Question

Let z be a complex number such that the imaginary part of z is nonzero and A = z^2 + z + 1 is real. Then A cannot take the value of what?

Answer 1

What have you tried?

Answer 2

i wrote z as a+bi and then plugged that into z^2+z+1 @Zarkon

Answer 3

i got \[a^2+2abi-b^2\]

Answer 4

so how do i find it what it can;t take the value of?

Answer 5

if a^2+2abi-b^2 is real?

Answer 6

you get more than that...you didn't include the z+1 part

Answer 7

oh yah oopps

Answer 8

so its \[a^2+2abi-b^2+a+bi+1\]

Answer 9

which if i simplify is\[a^2+(2a+1)bi-b^2+a+1\]

Answer 10

i factored out a 2a+1 because 2a+1 must be real right?

Answer 11

so that means that (2a+1)bi is complex right?

Answer 12

if A is real then (2a+1)bi=0

Answer 13

@Zarkon

Answer 14

how?

Answer 15

oh yah there is no imaginary part right?

Answer 16

oh so it simplifies to:\[a^2-b^2+a+1\]

Answer 17

now what should i do?

Answer 18

\[(a+b)(a-b)+a+1\]
this?

Answer 19

ok i think i can do it from here @Zarkon

Answer 20

no actually i'm stuck :(

Answer 21

i also wrote it like:\[a(a+1)-b^2+1\]

Answer 22

well -b^2 will always be negative

Answer 23

if (2a+1)bi=0 and b is non zero then a=-1/2

Answer 24

thanks @Zarkon