Complex numbers problem - lots of Algebra

[DrummingAtom]

Homework Statement

Show that $$\sqrt{\frac{1} {2} (a + \sqrt {a^2+b^2})} + i \sqrt{\frac{1} {2} (-a + \sqrt {a^2+b^2})}= a+ib$$

Homework Equations

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?

Thanks for your help.

 

[Mark44]

Homework Statement

Show that $$\sqrt{\frac{1} {2} (a + \sqrt {a^2+b^2})} + i \sqrt{\frac{1} {2} (-a + \sqrt {a^2+b^2})}= a+ib$$

Homework Equations

The Attempt at a Solution

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

You lost the i on the left side.

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?

Thanks for your help.

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.

 

[DrummingAtom]

. Are you instead supposed to find values of a and b for which it is conditionally true?

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 (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?

 

[Mark44]

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.

 

[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.

 

[Mark44]

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.

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

 

[HallsofIvy]

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.

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

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

 

[Dickfore]

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

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