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Index problem [Solved!]

My question

What is the difference between (-1)^2/2 and ((-1)^2)^1/2?

Relevant page

2. Basic Operations in Complex Numbers

What I've done so far

I tried expanding each out using complex number laws, but I'm not sure what I'm doing.

X

What is the difference between (-1)^2/2 and ((-1)^2)^1/2?
Relevant page

<a href="/complex-numbers/2-basic-operations.php">2. Basic Operations in Complex Numbers</a>

What I've done so far

I tried expanding each out using complex number laws, but I'm not sure what I'm doing.

Re: Index problem

I'm not very clear what (-1)^2/2 means. (Please use the math input system - it makes it much easier for use to read your math.)

If we take the final 2 to mean "half of", then it probably means `((-1)^2)/2 = 1/2`

The second expression, `((-1)^2)^1/2` if written `((-1)^2)^(1/2)` means "the square root of `(-1)` squared".

In this case, `(-1)^2 = 1` and the square root of `1` is `1`.

Hope that helps.

X

I'm not very clear what (-1)^2/2 means. (Please use the math input system - it makes it much easier for use to read your math.)

If we take the final 2 to mean "half of", then it probably means `((-1)^2)/2 = 1/2`

The second expression, `((-1)^2)^1/2` if written `((-1)^2)^(1/2)` means "the square root of `(-1)` squared".

In this case, `(-1)^2 = 1` and the square root of `1` is `1`.

Hope that helps.

Re: Index problem

I found your solution very helpful,well part of it.

Actually what i meant was,is {(-1)^(1/2)}^2 a real number?

X

I found your solution very helpful,well part of it.

Actually what i meant was,is {(-1)^(1/2)}^2 a real number?

Re: Index problem

Hi again

Yes, it is a real number.

You have the equivalent of the square of the imaginary number, `j`. The answer is real, and it is `-`1.

See half-way down this page:

1. Basic Definitions of Complex Numbers

X

Hi again

Yes, it is a real number.

You have the equivalent of the square of the imaginary number, `j`. The answer is real, and it is `-`1.

See half-way down this page:

<a href="/complex-numbers/1-basic-definitions.php">1. Basic Definitions of Complex Numbers</a>

Re: Index problem

Hi,it's me again.

What I think is that `{(-1)^(1/2)}^2=sqrt(-1)^2`,which is `j^2=-1`.

So I want to ask this,does it mean complex numbers are also REAL?

X

Hi,it's me again.

What I think is that `{(-1)^(1/2)}^2=sqrt(-1)^2`,which is `j^2=-1`.

So I want to ask this,does it mean complex numbers are also REAL?

Re: Index problem

Hi again

Complex numbers consist of a real part and an imaginary part.

You can have the following and call them all complex.

`3 + 5j`
`3 + 0j`
`0 + 5j`

But really, the first is complex, the second is real, the 3rd is imaginary.

Good luck with it.

X

Hi again

Complex numbers consist of a real part and an imaginary part. 

You can have the following and call them all complex.

`3 + 5j`
`3 + 0j`
`0 + 5j`

But really, the first is complex, the second is real, the 3rd is imaginary.

Good luck with it.

Re: Index problem

re: melon sandbox It's interesting to see your approach to finding the square root of a complex number using polar coordinates. Your calculation for (2+3i)^2 = -5+12i seems correct, and your conversion into polar form with theta = 0.5880 is on point. Have you considered exploring the concept of principal square roots in complex numbers further? It might shed more light on how to determine the correct square root when dealing with complex numbers. Keep up the good work in unraveling the mysteries of complex arithmetic!

X

re: <a href="https://melon-sandbox.com/">melon sandbox</a> It's interesting to see your approach to finding the square root of a complex number using polar coordinates. Your calculation for (2+3i)^2 = -5+12i seems correct, and your conversion into polar form with theta = 0.5880 is on point. Have you considered exploring the concept of principal square roots in complex numbers further? It might shed more light on how to determine the correct square root when dealing with complex numbers. Keep up the good work in unraveling the mysteries of complex arithmetic!

Re: Index problem

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X

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