \sqrt{\frac{1} {2} (-3 + \sqrt {3^2+4^2})}[/tex]
    [tex]=\sqrt{\frac{1} {2} (3 + 5)} + i \sqrt{\frac{1} {2} (-3 + 5)} = 2 + i \neq 3 + 4i[/tex]

    So clearly the equation is not identically true for all values of a and b. Are you instead supposed to find values of a and b for which it is conditionally true? For two complex numbers to be equal, their real parts must be equal and their imaginary parts must be equal.
  4. May 25, 2010 #3
    I got this out of Penrose's Road to Reality book, and actually I think I misread it. It's equation 4.3 @ http://mclerc.cl/penrosebook/book/RP/09_4_Magical%20complex%20numbers.pdf [Broken] (Page 4 , half way down).

    So, is he saying that that long equation (4.3) is equal a+bi? Or just the bulk of that equation can be viewed in a general form of a+bi?
    Last edited by a moderator: May 4, 2017
  5. May 25, 2010 #4


    Staff: Mentor

    No, what Penrose is saying is that the square of the long expression (it's not an equation) is a + ib. To verify that this is so, square the left side and it should simplify to a + ib.
  6. May 25, 2010 #5
    This is known as the Lagrange's identity. Square the lhs and simplify and you should get the rhs.


    I think you should have a square root above the whole expression on the rhs.
  7. May 25, 2010 #6


    Staff: Mentor

    Not according to the reference. The actual statement is really
    [tex]\left(\sqrt{\frac{1} {2} (a + \sqrt {a^2+b^2})} + i \sqrt{\frac{1} {2} (-a + \sqrt {a^2+b^2})}\right)^2= a+ib[/tex]
  8. May 25, 2010 #7


    User Avatar
    Science Advisor

    Which is equivalent to
    [tex]\sqrt{\frac{1} {2} (a + \sqrt {a^2+b^2})} + i \sqrt{\frac{1} {2} (-a + \sqrt {a^2+b^2})}= \sqrt{a+ib}[/tex]
  9. May 25, 2010 #8
    I guess he meant there are 2 square roots of a complex number.
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