Question

Find all z∈ℂ (complex numbers) which satisfy (conjugate of z)=z^2

Answer 1

Write z = a +bi. Then put that in the equation and find two equations in a and b. Solve.

Answer 2

i.e., z conjugate, z* = a - bi; and
z^2 = a^2 + 2abi + b^2i^2 = ... something.

Now put z* = z^2. The real and imaginary parts must be equal, which will give you two equations in a and b.

Answer 3

OK

Answer 4

these repeats are coming fast and furious. where are you all from?

Answer 5

i just did this one a moment ago

Answer 6

yiazmat, bluebrandon also posting the same questions

Answer 7

y'all should form a study group

Answer 8

I already asked this question a little while ago lol

Answer 9

@bluebrandon, where are you all from?

Answer 10

so did yiazmat. where are you all from?

Answer 11

lol

Answer 12

university of waterloo

Answer 13

and what class is this?

Answer 14

this is for algebra

Answer 15

fancy algebra in that university

Answer 16

i'll say. complex variable algebra.

Answer 17

found the answer?

Answer 18

I got z=(-1+0i), (1+2i), (1-2i)

Answer 19

Well, if z = -1, then z^2 = 1 and z* = -1, so that can't be right.

Answer 20

You should verify your other solutions as well.

Answer 21

woops wrong question
for this one I got
(x^2-y^2) (2xy)
--------- - --------- i
(x^2+y^2) (x^2+y^2)

Answer 22

Also not right, I don't think.

Answer 23

I put
z* = z^2
(z)(z) = z*
z= z*/z

then I let z=(x+yi) and solved

Answer 24

If z = a + bi, then
z^2 = (a^2 - b^2) + 2abi
and
z* = a - bi, where * means conjugate.

Hence if z^2 = z*, then

(a^2 - b^2) + 2abi = a - bi ---- (**)

Now for this to be true, the real and imaginary components must be equal. Hence (**) is true if and only if

a^2 - b^2 = a ------ (1)
2ab = -b ------ (2)

Answer 25

From equation (2), if b is not zero, then 2a = -1 => a = -1/2. Now solve for b.

Notice that if b = 0, then you can show a = 0 or 1; i.e., z = 0 and z= 1 are solutions, which makes sense.

Answer 26

does the way I've done it look like it could make sense?

Answer 27

No, because you've written down an infinite family of solutions. But there are only a finite number.

Answer 28

ok I will try to do it your way then

Answer 29

how does
z=0,1,(-1/2 + (sqrt3)/2), (-1/2 - (sqrt3)/2) sound?

Answer 30

Yep. Those are the four solutions; just put i after the imaginary parts.

Answer 31

alright, thanks for your help!