How to find real part of complex number of complex number α ?
If #k>0# ,#|z|=|w|=k# and #alpha=(z-bar w)/(k^2+zbarw)# , then find the value of #Re(alpha)# .
If
3 Answers
The other two people did a much better job of answering this question than I. Any attempt to fix this answer would result in plagiarism of their answers. Therefore, I recommend that you read their answers.
See below.
Explanation:
Given a complex number
then
now assuming that
Explanation:
Suppose
# \ |z|=k => x^2+y^2=k^2 # ..... [A]
# |w|=k => u^2+v^2=k^2 #
Then, given the definition of
# alpha = (z - bar w)/(k^2+zbarw)#
# \ \ \ = ((x+iy) - (u-iv))/(k^2+(x+iy)(u-iv))#
# \ \ \ = (x+iy - u+iv) / (k^2+(xu-ivx+iuy-i^2vy))#
# \ \ \ = (x-u+(v+y)i) / ( (k^2+xu+vy)+(uy-vx)i )#
Now we multiply numerator and denominator by the complex conjugate of the denominator to rationalize it:
# alpha = (x-u+(v+y)i) / ( (k^2+xu+vy)+(uy-vx)i ) * ( (k^2+xu+vy)-(uy-vx)i ) / ( (k^2+xu+vy)-(uy-vx)i ) #
And if we expand we get:
# alpha = (-k^2u + i k^2 v + k^2 x + i k^2 y - u^2 x + i u^2 y + u x^2 + u y^2 - v^2 x + i v^2 y + i v x^2 + i v y^2) / (k^4 + 2 k^2 u x + 2 k^2 v y + u^2 x^2 + u^2 y^2 + v^2 x^2 + v^2 y^2) #
# \ \ \ = (-k^2u + i k^2 v + k^2 x + i k^2 y - u^2 x + i u^2 y + u x^2 + u y^2 - v^2 x + i v^2 y + i v x^2 + i v y^2) / (k^4 + 2 k^2 u x + 2 k^2 v y + u^2 x^2 + u^2 y^2 + v^2 x^2 + v^2 y^2) #
So if we separate out the real component of this expression we have:
# Re(alpha) = (-k^2u + k^2 x - u^2 x + u x^2 + u y^2 - v^2 x) / (k^4 + 2 k^2 u x + 2 k^2 v y + u^2 x^2 + u^2 y^2 + v^2 x^2 + v^2 y^2) #
Now we use the earlier result [A]:
# Re(alpha) = (-k^2u + k^2 x - x(u^2+v^2) + u( x^2 + u y^2) ) / (k^4 + 2 k^2 u x + 2 k^2 v y + u^2( x^2 + y^2) + v^2 (x^2 + y^2)) #
# \ \ \ \ \ \ \ \ \ \ = (-k^2u + k^2 x - k^2 x + k^2u ) / (k^4 + 2 k^2 u x + 2 k^2 v y + k^2u^2 + k^2v^2 ) #
# \ \ \ \ \ \ \ \ \ \ = (0) / (k^4 + 2 k^2 u x + 2 k^2 v y + k^2u^2 + k^2v^2 ) #
# \ \ \ \ \ \ \ \ \ \ = 0 #