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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1$abs(1-i)=\sqrt{1^2+(-1)^2}=\sqrt{2}\neq 1+i$. – Git Gud Jan 19 '13 at 10:18
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1Do 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
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$|a+bi|=\sqrt{a^2+b^2}=|a-bi|$
$|z|=\sqrt{z\cdot\overline{z}}=\sqrt{\overline{z}\cdot z}=|\overline{z}|$
Martin Sleziak
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Well, that's the definition I know of and Wikipedia gives the same definition. – Martin Sleziak Jan 19 '13 at 10:34
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@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
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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?
Adi Dani
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