Complex Quadratics

[dense notch]

How am i supposed to solve for z;

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

It says to use the following formula;

z^2 - 2Re(∆)z + ∆*Conjugate(∆) = 0

When z has the complex root z = ∆

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[twilit dirge]

I would let z = a + ib
Then set real = 0 img = 0
And solve for 2 variables 2 eqns

[dense notch]

how tho

[twilit dirge]

Like (a + ib)² - (a + ib) + (1 + i ) = 0 + 0i

Set an equation for real part.
Another for imaginary part.

[cerulean onyx]

$z^2-z+\frac{1}{4}=-1-i+\frac{1}{4} \ (z-\frac{1}{2})^2=-\frac{3}{4}-i \$ so there are two complex numbers z that fulfill the last equation

[grave spire]

Ludwig

[bitter moat]

or use quadratic formula

[fresh drift]

Yeah, the quadratic formula still holds for complex coefficients too

Though to the person who asked the question, you should watch out for cases where the discriminant (b^2 - 4ac) ends up being a complex value, which may make it more difficult to calculate the square root of said discriminant

In cases where the discriminant is a complex value, my preferred method is to set the square root of the discriminant to be equal to some z = a + ib where a and b are real numbers.

If you’d like to know why this works, click here : ||This works because the set of complex numbers is ‘algebraically closed’, basically meaning any algebraic operation on complex numbers will result in solution(s) in the set of complex numbers.||

Anyway, following that, this would mean that the discriminant would equal to z^2, which is (a + ib)^2.

After some algebraic manipulation, you should end up with a pair of simultaneous equations that you should solve to find values of a and b.

[bitter moat]

complex square root:
$±(\sqrt{\frac{r+a}{2}} + sgn(b) i\sqrt{\frac{r-a}{2}})$
where $sgn(b)$ is sign of b

[grave spire]

(xy)²=(yz)²=(xz)²=xyyzxz=-1

[bitter moat]

or use this if you dont want to solve equations