# Z^2 = z conjugate

Homework Statement
Find all complex solutions to
z^2 = z conjugate
i.e. (a+bi)^2 = a-bi

The attempt at a solution
First attempt:
factoring out (a+bi)^2 = a-bi

Second attempt:
r^2 (cos2v + isin2v) = r (cos-v + isin-v)
r must be 1.
2v = -v + 2∏n
3v = 2∏n
v= 2∏n/3
But that isn't the case.

lurflurf
Homework Helper
You both forgot z=0.

The trig way form cos(v)=-1/2 we see |v|=2∏/3

$$\left( a+bi \right)^{2}\; =\; a^{2}-b^{2}+2bi$$
$$a^{2}-b^{2}+2bi=a-bi$$
How do you, leon, go from there to find a?

I would claim
$$a\neq a^{2}-b^{2}$$

vela
Staff Emeritus
Homework Helper
It would help if you multiplied (a+bi)2 out correctly first.

The polar form would lead to:
$$r^2 = r, \; r \ge 0$$
and
$$2\theta = -\theta + 2 k \pi, \; k \in \mathbb{Z}$$
What is the solution to these equations?

It would help if you multiplied (a+bi)2 out correctly first.
Yes, I forgot the a after 2bi. But that is still not equal to a-bi, which someone posted (seems like that post is now removed)

The polar form would lead to:
$$r^2 = r, \; r \ge 0$$
and
$$2\theta = -\theta + 2 k \pi, \; k \in \mathbb{Z}$$
What is the solution to these equations?

I wrote:
Second attempt:
r^2 (cos2v + isin2v) = r (cos-v + isin-v)
r must be 1.
2v = -v + 2∏n
3v = 2∏n
v= 2∏n/3
But that isn't the case.

vela
Staff Emeritus
Homework Helper
Yes, I forgot the a after 2bi. But that is still not equal to a-bi, which someone posted (seems like that post is now removed)
$a=a^2-b^2$ is true if you require that $z^2 = \bar{z}$. That's the point of the problem, to find the values of a and b so that the relationships hold.

$$r^2 e^{2it}=re^{-it}$$

re-aranage, I get:

$$r=e^{-3it}$$

Now, the right side is the unit circle. When on the unit circle, is that going to be a positive real number?

I wrote:
Second attempt:
r^2 (cos2v + isin2v) = r (cos-v + isin-v)
r must be 1.
2v = -v + 2∏n
3v = 2∏n
v= 2∏n/3
But that isn't the case.

r is not necessarily 1. Also, n can be {0, 1, 2}, because everything else, when taken as an argument of a trigonometric function, gives one of these cases. Also:

$$\cos{0} + i \, \sin{0} = 1$$

$$\cos{\left( \frac{2 \pi}{3} \right)} + i \sin{\left( \frac{2 \pi}{3} \right)} = -\cos{\left( \frac{\pi}{3} \right)} + i \sin{\left( \frac{\pi}{3} \right)} = \frac{-1 + i \sqrt{3}}{2}$$

$$\cos{\left( \frac{4 \pi}{3} \right)} + i \sin{\left( \frac{4 \pi}{3} \right)} = -\cos{\left( \frac{\pi}{3} \right)} - i \sin{\left( \frac{\pi}{3} \right)} = \frac{-1 - i \sqrt{3}}{2}$$

r doesn't neccessarily have to be 1 but it needs to be a real positive number. this gives you a constraint for t. now by taking r exp(it), you have your solutions.

r doesn't neccessarily have to be 1 but it needs to be a real positive number.
Not true, it needs to be a non-negative number.

right