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))#