Find a complex number z which satisfies a particular condition

  • Thread starter Seydlitz
  • Start date
  • #1
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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
 

Answers and Replies

  • #2
tiny-tim
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Hello Seydlitz! :smile:

Have you been at the coffee? :wink:

1/(a+b) = 1/a + 1/b = (a + b)/ab … and solve!​
 
  • #3
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Hello Seydlitz! :smile:

Have you been at the coffee? :wink:

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

Strange, I've tried it before and it leads to ##ab=1##?
 
  • #4
tiny-tim
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?? :confused:

try it again …

you should get a/b = something :smile:
 
  • #5
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?? :confused:

try it again …

you should get a/b = something :smile:

Woops..I made bad mistake on the way. I cancel (a+b) blatantly.

Is ##\frac{a}{b}=\frac{-b}{a+b}##?
 
  • #6
tiny-tim
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nooo :redface:

show us your calculations :smile:
 
  • #7
D H
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nooo :redface:
Try that again, tiny-tim.

What Seydlitz posted is correct. It's just not very useful.
 
  • #8
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nooo :redface:

show us your calculations :smile:

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
tiny-tim
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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 :smile:
 
  • #10
761
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Hint: in the end substitute z=a/b to make the calculation simpler.
 
  • #11
D H
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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
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nooo :redface:

show us your calculations :smile:

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

Huzzzah!!

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

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