Question
Prove that the linear fractional transformation maps the circle
into itself. Also prove that f(z) is conformal in
.
Answer :
Word Count : 141
Let (\phi(z)=\dfrac{2z-1}{2-z}). For (z) on the circle (|z|=1), consider [ |\phi(z)|^2=\frac{|2z-1|^2}{|2-z|^2}. ] Now [ |2z-1|^2=(2z-1)(2\bar z-1)=4z\bar z-2z-2\bar z+1, ] and [ |2-z|^2=(2-z)(2-\bar z)=4-2z-2\bar ___ _________ _________ ______ _______ _____ __________ _____ __________ ______.
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Let (\phi(z)=\dfrac{2z-1}{2-z}). For (z) on the circle (|z|=1), consider [ |\phi(z)|^2=\frac{|2z-1|^2}{|2-z|^2}. ] Now [ |2z-1|^2=(2z-1)(2\bar z-1)=4z\bar z-2z-2\bar z+1, ] and [ |2-z|^2=(2-z)(2-\bar z)=4-2z-2\bar ___ _________ _________ ______ _______ _____ __________ _____ __________ ______.
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Related Questions
Evaluate where c is the circle
.
a) If is entire such that
in
then show that f has the form
where
are constants with
.
Expand in a Laurent series valid for
i) 0 < |z - 1| < 2 and ii) 0 < |z - 3| < 2.
Find all the singularities of the function
Find the zeros and singularities of the function in
. Also find the residue at the poles.
Evaluate , where c is the eight like figure shown in Fig. 1.
b) Consider and the closed circular region
. Find points in R where |f(z)| has its maximum and minimum values.
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