complex numbers problem (1 Viewer)

[bally24]

hey guys im having massive problems trying to solve this....been a while since i did this so could be something v obvious, but nevertheless: Find the complex solutions of z² + 2z(conjugate of z here) + 1 = 0.

tried subbing x + iy, got weird stuff, strange thing is there are 3 answers: -1, 1 + 2i, 1 - 2i. doesnt this contradict the whole a polynomia of degree n can have at most n solutions thing? anyway thanks for your help.

 

[Bank$]

i think u might need to use CIS@ then somthing like cos@=cos(-@) and go along those lines and yes i beleive that the answer is incorrect due to the "fundemental theorem of algebra".

this somtimes happens to me lol but did u maybey look at the wrong answer to the wrong question ?

 

[ssglain]

z<SUP>2</SUP> + 2(conjugate of z) + 1 = 0 is not a polynomial because z and conj.z do not represent the same pronumeral. The fundamental theorem of algebra does not apply here.

It works out very easily using the substitutions z= x + iy & conj.z = x - iy:
(x + iy)^2 + 2x(x - iy) +1 = 0
x^2 + 2ixy - y^2 +2x - 2iy + 1 = 0

Equating Re & Im parts gives:
x^2 - y^2 + 2x + 1 = 0 -- eq.1
2xy - 2y = 0 --eq.2

From eq.2: 2y(x - 1) = 0 --> y = 0 or x = 1

Put y=0 into eq.1:
x^2 - 0 + 2x + 1 = 0
(x + 1)^2 = 0 --> x = -1

Put x = 1 into eq.1:
1 - y^2 + 2 + 1 = 0
y^2 = 4 --> y = +-2

.: z = -1 + 0i, 1 + 2i, 1 - 2i confirms the answer in your book.

 

[bally24]

ahhhhhhh ok excellent thank you, ive been trying that subbing all day, think my algebra's got quite rusty since last year haha.