# Homework Help: Question about imaginary numbers

1. Jan 19, 2012

### hms.tech

1. The problem statement, all variables and given/known data

For a real number x, √√(-x) equals :

a) +x b) -x c) complex d) pure imaginary

2. Relevant equations

√-1 = i

3. The attempt at a solution

Here is what i did:
If x is a positive real number then the answer comes out to be x^0.25 * √i (now what is square root of i equal to?)

But if x is a negative real number then the solution would be :
x^0.25

I don't know what to do next, any help would be appreciated.

2. Jan 19, 2012

Have you learnt how to take the square roots of complex numbers?

3. Jan 19, 2012

### genericusrnme

do you know about the old $re^{i x} = r Cos[x]+r i Sin[x]$?
if so, I'd use that

4. Jan 19, 2012

### HallsofIvy

If $(a+bi)^2= i$ then $a^2- b^2+ (2ab)i= 0+ 1(i)$ so you must have $a^2- b^2= 0$ and 2ab= 1. Solve for the values of a and b.

Last edited by a moderator: Jan 20, 2012
5. Jan 19, 2012

### vela

Staff Emeritus
You mean |x|1/4.

Since this is a multiple-choice question, you should be able to deduce what the correct answer is. I hope you can see that you can rule out the first two choices, so the answer has to be complex or purely imaginary. Can you rule out one or the other?

6. Jan 20, 2012

### hms.tech

No, can teach me that ?
It would really help me

If $(a+bi)^2= i$
^^how is that TRUE ?

7. Jan 20, 2012

### hms.tech

I know about that identity, but I can't figure out how would ot apply to this situation

8. Jan 20, 2012

### Mentallic

What is the mod-arg form for i? Now, how can $$\sqrt{re^{i\theta}}$$ be simplified?

9. Jan 20, 2012

### hms.tech

Alright, so for x=1
Or another one, and perhaps i should have opened with this one, is 1

Complex

What do you think, am i right ?

10. Jan 20, 2012

### HallsofIvy

You asked for the square root of i, didn't you? If a+ bi is the square root of i, then $(a+ bi)^2$ is equal to i.

11. Jan 21, 2012

### hms.tech

Alright, understood.

I have a concern however, why do I get two different answers by
Using the two methods
1.which was stated by hallsofivy
2.the one which uses Taylor series representation to represent
a complex number in e^(ix) form.

12. Jan 21, 2012

### HallsofIvy

Well, what answers did you get?

13. Jan 23, 2012

### hms.tech

Alright, this one is by the taylor series representation method:

i = e^(i*π/2)
so, √i = e^(i*π/4)

now representing the complex number in the form x+yi :

cos(π/4)+isin(π/4) = √i

so, the answer is 1/√2 + i1/√2

and the answer by the other method :

let i = (a + bi)^2

is exactly the same as before.
My bad on post no. 11

anyways, thanks

14. Jan 23, 2012

### SammyS

Staff Emeritus
Of course, if x ≤ 0, then $\sqrt{\sqrt{-x}\ }$ is a real number.

15. Jan 23, 2012

### HallsofIvy

In the complex numbers, every number has two square roots.
$$\sqrt{i}= \frac{1}{\sqrt{2}}\pm i\frac{1}{\sqrt{2}}$$