Question

what does the question mean by:
Find all complex numbers z for
conjugate of z = z^2

Answer 1

\[z^2 = (x+iy)^2 = x^2-y^2 + i(2xy)\]

Answer 2

You can write the conjugate of z as \(z^*\) and what it means is to put -i everywhere you see i, like this:

\[z^*=(x+iy)^* = x-iy\]

So right now you're on the right track,

\[z^*=z^2\]
\[x-iy = x^2-y^2+i2xy\]

Now you can continue solving from here, since this is really two equations. The real part on the left is equal to the real part on the right, and complex part on the left is equal to the complex part on the right.

Answer 3

so the question is asking for the real and imaginary part of z^2?

Answer 4

Question is asking for \(z\), since that solves the equation \(z^*=z^2\).

But \(z\) is made up of a real and imaginary part, so we have to find those, since finding x and y allows us to find z:

\[z=x+iy\]

Answer 5

sorry i dont think i understand
if i "complete" the equation
\[x^2-x-y^2+i(2xy+y)=0\]
then what do i do with it?
\[x^2-x-y^2=0\]
\[2xy+y=0\]
this?

Answer 6

Now you solve for x and y, I am pretty sure you've solved equations like the first and second before.

Answer 7

so what i did in my previous post is what the question wanted?

Answer 8

Yep, but you're not done yet.

Answer 9

x=-1/2
y=i(0.75^0.5)

Answer 10

so i just sub in and thats the ans?

Answer 11

Yeah, basically although I think you have some mistakes when you solved for y, two things:

it's a quadratic so you should have a positive and negative root

and I think the number you got is wrong, try it again.

Answer 12

opps yeah i was careless
y=+-0.75^0.5?

Answer 13

Better but not completely right yet! Here I'll help:

\[x^2-x-y^2=0\]
We know \(x = \tfrac{1}{2}\) so plug that in we get:

\[\tfrac{1}{4} - \tfrac{1}{2} = y^2\]

Rewrite \(\tfrac{1}{2} = \tfrac{2}{4}\) and then this becomes


\[\tfrac{1}{4} - \tfrac{2}{4} = \tfrac{-1}{4}= y^2\]

Keep going and then after you find them, plug x and y back in for z. Then check to make sure \(z^* = z^2\) works (it should be pretty easy I think once you have the right answer to see that it's true).

Answer 14

wait x=-0.5,
y(2x+1)=0
2x+1=0
2x=-1
x=-0.5?

Answer 15

oh I guess you're right hmm something is fishy I just woke up and am tired but I know I have found two answers somehow haha.

Maybe you can sort it out while I go make coffee, but clearly

\[z=0\] and \[z=1\] are solutions since
\[0^*=0^2\] and \[1^* = 1^2\]

Actually my intuition on this tells me there's 2 more, since geometrically this makes sense to me:

\[z=\tfrac{-1}{2} + \tfrac{\sqrt{3}}{2}\]
\[z=\tfrac{-1}{2} - \tfrac{\sqrt{3}}{2}\]

(complex conjugate means reflect across the x-axis and square means rotate a number twice from 1, I can explain in a bit I think, but first I'm making coffee)

Answer 16

wow, such a thorough analysis. thank you for the writeup i think i understand the question better now.