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How to show that absolute value of a complex number is invariant under complex conjugation?

my solution: Counterexample. $f(abs(1-i))= \overline{abs(1-i)}=\overline{1+i}=1-i \neq abs(1-i)$? is there something I am missing?

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$abs(1-i)=\sqrt{1^2+(-1)^2}=\sqrt{2}\neq 1+i$. –  Git Gud Jan 19 '13 at 10:18
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Do you know the definition of absolute value of a complex number? Because then the invariance under conjugation follows almost immediately. –  Nils Matthes Jan 19 '13 at 10:20

2 Answers 2

up vote 3 down vote accepted

$|a+bi|=\sqrt{a^2+b^2}=|a-bi|$

$|z|=\sqrt{z\cdot\overline{z}}=\sqrt{\overline{z}\cdot z}=|\overline{z}|$

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So is absolute value of a complex number same as modulus? –  laovultai Jan 19 '13 at 10:32
    
Well, that's the definition I know of and Wikipedia gives the same definition. –  Martin Sleziak Jan 19 '13 at 10:34
    
@alvoutila: The absolute value, modulus or length of a complex number is different words for the same operation. –  Stefan Hansen Jan 19 '13 at 10:35

You write $$f(abs(1-i))= \overline{abs(1-i)}=\overline{1+i}=1-i \neq abs(1-i)$$ your notation is strange you should write $$f(|1-i|)= \overline{|1-i|}=\overline{|1+i|}\neq1-i \neq |1-i|$$ you must know that $|a+bi|$ or absolute value of complex number is real number you miss that?

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