Real part of a complex number

1. Sep 25, 2006

sjmacewan

Hello there,
I've been given the task of find the real part for the following expression

$$\sqrt{x+iy}$$

And I'm a bit stuck. I figure that I'll just say that that equation is equal to some other imaginary number a+bi where 'a' is the real part and 'b' is the imaginary part, and try to solve for a. But after squaring both sides i get stuck immediately...

$$x+iy = a^2 + 2abi - b^2$$

And i don't know where to go. Perhaps I'm going the wrong way with this one, any help would be appreciated.

Edit: Ok, i've made some progress...

I know then that

$$x = a^2 - b^2$$
and
$$iy = 2abi$$

So i try to get rid of the b term in the real one, but the only substitution I can make results in a y term being introduced into the real part, which is just adding another imaginary number in there...

Last edited: Sep 25, 2006
2. Sep 25, 2006

Hurkyl

Staff Emeritus
Aren't you trying to express a and b in terms of x and y?

No it's not -- y is not imaginary.

3. Sep 25, 2006

sjmacewan

ooh, after seeing iy i just labeled both as being imaginary, but i suppose that's not true...i'll keep working then!

edit: i still seem to get stuck pretty quickly...

making the substitution $$b=\frac{y}{2a}$$ and plugging that into $$x=a^2-b^2$$ gets me to:
$$x=a^2- \frac{y^2}{4a^2}$$
And again, i feel stuck. I've tried putting them over a common denominator, but that doesn't seem to lead anyplace useful...

Last edited: Sep 25, 2006
4. Sep 25, 2006

HallsofIvy

Staff Emeritus
Getting a common denominator is one way but the better way to handle equations with fractions is to multiply the entire equation by the "common" denominator- here just 4a2. If you do that you get the quartic equation $4a^2= 4a^4- y^2$ or $4a^4- 4a^2- y^2= 0$. That's actually a quadratic equation in a2.

Last edited: Sep 26, 2006
5. Sep 25, 2006

sjmacewan

OK...tell me if this looks ok...it's really ugly and there's step or two which i'm not confident about:

$$0=a^2-\frac{y^2}{4a^2}-x$$
$$0=\frac{4a^4 - 4xa^2 - y^2}{4a^2}$$
$$0=4a^4 - 4xa^2 - y^2$$

$$a^2 = \frac{4x \pm \sqrt{16x^2 - (4*4*-y^2)}}{2*4}$$

$$a^2 = \frac{4x \pm \sqrt{16x^2 + 16y^2}}{8}$$

$$a^2 = \frac{4x \pm \sqrt{16(x^2+y^2)}}{8}$$

$$a^2 = \frac{4x \pm 4\sqrt{x^2 + y^2}}{8}$$

$$a^2 = \frac{x \pm \sqrt{x^2 + y^2}}{2}$$

$$a=\sqrt \frac{x \pm \sqrt{x^2 + y^2}}{2}}$$

Can i leave it like that? (you may need to refresh the screen, i've fixed the tex)

Last edited: Sep 25, 2006
6. Sep 26, 2006

HallsofIvy

Staff Emeritus
Yes, that looks good- except of course that you should have a $\pm$ on the outside and you don't really need the one inside the square root, only +. Taking the negative would make a imaginary and it must be real.

You can check by looking at some simple examples. Suppose x is positive, y= 0. What does that formula give? Suppose x is negative, y=0. Suppose x= 0, y= 1. (The square root of i is $\frac{\sqrt{2}}{2}(1+ i)$ and $-\frac{\sqrt{2}}{2}(1+ i)$.)

Last edited: Sep 26, 2006
7. Sep 26, 2006

sjmacewan

alright, thanks for the pointers there, i can't say i've run accross many solutions with a root inside a root, seemed a bit odd to me...thanks again!