Solutions of $2z^2+(2+2i)z+5i=0$?

Question

How do you find all solutions where $z$ is a complex number? $$2z^2+(2+2i)z+5i=0$$ I know how to do this with real numbers but not complex numbers.

Answer 1

Suppose $z \in \mathbb{C}$ satisfies $$2z^2+(2+2i)z+5i=0$$ Then $z = x + yi$, for some $x,y \in \mathbb{R}$.

Replace $z$ by $x+yi$, expand, and regroup. Then the equation becomes $$(2x^2-2y^2 + 2x-2y) + (4xy+2x+2y+5)i = 0$$ which yields the system $$2x^2-2y^2 + 2x-2y = 0$$ $$4xy+2x+2y+5=0$$ Noting that $$2x^2-2y^2 + 2x-2y = (x-y)(x+y+1)$$ it follows that either $y = x$, or $y = -x - 1$.

For the case $y=x$, you get the quadratic equation $$4x^2+4x+5=0$$ but since discriminant is negative, and $x$ must be real, that case can be excluded.

For the case $y=-x-1$, you get the quadratic equation $$(2x-1)(2x+3)=0$$ which yields the solutions $x = {\large{\frac{1}{2}}},\;x = -{\large{\frac{3}{2}}}$.

Then from the equation $y=-x-1$, the corresponding $y$-values are $y=-{\large{\frac{3}{2}}},\;\;y = {\large{\frac{1}{2}}}$, hence the solutions for $z$ are $$z = \frac{1}{2}-\frac{3}{2}i,\;\;z = -\frac{3}{2}+\frac{1}{2}i$$

Answer 2

Hint. Notice that by completing the square we have that $$2z^2+(2+2i)z+5i=2\left(z+\frac{1+i}{2}\right)^2+4i.$$ Can you take it from here? Recall how to find the square roots of a complex number.

Answer 3

You do it the exact same way for complex numbers as you do for real. The quadratic formula doesn't suddenly stop working (unless you think too hard about whether the square root makes sense).

Answer 4

You can use the (complex variant) of the quadratic formula, or rewrite by completing the square: $$2z^2+(2+2i)z+5i=0 \iff 2\left(z+\left(\frac{1}{2}+\frac{i}{2}\right)\right)^2+4i=0 \iff \ldots$$

Answer 5

your equation can factorized into $$((1-i) z+(1+2 i)) ((1+i) z+(2+i))=0$$

Answer 6

Let apply the ordinary formula for roots of quadratic equation

$$ az^2+bz+c=0\implies z_1,z_2=\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$

or as an alternative note that

$$2z^2+(2+2i)z+5i=0\iff z^2+(1+i)z+\frac52i=0$$

thus

• $z_1\,z_2=\frac52 i$
• $z_1+z_2=-1-i$

indeed

$$(z-z_1)(z-z_2)=z^2-(z_1+z_2)z+z_1z_2=0$$