Solving z^2 + az+ (1+i)=0: Finding a Complex Solution

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SUMMARY

The discussion focuses on solving the quadratic equation z² + az + (1+i) = 0, specifically finding the complex number 'a' given that 'i' is a root. Participants clarify that since 'i' is a root, (z-i) is a factor of the equation. By performing polynomial long division, it is established that the remainder must equal zero, leading to the conclusion that 'a' must satisfy the condition derived from the remainder (1+a)i = 0.

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UnD
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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.
 
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UnD said:
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!:biggrin:
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 ____.
 

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