Homework Help: Complex numbers problem - lots of Algebra

1. May 24, 2010

DrummingAtom

1. The problem statement, all variables and given/known data
Show that $$\sqrt{\frac{1} {2} (a + \sqrt {a^2+b^2})} + i \sqrt{\frac{1} {2} (-a + \sqrt {a^2+b^2})}= a+ib$$

2. Relevant equations

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

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

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

2. May 25, 2010

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

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

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.

3. May 25, 2010

DrummingAtom

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
4. May 25, 2010

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.

5. May 25, 2010

Dickfore

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.

6. May 25, 2010

Staff: Mentor

Not according to the reference. The actual statement is really
$$\left(\sqrt{\frac{1} {2} (a + \sqrt {a^2+b^2})} + i \sqrt{\frac{1} {2} (-a + \sqrt {a^2+b^2})}\right)^2= a+ib$$

7. May 25, 2010

HallsofIvy

Which is equivalent to
$$\sqrt{\frac{1} {2} (a + \sqrt {a^2+b^2})} + i \sqrt{\frac{1} {2} (-a + \sqrt {a^2+b^2})}= \sqrt{a+ib}$$

8. May 25, 2010

Dickfore

I guess he meant there are 2 square roots of a complex number.