# Solutions to this equation - complex numbers

• Jan 7th 2012, 01:33 AM
cosmonavt
Solutions to this equation - complex numbers
1. $\displaystyle x^3-y^3=27$
The question says, what is the number of solutions to this equation in the field of complex numbers?

The choices are

a) 1
b) 2
c) 3
d) infinitely many

I have no clue to this one.

2. And here's another one:

For any real number x, $\displaystyle \sqrt{\sqrt{-x}}$ is equal to

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

Writing it in the form $\displaystyle ({-x}^{1/2})^{1/2}$, we can simplify it to $\displaystyle \sqrt[4]{-x}$

Which shows that the number is not real (since no number to the power four would give -x). So +x and -x are eliminated. But what about the last two choices? What is the difference between complex and pure imaginary?
• Jan 7th 2012, 02:23 AM
Siron
Re: Solutions to this equation - complex numbers
2) You can write: $\displaystyle (-x)^{\frac{1}{4}}=(-1)^{\frac{1}{4}}\cdot x^{4}$
What about $\displaystyle (-1)^{4}$?

You know every complex number has a real part and an imaginary part, if the real part is equal to zero (so if there's no real part, only an imaginary part) then we call it pure imaginary.
• Jan 12th 2012, 09:32 AM
cosmonavt
Re: Solutions to this equation - complex numbers
Quote:

Originally Posted by Siron
2) You can write: $\displaystyle (-x)^{\frac{1}{4}}=(-1)^{\frac{1}{4}}\cdot x^{4}$
What about $\displaystyle (-1)^{4}$?

You know every complex number has a real part and an imaginary part, if the real part is equal to zero (so if there's no real part, only an imaginary part) then we call it pure imaginary.

Thanks for the reply!!

Complex numbers are of the form a + b i (plus sign in between the complex and imaginary part). The number $\displaystyle (-x)^{\frac{1}{4}}=(-1)^{\frac{1}{4}}\cdot x^{4}$ has a multiply sign in between so even if $\displaystyle x^{4}$ is real, this does not qualify as a complex. Does this mean that this would be pure imaginary?

Also, any clue about the first question?
• Jan 12th 2012, 03:56 PM
HallsofIvy
Re: Solutions to this equation - complex numbers
Every imaginary number is a complex number: 0+ bi.
• Jan 12th 2012, 05:42 PM
Charney
Re: Solutions to this equation - complex numbers
2. As HallsofIvy has already pointed out, the imaginary numbers are in fact a subset of the complex numbers.

We can easily rewrite this expression as the following:
$\displaystyle \sqrt{\sqrt{-x}} = \sqrt{i\sqrt{x}} = \sqrt[4]{x}\sqrt{i}$

Sadly I was unable to come up with a solution of my own after two attempts, however a nice solution is presented here.