How to Solve (a+bi)/(x-i)=0 to Find X?

[Wi_N]

Homework Statement

so the equation is 5+3i/(a-i) =0

i need to find a REAL A so it becomes 0

Homework Equations

The Attempt at a Solution

i tried multiplying with its conjugate but it won't take. I am completely clueless.

 

[BvU]

i need to find a REAL A

No way. Please post the complete problem statement . And your attempt at solution in detail. You know how things go at PF

 

[Wi_N]

I need to find a REAL "a" that makes this function 5+3i/(a-i) =0

maybe i can use trigonometry...

 

[SrayD]

There is no single value 'a' can take to give zero. If you multiply, numerator & denominator, by complex conjugate of the denominator to simplify, then will get two equations one for real part & other imaginary. Each requires 'a' to be different values; in fact they are negative reciprocals of each other. So, I'm with BvU in that this cannot be entire problem statement, or there is some other error.

 

[SammyS]

Homework Statement

so the equation is 5+3i/(a-i) =0

i need to find a REAL A so it becomes 0

Homework Equations

The Attempt at a Solution

i tried multiplying with its conjugate but it won't take. I am completely clueless.

Do you perhaps mean

(5+3i)/(a-i) =0 ?

 

[Ray Vickson]

Homework Statement

so the equation is 5+3i/(a-i) =0

i need to find a REAL A so it becomes 0

Homework Equations

The Attempt at a Solution

i tried multiplying with its conjugate but it won't take. I am completely clueless.

Do you mean
$$\frac{5 + 3i}{a-i} = 0,$$
or do you mean
$$5 + \frac{3i}{a-i}=0?$$
It makes a great difference.

Actually, if I read your expression using standard rules for parsing mathematical expressions, what you wrote really is the second one.

 

[Mark44]

Homework Statement

so the equation is 5+3i/(a-i) =0
i need to find a REAL A so it becomes 0

I am assuming from what you wrote in the title, "(a+bi)/(x-i)=0 finding x", your equation is really this: (5 + 3i)/(a - i) = 0.

The only way for a fraction or other rational expression to be equal to zero is when the numerator is zero.

 

[Ray Vickson]

There is no single value 'a' can take to give zero. If you multiply, numerator & denominator, by complex conjugate of the denominator to simplify, then will get two equations one for real part & other imaginary. Each requires 'a' to be different values; in fact they are negative reciprocals of each other. So, I'm with BvU in that this cannot be entire problem statement, or there is some other error.

No, there is a unique $a$ that solves
$$5 + \frac{3i}{a-i} = 0,$$
but it is not real.

 

[Wi_N]

(5+3i)/(a-i) =0

"a" has to be a real number.

 

[Ray Vickson]

(5+3i)/(a-i) =0

"a" has to be a real number.

You can spend the next thousand years looking for a solution, but you will not find one.

 

[epenguin]

I'm guessing the real question was ... = 1

 

[Mark44]

(5+3i)/(a-i) =0

"a" has to be a real number.

You can spend the next thousand years looking for a solution, but you will not find one.

I agree completely with Ray here. The only way a fraction can be zero is if the numerator is zero. Are you positive that what you have written is the problem that is to be solved?

 

[Wi_N]

nevermind. missunderstood the question i guess. the answer is just solving the a in a+bi and ignoring the i part. trivial question.

 

[BvU]

Is not what I see in post #1. By accident ?

 

[Wi_N]

Is not what I see in post #1. By accident ?

not an accident. but same kind of problem.

 

[BvU]

repeat my post #2. What do you want to work on ?