Thread: Complex numbers
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HallsofIvy
HallsofIvy is offline
#2
Oct28-05, 09:25 PM
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Quote Quote by UnD
Lol sry got stuck on this past quesiton
consider z^2 + az+ (1+i)=0 , find the complex number a, given that i is a root of the equation
so (x-i) is a factor thing
z= -a +_ sqroot a^2 -4 -4i
----------------------------
2
then let sqroot a^2 -4 - 4I =c+ib
lol i don't know where to go after that.
Well actually, z- i is a factor, not x-i!
Given that z-i is a factor we can divide by it:
z-i divides into z2+ az+ (1+i)
z+ (a+i) times with a remainder of (1+a)i . If i really is a root of the equation, then that remainder must be 0. So a must be ____.