- #1

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## Homework Statement

So it is pretty straight forward, solve this.

z

^{2}+2(1-i)z+7i=0

## Homework Equations

z

^{2}+2(1-i)z+7i=0

(-b±√(b

^{2}-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 z

^{2}-2iz+7i+2z=0

Now I don't really know what to do because my textbook has two examples, in both z

^{2}is ignored.

first it has z

^{2}+2iz-1-i=0 and used a=1, b=2i and c=-1-i

the second example shows z

^{2}+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 doesnt help me understand what it should be and why, which is really what I want to know and not the solution.