A community for students.
Here's the question you clicked on:
 0 viewing
idku
 one year ago
Maybe it is too early for me to ask this question, but why .... ?
idku
 one year ago
Maybe it is too early for me to ask this question, but why .... ?

This Question is Closed

idku
 one year ago
Best ResponseYou've already chosen the best response.0\(\large\color{black}{ \displaystyle \left1+i\right=\sqrt{2} }\)

idku
 one year ago
Best ResponseYou've already chosen the best response.0(and so \(\large\color{black}{ \displaystyle \left1i\right=\sqrt{2} }\) )

idku
 one year ago
Best ResponseYou've already chosen the best response.0can someone explain this?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0both \(1+i\) and\(1i\) are \(\sqrt2\) units away from \(0\)

idku
 one year ago
Best ResponseYou've already chosen the best response.0Is that because i is a rotation?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1442713983303:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0just like both \((1,1)\) and \((1,1)\) have distance \(\sqrt2\) from \((0,0)\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0or if you prefer, just like the diagonal of any square with side 1 has length \(\sqrt2\)

idku
 one year ago
Best ResponseYou've already chosen the best response.0Wait, so i is a diagonal?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1442714099587:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0ok 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
 one year ago
Best ResponseYou've already chosen the best response.0and that would mean that \(\large\color{black}{ \displaystyle \lefta+i\right=a\sqrt{2} }\) \(\forall a\in\mathbb{R}\)

idku
 one year ago
Best ResponseYou've already chosen the best response.0oh, that is not right I guess. (I just heard someone said that i is sort of a rotation)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[a+i=\sqrt{a^2+1}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0what is true is that \[a+bi=abi=\sqrt{a^2+b^2}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0it has no bearing on the sign of \(a\) or \(b\) since you are going to square them in any case

idku
 one year ago
Best ResponseYou've already chosen the best response.0oh, ok.... I guess I will explore that a little more. For now I will just have the rules at least;)

idku
 one year ago
Best ResponseYou've already chosen the best response.0yes as long as a, b are Real...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0in \(a+bi\) \(a\) and \(b\) are always real
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.