Thread: Complex numbers

Hi!
I'm soon having my final examination in Algebra, so I'm sitting here reahearsing old tests. In one of them I encountered the following problem:

"Let . Show that z satisfies the equation and use this to determine what z is."

The right answer is 1+3i

So, my first question is: Can you actually define a problem like that?
is negative, so is not defined.

My attempt at a solution: I simply took the cube of z, and found it to be equal to
. It would be nice if we could just factorize this by letting (-1)^(1/3) be equal to -1, thus obtaining the mentioned equation. The only problem is that (-1)^(1/3) is not defined, and could be -1 as well as . How do I proceed from here?

My third question is: I can solve the mentioned third-degree equation, but it has three solutions. I managed to "pick" the right one (which is 1+3i) by "seeing" that z has a positive imaginary part, and a positive real part. However, I'm not very confident with my method, so I wonder if there's any better way.

I'd be grateful for your help!

Originally Posted by thyristor

Hi!
I'm soon having my final examination in Algebra, so I'm sitting here reahearsing old tests. In one of them I encountered the following problem:

"Let . Show that z satisfies the equation and use this to determine what z is."

The right answer is 1+3i

So, my first question is: Can you actually define a problem like that?
is negative, so is not defined.

There is no problem with defining the cube root of a negative number. In fact there is no problem with the cube root of any real number.



You run into problems with the even-numbered root of a negative real number (if you are looking for real roots) and with the fact that even-numbered roots of positive real numbers are not unique (there are positive and negative roots). But odd roots exist and are unique (in the real numbers).

Originally Posted by thyristor

My attempt at a solution: I simply took the cube of z, and found it to be equal to
. It would be nice if we could just factorize this by letting (-1)^(1/3) be equal to -1, thus obtaining the mentioned equation. The only problem is that (-1)^(1/3) is not defined, and could be -1 as well as . How do I proceed from here?

My third question is: I can solve the mentioned third-degree equation, but it has three solutions. I managed to "pick" the right one (which is 1+3i) by "seeing" that z has a positive imaginary part, and a positive real part. However, I'm not very confident with my method, so I wonder if there's any better way.

I'd be grateful for your help!

which you can see by noting that -2 is a root and dividing by z+2.

So, if you show that z satisfies the given equation, you can read off the possible values for z and then select the correct one, from the original expression for z, say by noting that z clearly has positive imaginary part.

Originally Posted by DrRocket

Originally Posted by thyristor

Hi!
I'm soon having my final examination in Algebra, so I'm sitting here reahearsing old tests. In one of them I encountered the following problem:

"Let . Show that z satisfies the equation and use this to determine what z is."

The right answer is 1+3i

So, my first question is: Can you actually define a problem like that?
is negative, so is not defined.

There is no problem with defining the cube root of a negative number. In fact there is no problem with the cube root of any real number.

Thanks for your help!
However, I still don't understand how to define (-1)^(1/3).
Let's say that
Now let . Then . Thus z=w satisfies the equation -1=z^3, but z=-1 also satisfies the equation. This is just like when we solve the equation
1^(1/2)=a =>1=a^2, where a could be both 1 and -1. However, a is 1 because we want the sqrt to be unique.

So my question is now, if we want a unique solution to the equation (-1)^(1/3)=z, how can we say that is a false root?[/tex]

Originally Posted by thyristor

Thanks for your help!
However, I still don't understand how to define (-1)^(1/3).
Let's say that
Now let . Then . Thus z=w satisfies the equation -1=z^3, but z=-1 also satisfies the equation. This is just like when we solve the equation
1^(1/2)=a =>1=a^2, where a could be both 1 and -1. However, a is 1 because we want the sqrt to be unique.

So my question is now, if we want a unique solution to the equation (-1)^(1/3)=z, how can we say that is a false root?[/tex]

In the complex numbers is not well-defined since, given any complex number there are three complex numbers such that . But given any real number there is a unique real number such that . So is clearly and uniquely defined as a function on the real numbers. It is that function that you need in order to solve your problem. So, for instance where the cube root function is being viewed as a real-valued function of a real variable.

It is not that is a false root. It is just not the real cube root of -1.

There are two questions here. 1) Can you define the cube root function unambiguously as a real-valued function of a real variable ? -- Ans Yes

2) Can you define the cube root function unambiguously as a complex-valued function of a complex variable -- Ans No (you will later learn a way to define it analytically after your "choose" a branch of the logarithm, but that is another separate topic).

Allright, but what says that if z is given by , then z is real?

Originally Posted by thyristor

Allright, but what says that if z is given by , then z is real?

It all depends on the application.

This stuff is dependent on what definitions one adopts, and is subject only to the requirements of logic. Beyond that there are no rules. So it is perfectly acceptable to DEFINE to be that real number z such that for all real numbers .

Okay, then I think that I understand. Thank you for your help.