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December 16th, 2012, 12:42 AM   #1
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Complex numbers;

Help on this, if z=1+i
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December 16th, 2012, 01:07 AM   #2
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Re: Complex numbers;

If , then what is ?
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December 16th, 2012, 01:11 AM   #3
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Re: Complex numbers;

Not listed I think that it changes the - to + and + to -
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December 16th, 2012, 01:12 AM   #4
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Re: Complex numbers;

Conjugation
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December 16th, 2012, 01:14 AM   #5
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Re: Complex numbers;

Yes, the bar over a complex value represents the conjugate, in this case . So, make the substitutions, then simplify. What do you get?
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December 16th, 2012, 01:21 AM   #6
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Re: Complex numbers;

Are : (1+i)^2 = 1-1= 0?
Or: (1-i)*(1+i)= 1+i-i-i^2= 1+1=2?
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December 16th, 2012, 02:16 AM   #7
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Re: Complex numbers;



You won't need the second product.
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December 16th, 2012, 02:20 AM   #8
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Re: Complex numbers;

When it is (1-i)^2, what should I do?
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December 16th, 2012, 02:24 AM   #9
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Re: Complex numbers;

That won't show up either, but you use the square of a binomial as I did above.
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December 16th, 2012, 02:32 AM   #10
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Re: Complex numbers;

I get this: 2i(1+i-1+i)i+2-2i
= 2i(2i)i+2-2i
= 2i^2 i +2-2i
= 2(-1)i+2-2i
= -2i+2-2i
= 2-4i

Correct??
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