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Z^2 = z conjugate

  • Thread starter vilhelm
  • Start date
  • #1
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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.
 

Answers and Replies

  • #2
lurflurf
Homework Helper
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You both forgot z=0.

The trig way form cos(v)=-1/2 we see |v|=2∏/3
 
  • #3
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[tex]\left( a+bi \right)^{2}\; =\; a^{2}-b^{2}+2bi[/tex]
[tex] a^{2}-b^{2}+2bi=a-bi[/tex]
How do you, leon, go from there to find a?

I would claim
[tex] a\neq a^{2}-b^{2} [/tex]
 
  • #4
vela
Staff Emeritus
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It would help if you multiplied (a+bi)2 out correctly first.
 
  • #5
2,967
5
The polar form would lead to:
[tex]
r^2 = r, \; r \ge 0
[/tex]
and
[tex]
2\theta = -\theta + 2 k \pi, \; k \in \mathbb{Z}
[/tex]
What is the solution to these equations?
 
  • #6
18
0
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:
[tex]
r^2 = r, \; r \ge 0
[/tex]
and
[tex]
2\theta = -\theta + 2 k \pi, \; k \in \mathbb{Z}
[/tex]
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.
 
  • #7
vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
14,554
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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)
[itex]a=a^2-b^2[/itex] is true if you require that [itex]z^2 = \bar{z}[/itex]. That's the point of the problem, to find the values of a and b so that the relationships hold.
 
  • #8
1,796
53
How about writing it as:

[tex]r^2 e^{2it}=re^{-it}[/tex]

re-aranage, I get:

[tex]r=e^{-3it}[/tex]

Now, the right side is the unit circle. When on the unit circle, is that going to be a positive real number?
 
  • #9
2,967
5
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:

[tex]
\cos{0} + i \, \sin{0} = 1
[/tex]

[tex]
\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}
[/tex]

[tex]
\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}
[/tex]
 
  • #10
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.
 
  • #11
2,967
5
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.
 
  • #12
right
 

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