1. ## Solving quadratic equations with Complex numbers

Hey guys,
The problem in question here is- "Solve this equation over C: Z2+Z+(1-i) "

I have tried completing the square, and ended up with: (Z+1/2)2 - (-3+4i)/4 = 0
This doesn't seem very helpful, because when I go to factorise, I end up with sqrt(-3+4i)/2, which seems unable to be simplified.

Using the quadratic equation yields the same result :/

The solution answers are: Z=-i and Z=-1-i

Any help much appreciated, thank-you
Smeato

2. ## Re: Solving quadratic equations with Complex numbers

You need to learn a little more about complex numbers. The complex number (-3+ 4i)/4 can be written in the "polar form" $\frac{5}{4}(cos(53.13)- i sin(53.13))$. The square roots are $\frac{\sqrt{5}}{2}(cos(53.13/2)- i sin(53.13/2))= \frac{\sqrt{t}}{2}(cos(26.575)+ i sin(26.575))$ and $\frac{\sqrt{5}}{2}(cos(206.575)- i sin(206.575))$ where the angles are in degrees.

3. ## Re: Solving quadratic equations with Complex numbers

Why aren't any of my replies showing up in the thread?

4. ## Re: Solving quadratic equations with Complex numbers

$z^2+z+(1-i) = 0$

$(z^2+1) + (z-i) = 0$

$(z-i)(z+i) + (z-\color{red}{i}) = 0$

$(z-i)(z+i+1) = 0$

$z = i$ or $z = -(1+i)$

5. ## Re: Solving quadratic equations with Complex numbers

Thanks skeeter, but how did you get from line 3 to line 4?

6. ## Re: Solving quadratic equations with Complex numbers

Originally Posted by Smeato
Thanks skeeter, but how did you get from line 3 to line 4?
line 3 should have $(z-i)$ as the last term ... accidentally hit 1 instead. Following is correct ...

$\color{red}{(z-i)}(z+i) + \color{red}{(z-i)} = 0$

pull out the common factor ...

$\color{red}{(z-i)}[(z+i) + 1] = 0$

7. ## Re: Solving quadratic equations with Complex numbers

Ahhhh, makes much more sense! Thanks a bunch