Complex numbers: adding two fractions and solving for z

In summary, if you have a equation that looks like this, it might be a good idea to try and solve for the unknowns first.
  • #1
14
3

Homework Statement


$$\frac{1}{z}+\frac{1}{2-z}=1$$

Homework Equations


Quadratic-formula and algebra

The Attempt at a Solution



Been struggling with this one.. I keep getting the wrong answer, but that isn't the worst part, I can live with a wrong answer as long as the math behind it is correct(formulas etc.) .

So this is the culprit:

$$\frac{1}{z}+\frac{1}{2-z}=1
\Rightarrow \frac{z}{z(2-z)}+\frac{z-2}{z(2-z)}=1
\Rightarrow \frac{z+z-2}{2z-z^2+2z-z^2}=1
\Rightarrow \frac{-2}{2z-2z^2}=1
\Rightarrow -2z^2+2z+2=0
\Rightarrow \frac{-2\pm\sqrt{s^2-4*-2*2}}{2*-2}
\Rightarrow \frac{-2\pm\sqrt{-12}}{-4}
\Rightarrow \frac{2\pm4i\sqrt{3}}{4}
\Rightarrow \frac{1\pm2i\sqrt{3}}{2}
\Rightarrow \frac{1}{2}\pm i\sqrt{3}$$

And this is far from correct.. I should have gotten ## 1 \pm i##.
So I must have done some illegal operation(sick with the flu so my brain isn't working 100%..)

Cheers!
 
Last edited:
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  • #2
lamefeed said:

Homework Statement


$$\frac{1}{z}+\frac{1}{2-z}=1$$

Homework Equations


Quadratic-formula and algebra

The Attempt at a Solution



Been struggling with this one.. I keep getting the wrong answer, but that isn't the worst part, I can live with a wrong answer as long as the math behind it is correct(formulas etc.) .

So this is the culprit:$$\frac{1}{z}+\frac{1}{2-z}=1
\Rightarrow \frac{z}{z(2-z}+\frac{z-2}{z(2-z)}=1
\Rightarrow \frac{z+z-2}{2z-z^2+2z-z^2}=1
\Rightarrow \frac{-2}{2z-2z^2}=1
\Rightarrow -2z^2+2z+2=0
\Rightarrow \frac{-2\pm\sqrt{s^2-4*-2*2}}{2*-2}
\Rightarrow \frac{-2\pm\sqrt{-12}}{-4}
\Rightarrow \frac{2\pm4i\sqrt{3}}{4}
\Rightarrow \frac{1\pm2i\sqrt{3}}{2}
\Rightarrow \frac{1}{2}\pm i\sqrt{3}$$
And this is far from correct.. I should have gotten ## 1 \pm i##.
So I must have done some illegal operation(sick with the flu so my brain isn't working 100%..)

Cheers!
I noticed a couple of errors.

##\displaystyle \frac{z}{z(2-z)}+\frac{z-2}{z(2-z)} \ ## is not equivalent to ##\displaystyle \ \frac{z+z-2}{2z-z^2+2z-z^2} \ ##

The two terms have a common denominator. Don't add the denominators.

Also;
What is the square root of ##\ -12 \ ##?
 
  • Like
Likes lamefeed
  • #3
Oh, my. I feel so embarrassed.. I'll get right to it.

And when it comes to the root of ##-12## I forgot to do ## \sqrt{-12} \Rightarrow i\sqrt{4}\sqrt{3} = 2i\sqrt{3}##
 
  • #4
lamefeed said:
Oh, my. I feel so embarrassed.. I'll get right to it.

And when it comes to the root of ##-12## I forgot to do ## \sqrt{-12} \Rightarrow i\sqrt{4}\sqrt{3} = 2i\sqrt{3}##
Yup. Of course, once you correct the first error, you won't have a square root of ##\ -12## .
 
  • #5
Consider it solved!

$$
\frac{z+(-z)+2}{2z-z^2}=1 \Rightarrow
\frac{2}{2z-z^2} \Rightarrow
2=2z-z^2 \Rightarrow
z^2-2z+2=0 \Rightarrow
\frac{2\pm\sqrt{2^2-4*1*2}}{2} \Rightarrow
\frac{2\pm\sqrt{-4}}{2} \Rightarrow
\frac{2\pm2i}{2} \Rightarrow \\
1 \pm i
$$

I got such a good feeling when I solved it. Feels weird, I'm 100% super excited! Thanks SammyS and fresh_42(reply in the old thread).
 
  • #6
You can optimise
[tex]
\frac{1}{z} + \frac{1}{2-z} = 1 \iff \frac{1}{z} = 1 + \frac{1}{z-2} \iff \frac{1}{z} = \frac{z-1}{z-2} \iff z^2 -2z +2 =0.
[/tex]
and you get solutions ##1\pm i ##. If you have something of the form ##\frac{a}{f(x)} + \frac{b}{g(x)} = c ##, perhaps it's not such a hot idea to immediately mess with the common denominators and such.
 

1. What are complex numbers?

Complex numbers are numbers that contain both a real and imaginary component. They are written in the form a + bi, where a is the real part, b is the imaginary part, and i is the imaginary unit (equal to the square root of -1).

2. How do you add two complex numbers?

To add two complex numbers, you simply add the real parts and the imaginary parts separately. For example, (2 + 3i) + (4 + 5i) = (2+4) + (3i+5i) = 6 + 8i.

3. What is the process for solving for z in a complex number equation?

To solve for z in a complex number equation, you need to isolate the complex number on one side of the equation. This may involve using algebraic operations to move terms around. Once the complex number is by itself, you can then equate the real and imaginary parts to find the values of a and b in the form a + bi.

4. Can you add two fractions with complex number denominators?

Yes, you can add two fractions with complex number denominators. To do so, you need to find a common denominator and then add the numerators as you would with any other fraction. For example, (1/2 + 3/4i) + (1/3 + 2/3i) = (3/6 + 6/12i) + (4/12 + 8/12i) = (7/6 + 14/12i).

5. Are there any special rules for adding complex numbers?

Yes, there are two special rules for adding complex numbers. First, the commutative property holds, meaning that the order in which you add two complex numbers does not matter. Second, the distributive property holds, meaning that you can distribute a number to both the real and imaginary parts of a complex number separately. For example, a(2 + 3i) = 2a + 3ai.

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