How to find real part of complex number of complex number α ?

If k>0k>0,|z|=|w|=k|z|=|w|=k and alpha=(z-bar w)/(k^2+zbarw)α=z¯¯¯wk2+z¯¯¯w, then find the value of Re(alpha)Re(α).

3 Answers
Jan 7, 2018

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.

Jan 7, 2018

See below.

Explanation:

Given a complex number zz and being bar z¯z it's conjugate then

"Re"(z) = 1/2(z+bar z)Re(z)=12(z+¯z)
"Im"(z)=1/(2i)(z-bar z)Im(z)=12i(z¯z)

then

"Re"(alpha) = 1/2((z-bar w)/(k^2+z bar w)+(bar z-w)/(k^2+bar z w))=0Re(α)=12(z¯¯¯wk2+z¯¯¯w+¯zwk2+¯zw)=0

"Im"(alpha) = 1/(2i)((z-bar w)/(k^2+z bar w)-(bar z-w)/(k^2+bar z w)) = 1/(2i)((2k^2(z-barz+w-barw))/(2k^4+k^2(z bar w+bar z w))) = 1/(2i)((2(z-barz+w-barw))/(2k^2+(z bar w+bar z w))) Im(α)=12i(z¯¯¯wk2+z¯¯¯w¯zwk2+¯zw)=12i(2k2(z¯z+w¯¯¯w)2k4+k2(z¯¯¯w+¯zw))=12i(2(z¯z+w¯¯¯w)2k2+(z¯¯¯w+¯zw))

now assuming that z = k e^(i phi)z=keiϕ and w = k e^(i psi)w=keiψ we have

"Im"(alpha) = 1/(2i)((2k(e^(i phi)-e^(-i phi)+e^(i phi)-e^(-ipsi)))/(2k^2+k^2(e^(i(phi-psi))-e^(-i(phi-psi)))))=Im(α)=12i(2k(eiϕeiϕ+eiϕeiψ)2k2+k2(ei(ϕψ)ei(ϕψ)))=

=1/(ik)((e^(i phi)-e^(-i phi)+e^(i phi)-e^(-ipsi))/(2+(e^(i(phi-psi))+e^(-i(phi-psi))))) = 1/(ik)((2i(sin phi+sin psi))/(2+2cos (phi-psi))) = 1/k((sin phi+sin psi)/(1+cos(phi-psi)))=1ik(eiϕeiϕ+eiϕeiψ2+(ei(ϕψ)+ei(ϕψ)))=1ik(2i(sinϕ+sinψ)2+2cos(ϕψ))=1k(sinϕ+sinψ1+cos(ϕψ))

Jan 7, 2018

Re(alpha)=0Re(α)=0

Explanation:

Suppose z=x+iyz=x+iy and w=u+ivw=u+iv where x,y,u,v in RR

\ |z|=k => x^2+y^2=k^2 ..... [A]
|w|=k => u^2+v^2=k^2

Then, given the definition of alpha, we have:

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