How do I solve $z^2+(3+4i)z-1+5i=0$

Question

Right, I asked yesterday about the explanation for the roots of quadratic equations, now I'm trying to apply these concepts.

As stated in the title, we start with:

$$z^2+(3+4i)z-1+5i=0$$

If we calculate the terms in the parenthesis we get:

$$z^2+3z+4zi -1 + 5i=0$$

Now, ordinarily I would convert $z^2+4zi$ into $(z+2i)^2+4$, move all of the term apart from the square to the other side and set $(z+2i)^2=w^2$, then I'd set $w=a+bi$ and work the problem from there.

But how do I deal with that $3z$?

I can't very well set $z$ to $a+bi$ since those would't be the same $a$s and $b$s as those in $w$....

Answer 1

Why convert $z^2+4zi$ and leave out the factor $3z$? You can write $$z^2+(3+4i)z=\left(z+\frac{3+4i}{2}\right)^2-\left(\frac{3+4i}{2}\right)^2.$$

Answer 2

By the quadratic formula we get $$z_ {1,2}=-\frac{3+4i}{2}\pm\sqrt{\left(\frac{3+4i}{2}\right)^2+1-5i}$$ the radicand simplifies to $$-\frac{3}{4}+i$$

Answer 3

The discriminant of quadratic equation is $$-3+4i = (2i+1)^2$$

so $$ z_{1,2} = {-3-4i\pm (2i+1)\over 2}$$

Answer 4

My first suggestion is that it is not a good thing to expand out the equation.

My hint is the following:

$$z^2+bz+c=(z+b/2)^2-(b/2)^2+c$$

and note that $b=3+4i$.