# Find a complex number z which satisfies a particular condition

1. Nov 18, 2013

### Seydlitz

Hello guys!

1. The problem statement, all variables and given/known data
The question is like this:

If $z=\frac{a}{b}$ and $\frac{1}{a+b}=\frac{1}{a}+\frac{1}{b}$, find $z$.

3. The attempt at a solution
This question is challenging for me because I don't know exactly where to start. The latter condition stated, the sum of the reciprocal of two number is equal to the sum of their reciprocal is already intuitively false for me. You just don't do that kind of operation everyday, and I think it's false for real number. I tried several complex numbers by putting $a$ and $b$ conjugate to each other and see if the condition is true. It's just not true.

I try expressing $a$ and $b$ in rectangular form.

If for example $a=x+iy$, and $b=r+is$. (1)
$$\frac{1}{x+iy+r+is}=\frac{(x+r)-i(y+s)}{(x+r)^2+(y+s)^2}$$

On the other hand.
$$\frac{1}{x+iy}=\frac{x-iy}{x^2+y^2}$$
$$\frac{1}{r+is}=\frac{r-is}{r^2+s^2}$$

To sum them up: (2)
$$\frac{(r^2+s^2)(x-iy)+(x^2+y^2)(r-is)}{(x^2+y^2)(r^2+s^2)}$$

I don't see how (1) and (2) can be equal somehow by cleverly choosing the right $a$ and $b$. On the denominator of (1) you already get addition, but then in (2) you have multiplication. Maybe if $(x^2+y^2)$ and $(r^2+s^2)$ are both $2$, we can have the point where addition is equal to multiplication. I am not certain and I hope you can enlighten me on this.

Thank You

2. Nov 18, 2013

### tiny-tim

Hello Seydlitz!

Have you been at the coffee?

1/(a+b) = 1/a + 1/b = (a + b)/ab … and solve!​

3. Nov 18, 2013

### Seydlitz

Strange, I've tried it before and it leads to $ab=1$?

4. Nov 18, 2013

### tiny-tim

??

try it again …

you should get a/b = something

5. Nov 18, 2013

### Seydlitz

Is $\frac{a}{b}=\frac{-b}{a+b}$?

6. Nov 18, 2013

### tiny-tim

nooo

7. Nov 18, 2013

### D H

Staff Emeritus
Try that again, tiny-tim.

What Seydlitz posted is correct. It's just not very useful.

8. Nov 18, 2013

### Seydlitz

So the only equation is this right?
$\frac{1}{a+b}=\frac{a+b}{ab}$
$\frac{a}{a+b}=\frac{a+b}{b}$
$\frac{a}{a+b}=\frac{a}{b}+1$
$\frac{a}{a+b}-1=\frac{a}{b}$
$\frac{a-(a-b)}{a+b}=\frac{a}{b}$
$\frac{-b}{a+b}=\frac{a}{b}$

Are we expecting a numerical or symbolic answer to this question?
(I haven't eaten yet..)

9. Nov 18, 2013

### tiny-tim

ahh!

Seydlitz, 1/a+b = (a+b)/ab

=> ab = (a+b)2

carry on from there

10. Nov 18, 2013

### dirk_mec1

Hint: in the end substitute z=a/b to make the calculation simpler.

11. Nov 18, 2013

### D H

Staff Emeritus
One more hint: a/b = ab * what? Now multiply both sides of ab=something by this what to get an expression for a/b=z.

12. Nov 18, 2013

### Seydlitz

$$z=\frac{-1\pm i\sqrt{3}}{2}$$

Huzzzah!!

Thank you everybody, especially tiny-tim for being particularly patient with me. I'm that clunky sometimes haha.