Solving Complex Equations: z2+2(1-i)z+7i=0

[KUphysstudent]

Homework Statement

So it is pretty straight forward, solve this.
z²+2(1-i)z+7i=0

Homework Equations

z²+2(1-i)z+7i=0
(-b±√(b²-4ac))/2a

The Attempt at a Solution

So what I would do first is solve 2(2-1)z, I get (2-2i)z=2z-2iz
we now have z²-2iz+7i+2z=0
Now I don't really know what to do because my textbook has two examples, in both z² is ignored.
first it has z²+2iz-1-i=0 and used a=1, b=2i and c=-1-i
the second example shows z²+2z+4=0 and has 2z=b and ac =4*1

the problem with the two examples is I cannot deduce what will be a, b, and c in my problem.
I mean following the logic of the first example I get 2z = b or -2iz = b and then c = either 7i or 7i + 2z or something completely different.
I tried plugging in the numbers
so:
(-2i ±√(2i^2-4*7i))/2 = (-2i±√(-4-28i))/2

then I tried 2z = b instead of 2i = b
(-2 ±√(2^2-4*7i))/2 = (-2±√(4-28i))/2

I mean this is just guessing and I kept going, what if 7i = b, etc. but it doesn't help me understand what it should be and why, which is really what I want to know and not the solution.

 

[member 587159]

We consider equations ##az^2 + bz + c = 0##.

That is, we identify the coefficients in a specific equation with the numbers ##a,b,c.##

We have:

##z^2+2(1-i)z+7i=0 \implies a = 1, b = 2(1-i), c = 7i##.

##z^2+2z+4=0 \implies a = 1, b = 2, c = 4##

and in the last example ##b = 2z## is wrong.

 

[FactChecker]

The ##a## and ##b## of the quadratic equation are the complex coefficients of ##z^2## and ##z## of the equation. ##c## is just the complex constant term. So you can just read them off from the equation. They don't include ##z## or any power of ##z##.

 

[KUphysstudent]

oh, that explains a lot. Thanks you two I was getting frustrated :)