Complex numbers linear algebra

[karnten07]

[SOLVED] Complex numbers linear algebra

Homework Statement

Let w, z be complex numbers. Solve the linear equation wx=z; in other words find all x (of the set of complex numbers) such that wx=z. (hint: You need to distinguish 3 cases)

Homework Equations

The Attempt at a Solution

If w and z are complex numbers equal to i then if x was i the the power 0 it would equal one and still satisfy the equation. A second case is if x = -i^2 as this also equals one. I can't think of a third case, but to be honest i really don't think i have understood this question.

Can anyone clarify what i have to do here please??/

 

[NateTG]

I suspect there are really 4 cases to consider:
z \neq 0, w \neq 0
z = w = 0
z =0, w \neq 0
z \neq 0, w=0

 

[HallsofIvy]

For this particular problem, there is really nothing special about complex numbers. The answer is essentially the same as if they were asking about real numbers.

 

[karnten07]

For this particular problem, there is really nothing special about complex numbers. The answer is essentially the same as if they were asking about real numbers.

So i have to identify 3 values of x that make wx=z. I don't see how i do this though because all i know about w and z is that they are complex. Any hints please?

 

[Dick]

You must be tempted to write that the solution set to wx=z is just {z/w}, right? That's great if w and z are non-zero. Now just start thinking about how you have to qualify that statement if for example w is zero and z is nonzero. Or if both w and z are zero.

 

[karnten07]

You must be tempted to write that the solution set to wx=z is just {z/w}, right? That's great if w and z are non-zero. Now just start thinking about how you have to qualify that statement if for example w is zero and z is nonzero. Or if both w and z are zero.

if w and z are zero then x must be zero, the question states that x is in the set of complex numbers though so does x =0 count as a solution? If either w or z are zero, then that would also make x zero in that case too? I am confused, possibly because i am new to linear algebra and haven't seen a question like this before. Am i missing something i need to know to answer the question? thanks

 

[Dick]

If w AND z are zero, then 0*x=0 is true for ANY x, right? This isn't really a 'linear algebra' problem. As Halls pointed out the complex numbers in a problem like this are just like the ordinary real numbers. What about trying to solve 0*x=z for z not equal to zero?

 

[karnten07]

If w AND z are zero, then 0*x=0 is true for ANY x, right? This isn't really a 'linear algebra' problem. As Halls pointed out the complex numbers in a problem like this are just like the ordinary real numbers. What about trying to solve 0*x=z for z not equal to zero?

Ok so a solution would be x is a set of the complex numbers (indicating it has any value) when w and z are zero. If z is \neq 0, then i would think 0*x is inconsistent?

 

[karnten07]

Ok so a solution would be x is a set of the complex numbers (indicating it has any value) when w and z are zero. If z is \neq 0, then i would think 0*x is inconsistent?

What am i missing here anyone, please. Thanks

 

[Dick]

Yes, it's inconsistent. So there are no solutions. So w=z=0 has all complex numbers as a solution, w and z not equal to zero has a single solution and w=0, z not equal to zero has no solutions. Any other cases you need to worry about?

 

[karnten07]

Yes, it's inconsistent. So there are no solutions. So w=z=0 has all complex numbers as a solution, w and z not equal to zero has a single solution and w=0, z not equal to zero has no solutions. Any other cases you need to worry about?

No, that's a great help Dick for clarifying that. The next part of the question is to solve the following system of linear equations:

(5i)x1 + 3x2 = 12+i (1)

x1 + (2i)x2 = 3-i (2)

I treated it just like simultaneous equations, by multiplying (2) by 5i and cancelling i^2 terms to make (3). Then Doing (1)-(3) to get

-7x2+14i-7=0 to find x2=2i-1. Then substituting to get x1=(3/i) -1

 

[Dick]

That's the right idea. But what did you get for (3)? I get 13x2+14i-7=0. Check the work for simple mistakes.

 

[karnten07]

Yes, it's inconsistent. So there are no solutions. So w=z=0 has all complex numbers as a solution, w and z not equal to zero has a single solution and w=0, z not equal to zero has no solutions. Any other cases you need to worry about?

Also for w and z not zero i put x is the set of complex numbers excluding zero as a solution. When writing either w or z = 0, but not both, can i write this:

when exclusively w\cupz=0 then there are no solutions.

 

[karnten07]

That's the right idea. But what did you get for (3)? I get 13x2+14i-7=0. Check the work for simple mistakes.

Yes, you're right, thanks again. I then get x2= 7/13 - 14i/13 and x1 = 11/13 - 27i/13

 

[Dick]

Why would you exclude x=0 from the solutions in the case w,z not zero? And the cases w=0 or z=0 with the other being nonzero are quite different. I think you should think this over with a clear head. Because I'm not sure what your "when exclusively" phrase even means.

 

[Dick]

The only possibilities are everything is a solution, nothing is a solution and there is one unique solution. Figure how to split these possibilities up depending on whether w and z are or are not zero. If you are solving linear equations in two variables you can probably handle that.

 

[karnten07]

Why would you exclude x=0 from the solutions in the case w,z not zero? And the cases w=0 or z=0 with the other being nonzero are quite different. I think you should think this over with a clear head. Because I'm not sure what your "when exclusively" phrase even means.

I thought that if w and z were not zero then x couldn't be either?

For the case when w or z are zero but not both, (that is what i mean the exclusive union of the set of w and z) then there are no solutions. I think iwill just write it more simlply as my answer. Thanks again

 

[karnten07]

The only possibilities are everything is a solution, nothing is a solution and there is one unique solution. Figure how to split these possibilities up depending on whether w and z are or are not zero. If you are solving linear equations in two variables you can probably handle that.

Yes i think I've got it now. Thanks

 

[yopeace92]

Hey, i got the same questions for my exercise, but when solving part b i got x1= -1-3i and x2=-1+2i. Not sure if that's right but i substituted it into each of the equations. But thanks for the help on part a! I'm new to linear algebra too

 

[Dick]

Hey, i got the same questions for my exercise, but when solving part b i got x1= -1-3i and x2=-1+2i. Not sure if that's right but i substituted it into each of the equations. But thanks for the help on part a! I'm new to linear algebra too

If you are talking about the problem in post 11, that's not the right solution.

 

[yopeace92]

I've just rechecked the equations after your reply, and one of mine is slightly different which would explain it. Sorry!