1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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

    Mark44

    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

    Mark44

    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.

    EDIT:

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

    Mark44

    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

    HallsofIvy

    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.
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook