idku
  • idku
Maybe it is too early for me to ask this question, but why .... ?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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idku
  • idku
\(\large\color{black}{ \displaystyle \left|1+i\right|=\sqrt{2} }\)
idku
  • idku
(and so \(\large\color{black}{ \displaystyle \left|1-i\right|=\sqrt{2} }\) )
idku
  • idku
can someone explain this?

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More answers

anonymous
  • anonymous
both \(1+i\) and\(1-i\) are \(\sqrt2\) units away from \(0\)
idku
  • idku
Is that because i is a rotation?
idku
  • idku
A 45 degree rotation?
anonymous
  • anonymous
|dw:1442713983303:dw|
anonymous
  • anonymous
just like both \((1,1)\) and \((1,-1)\) have distance \(\sqrt2\) from \((0,0)\)
anonymous
  • anonymous
or if you prefer, just like the diagonal of any square with side 1 has length \(\sqrt2\)
idku
  • idku
Wait, so i is a diagonal?
anonymous
  • anonymous
|dw:1442714099587:dw|
idku
  • idku
|dw:1442714123626:dw|
anonymous
  • anonymous
ok i didn't mean to confuse you \(i\) is not the diagonal the absolute value of a complex number is its distance from 0
idku
  • idku
and that would mean that \(\large\color{black}{ \displaystyle \left|a+i\right|=a\sqrt{2} }\) \(\forall a\in\mathbb{R}\)
idku
  • idku
wait, sorry for a>0 too
anonymous
  • anonymous
oh no
idku
  • idku
oh, that is not right I guess. (I just heard someone said that i is sort of a rotation)
anonymous
  • anonymous
\[|a+i|=\sqrt{a^2+1}\]
anonymous
  • anonymous
what is true is that \[|a+bi|=|a-bi|=\sqrt{a^2+b^2}\]
anonymous
  • anonymous
it has no bearing on the sign of \(a\) or \(b\) since you are going to square them in any case
idku
  • idku
oh, ok.... I guess I will explore that a little more. For now I will just have the rules at least;)
idku
  • idku
yes as long as a, b are Real...
anonymous
  • anonymous
in \(a+bi\) \(a\) and \(b\) are always real
idku
  • idku
ok...

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