Question about imaginary numbers

[hms.tech]

Homework Statement

For a real number x, √√(-x) equals :

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

Homework Equations

√-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.

 

[Bread18]

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

 

[genericusrnme]

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

 

[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.

 

[vela]

Homework Statement

For a real number x, √√(-x) equals :

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

Homework Equations

√-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?

 

[hms.tech]

Have you learned 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 [itex]a^2- b^2= 0[/tex] and 2ab= 1. Solve for the values of a and b.

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

 

[hms.tech]

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]

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?

 

[hms.tech]

Alright, so for x=1
The answer is e^[2∏i]
Or another one, and perhaps i should have opened with this one, is 1


So the answer is probably
Complex



What do you think, am i right ?

 

[HallsofIvy]

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.

 

[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.

 

[HallsofIvy]

Well, what answers did you get?

 

[hms.tech]

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]

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

 

[HallsofIvy]

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