- #1

m_s_a

- 91

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hi,

let

z=x+iy

z^2=z.zpar=(x+iy)(x-iy)=x^2+y^2

or

z^2=(x+iy)(x+iy)=(x^2-y^2)

let

z=x+iy

z^2=z.zpar=(x+iy)(x-iy)=x^2+y^2

or

z^2=(x+iy)(x+iy)=(x^2-y^2)

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- Thread starter m_s_a
- Start date

- #1

m_s_a

- 91

- 0

hi,

let

z=x+iy

z^2=z.zpar=(x+iy)(x-iy)=x^2+y^2

or

z^2=(x+iy)(x+iy)=(x^2-y^2)

let

z=x+iy

z^2=z.zpar=(x+iy)(x-iy)=x^2+y^2

or

z^2=(x+iy)(x+iy)=(x^2-y^2)

- #2

Hootenanny

Staff Emeritus

Science Advisor

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- #3

m_s_a

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Thank you for you on your response

And on the new information for me

- #4

Dick

Science Advisor

Homework Helper

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hi,

let

z=x+iy

z^2=z.zpar=(x+iy)(x-iy)=x^2+y^2

or

z^2=(x+iy)(x+iy)=(x^2-y^2)

Take note that (x+iy)(x+iy) is NOT equal to x^2-y^2. It's x^2-y^2+2ixy. Your first 'z^2' is the modulus (size) of the complex number squared. The second is the complex function z*z. They are quite different. A physicist who refers to the first operation as 'squaring' is being pretty sloppy. The proper term is 'modulus squared' and the proper notation is |z|^2.

- #5

Hootenanny

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Nice catch Dick, didn't even see itTake note that (x+iy)(x+iy) is NOT equal to x^2-y^2. It's x^2-y^2+2ixy.

- #6

m_s_a

- 91

- 0

And thank you on the information that you presented

But this is a question in one of the issues

Thanks

- #7

Hootenanny

Staff Emeritus

Science Advisor

Gold Member

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Then I would suggest that,But this is a question in one of the issues

[tex]z^2 = x^2 +2ixy - y^2[/tex]

- #8

m_s_a

- 91

- 0

Then I would suggest that,

[tex]z^2 = x^2 +2ixy - y^2[/tex]

Thank you a lot

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