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## Homework Statement

The question is like this:

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

## 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