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π (π) (π§) = π! /2ππ β«ππ· π(π€)ππ€ / (π€βπ§)π+1 . using the generalized Cauchy integral formula (GCIF) expressed in
Prove it with the Mathematical Induction Method.(As is known, the formula for = 1 is Cauchy Integral
It becomes the Cauchy Integral Formula (CIF), which is the result of the theorem and it is correct.)
Show that
a)z+z*=2 Re z=2x
b)z-z*=2i Im z=2iy
c)z/z*={x^2-y^2/x^2+y^2}+i{2xy/x^2+y^2}
Β π΄ = { π§ β β:|π§| < 1 π£π |π§ β 1/ 2 | > 1 /2 } βͺ { 1/ 2 } denote the set in the complex plane and
π΄ β² and ππ΄ = π΄Μ \ π΄π
Write the set.
If f is analytic in a convex domain D such that Re f'(z) is not equal to 0 for all z belongs to D, then f is univalent in D
Let f(z) = sin z/z
and f(0) = 0. Explain why f is analytic at z = 0. Find the MaclaurianΒ
series for f(z) and g(z) = β« f(ΞΎ)dΞΎ from 0 to z
. Does there exist a function f with anΒ
isolated singularity at 0 and such that |f(z)|~ exp( 1/|z|) near z= 0?Β
2
, z + z = 2Re(z) and Re(z) β€ |z|. Hence
show that
i. |z1 + z2|
2 = |z1|
2 + |z2|
2 + 2Re(z1z2),
ii. |z1 + z2| β€ |z1| + |z2|,
where Re(z) is the real part of z and z the conjugate of z.
2
, z + z = 2Re(z) and Re(z) β€ |z|. Hence
show that
i. |z1 + z2|
2 = |z1|
2 + |z2|
2 + 2Re(z1z2),
ii. |z1 + z2| β€ |z1| + |z2|,
where Re(z) is the real part of z and z the conjugate of z. [26 marks]
(b) If z1 = 1 + 2i, find the set of values of z2 for which
(i) |z1 + z2| = |z1| + |z2| (ii) |z1 + z2| = |z1| β |z2|.
#339914 on Dec 2023