Graph complex numbers to verify z^2 = (conjugate Z)^2

  • #1
31
7

Homework Statement:

Find the complex numbers that verify the equation

Relevant Equations:

z^2 = (conjugate Z)^2
Hello! :smile:
I am locked in an exercise.
I must find (and graph) the complex numbers that verify the equation:
##z^2=\bar z^2 ##​

If ##z=x+iy## then:

##(x+iy)^2=(x-iy)^2 ##
and operating and simplifying,

##4.x.yi=0 ##
and here I don't know how to continue...
can you help me with ideas?
thanks!!
 

Answers and Replies

  • #2
BvU
Science Advisor
Homework Helper
2019 Award
13,682
3,335
Well, when is a product of two real numbers equal to zero ?
After all, ##x## and ##y## are real numbers !
 
  • #3
FactChecker
Science Advisor
Gold Member
5,840
2,197
A correction of your terminology: You should say they are values of z that satisfy the equation, not verify.
 
  • Like
Likes il postino
  • #4
31
7
Well, when is a product of two real numbers equal to zero ?
After all, xxx and yyy are real numbers !

Then the answer is: It is true for pure complex numbers, and for pure imaginary complex numbers?
how can i graph them?
thanks for answering!
 
  • Like
Likes BvU
  • #5
PeroK
Science Advisor
Homework Helper
Insights Author
Gold Member
15,033
7,213
Then the answer is: It is true for pure complex numbers, and for pure imaginary complex numbers?
how can i graph them?
thanks for answering!
Whenever you are asked to draw a graph of complex numbers, you should consider looking at the polar form.
 
  • #6
FactChecker
Science Advisor
Gold Member
5,840
2,197
Then the answer is: It is true for pure complex numbers, and for pure imaginary complex numbers?
how can i graph them?
Are you familiar with the complex plane and how it is drawn in two dimensions?
 
  • #7
666
204
When does multiplying two real numbers yield zero? When either or both are equal to zero.
It is true for pure complex numbers, and for pure imaginary complex numbers?
Indeed. Do you know where these numbers, i.e ##z=x## and ##z=iy##, lie in the complex plane?
 
  • #8
BvU
Science Advisor
Homework Helper
2019 Award
13,682
3,335
Then the answer is: It is true for pure complex numbers, and for pure imaginary complex numbers?
how can i graph them?
thanks for answering!
Small correction: you represented ##{\bf z} = x + iy## and found that ##{\bf z} ^2 = \bar {\bf z} ^2\ \ ## if ##\ x=0\ ## or ##\ y = 0\ ##.

##\ y = 0\ ## for real numbers, ##\ x = 0\ ## for purely imaginary numbers.

Real numbers are a subset of the complex numbers.
The term 'pure complex numbers' is meaningless.

And you know how to draw ##\ x=0\ ## and how to draw ##\ y = 0\ ##.

I warmly recommend @LCKurtz ' excellent essay on complex numbers. It is aimed at teachers but contains a treasuretrove of guidance for students as well.

[edit]
Oh, and: an alternative way to look at ##{\bf z} ^2 = \bar {\bf z} ^2\ \ ## is to conclude that this is satisfied if ##{\bf z} = \bar {\bf z} \ ## and also if ##{\bf z} = -\bar {\bf z} \ ## so you don't even need to work out the complex squares :wink:
 
  • Like
Likes FactChecker and il postino
  • #9
FactChecker
Science Advisor
Gold Member
5,840
2,197
It's not clear how much you know about complex numbers. Their polar form, how multiplication works in polar form, and how conjugation works in polar form, would all make the answer to this problem much more intuitive.
 
  • Like
Likes BvU
  • #10
31
7
[QUOTE = "FactChecker, publicación: 6322397, miembro: 500115"]
No está claro cuánto sabes sobre números complejos. Su forma polar, cómo funciona la multiplicación en forma polar y cómo funciona la conjugación en forma polar, haría que la respuesta a este problema sea mucho más intuitiva.
[/CITAR]

Planteo el problema en forma polar y te muestro
 

Related Threads on Graph complex numbers to verify z^2 = (conjugate Z)^2

  • Last Post
Replies
1
Views
922
Replies
8
Views
1K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
7
Views
3K
  • Last Post
Replies
4
Views
4K
  • Last Post
Replies
2
Views
53K
  • Last Post
Replies
17
Views
1K
Replies
26
Views
574
Top