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

  • Thread starter Thread starter UnD
  • Start date Start date
  • Tags Tags
    Complex
AI Thread Summary
To solve the equation z^2 + az + (1+i) = 0 with i as a root, it is established that z - i is a factor. Dividing the polynomial by this factor yields z + (a+i) with a remainder of (1+a)i. For i to be a root, the remainder must equal zero, leading to the conclusion that a must be -1. The discussion highlights the steps to find the complex number a and the importance of correctly identifying factors in polynomial equations. The solution process emphasizes the relationship between the remainder and the roots of the equation.
UnD
Messages
21
Reaction score
0
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.
 
Physics news on Phys.org
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 ____.
 
I picked up this problem from the Schaum's series book titled "College Mathematics" by Ayres/Schmidt. It is a solved problem in the book. But what surprised me was that the solution to this problem was given in one line without any explanation. I could, therefore, not understand how the given one-line solution was reached. The one-line solution in the book says: The equation is ##x \cos{\omega} +y \sin{\omega} - 5 = 0##, ##\omega## being the parameter. From my side, the only thing I could...
Back
Top