# Find a complex number z which satisfies a particular condition

Hello guys!

## 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

tiny-tim
Homework Helper
Hello Seydlitz!

Have you been at the coffee?

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

Hello Seydlitz!

Have you been at the coffee?

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

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

tiny-tim
Homework Helper
??

try it again …

you should get a/b = something

??

try it again …

you should get a/b = something

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

tiny-tim
Homework Helper
nooo

D H
Staff Emeritus
nooo
Try that again, tiny-tim.

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

nooo

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..)

tiny-tim
Homework Helper
Try that again, tiny-tim.

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

ahh!

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

=> ab = (a+b)2

carry on from there

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

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.

nooo

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