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anonymous
 one year ago
Question with Complex Numbers
anonymous
 one year ago
Question with Complex Numbers

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0If \(z = x + iy\) where \(y ≠ 0\) and \(1+ z^2 ≠ 0\), show that the number \(w=\dfrac{z}{1+z^2}\)is real only if \(z = 1\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0would it help if i prove the converse?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0What is the converse, @divu.mkr?

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0it's the reversed version of a statement

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0that if z = 1 then 'w' is real?

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0for example. if A then B is the original statement the converse would be If B then A and I'm going in and out of the question, my openstudy is real bad

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Sure, if that works @divu.mkr

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I dont know how to use latex... so try to understand just use this (z)(z bar) = z^2 so z=1/zbar and if a complex number is real then w= w bar take bar on both the sides replace 'z bar' by z they will become same

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0So, \(z\times \overline{z}=z^2\) Therefore, \(z=\dfrac{1}{\overline{z}}\) I don't get what you mean if a complex number is real, then w = w bar

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0when we take a bar over a complex number the sign of the complex quantity reverses so if the complex number is purely real then there will be no change in the complex number as the complex quantity is absent

UsukiDoll
 one year ago
Best ResponseYou've already chosen the best response.0latex is \bar{insertsomethinghere} for the bar over the area

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Ahh, I see. So, \(w\) is a complex number (cause you are dividing a complex by a complex), but since we have established its real, its conjugate should be the same..?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Cause there is no imaginary component?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Great :) So, we turn: \[\huge{w=\frac{z}{1+z^2}}\] To \[\huge{\overline{w}=\frac{\overline{z}}{\overline{1+z^2}}}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0yep now just one of them

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\huge{w}=\frac{\frac{1}{z}}{1+(\frac{1}{z})^2}\]?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0but it will be w bar

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Didn'w we say w = w bar tho?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0you substituted in wrong eqn ...just saying that
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