Solving z^2 = z Conjugate Complex Equation

[vilhelm]

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
leads nowhere.

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]

You both forgot z=0.

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

 

[vilhelm]

\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]

It would help if you multiplied (a+bi)² out correctly first.

 

[Dickfore]

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

 

[vilhelm]

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:
<br /> r^2 = r, \; r \ge 0<br />
and
<br /> 2\theta = -\theta + 2 k \pi, \; k \in \mathbb{Z}<br />
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]

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.

 

[jackmell]

How about writing it as:

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?

 

[Dickfore]

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:

<br /> \cos{0} + i \, \sin{0} = 1<br />

<br /> \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}<br />

<br /> \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}<br />

 

[susskind_leon]

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.

 

[Dickfore]

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.

 

[susskind_leon]

right