Question

complex numbers, znot=conjugate
z(znot)+z+znot=0

Answer 1

I found a^2+2a+b^2=0

Answer 2

maybe we can use this
\[z=a+bi =r e^{i \theta} \\ z'=a-bi=r e^{-i \theta}\]

Answer 3

they are asking for the graph of z's that satisfy the equation

Answer 4

doesn't matter what r is
any r will work
that equation you can either try to solve as a quadratic or you can try to use
\[\cos(\theta)=\frac{1}{2}(e^{i \theta} +e^{- i \theta})\]

Answer 5

well, I haven't seen this expression yet. there is no other way to solve it?

Answer 6

wait which expression?

Answer 7

e^(i*theta)

Answer 8

oh so no euler formula at all ?

Answer 9

nope

Answer 10

I started with z=a+bi

Answer 11

and expended the equation

Answer 12

what ab out cos and sin?

Answer 13

I didn't try the trigonometric form yet

Answer 14

\[z=r \cos(\theta)+i r \sin(\theta) \\ z'=r \cos(\theta)-i r \sin(\theta)\]
can we use this ?

Answer 15

well you will see something cool happen if you do

Answer 16

and what I mean is you will get to use pythagorean identity

Answer 17

ok, I will do it. thanks

Answer 18

you closed it
does that mean you got it?

Answer 19

yeah =]

Answer 20

cool stuff