## Homework Statement

For a real number x, √√(-x) equals :

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

√-1 = i

## 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.

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

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

HallsofIvy
Homework Helper
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.

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vela
Staff Emeritus
Homework Helper

## Homework Statement

For a real number x, √√(-x) equals :

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

√-1 = i

## 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
You mean |x|1/4.

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

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

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

If $(a+bi)^2= i$ then $a^2- b^2+ (2ab)i= 0+ 1(i)$ so you must have $a^2- b^2= 0[/tex] and 2ab= 1. Solve for the values of a and b. If [itex](a+bi)^2= i$
^^how is that TRUE ?

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

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

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

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

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 ?

HallsofIvy
Homework Helper
No, can teach me that ?
It would really help me

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

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.

HallsofIvy
Homework Helper
Well, what answers did you get?

Well, what answers did you get?

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

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

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

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

anyways, thanks