How do you solve the quadratic $a^2-(1-4i)a-5+i=0$ using any method?

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Answer 1 · 2016-10-29T13:24:45

With $a=x+iy$ the solutions are
$((x = 0, y = 1),(x=1/2 - 3 i, y= -2 - i/2),(x = 1/2 + 3 i, y = -2 + i/2),(x = 1, y= -5))$

Calling $p(a)=a^2-(1-4i)a-5+i=0$

and making $a = x+iy$ we have

$p(x+iy)=x^2 - 4 y - y^2-x-5 + i (4 x - y + 2 x y+1)=0$

Solving now for $x,y$

${(x^2 - 4 y - y^2-x-5=0),(4 x - y + 2 x y+1=0):}$

we obtain the solutions

$((x = 0, y = 1),(x=1/2 - 3 i, y= -2 - i/2),(x = 1/2 + 3 i, y = -2 + i/2),(x = 1, y= -5))$

How do you solve the quadratic a^2-(1-4i)a-5+i=0a^2-(1-4i)a-5+i=0 using any method?