cauchy-Riemann Equations in polar coordinates?-

cauchy-Riemann Equations in polar coordinates?-
suppose f(z)=f(re^i? )=u(r,?)+iv(r,?)
where z=re^i?
suppose f is differentiable at z=z(r,?)we have.
f(re^i? )=u(r,?)+iv(r,?)
then differentiating w.r.to r we get
e^i? f ?(re^i? )=u_r+iv_r…(1)
differentiating it w.r.to ? we have
e^i?.ir.f ?(re^i? )=u_?+iv_?…(2)
multiplying (1)by ir we get
?ir.e?^i?.f ?(re^i? )=ir u_r-r v_r
by (2)we get
ir u_r-r v_r=u_?+iv_?
equaling the real and imaginary parts one can have
ru_r=v_?
-rv_r=u_?
this means
u_r=1/r v_?
v_r=(-1)/r u_? (*)
which are called the cauchy-Riemann Equations in polar coordinates
,here we have
f ?(z)=e^(-i?) [ u_r+iv_r]
from (*) we get
u_rr=(-1)/r^2 v_?+1/r v_(r? )…(**)
differentiating
v_r=(-1)/r u_(? ) w.r.to ? we get
v_?r=(-1)/r u_??
but then
u_rr=(-1)/r^2 v_?+1/r[(-1)/r u_??]
=(-1)/r.1/r v_?-1/r^2 u_??
so,? u?_rr=(-1)/r u_r-1/r^2 u_??
?? u?_rr+1/r u_r+1/r^2 u_??=0
so u satisfies the polar form of laplacian equation similarly we can have.
(?^2 v)/(?r^2 )+1/r ?v/?r+i/r^2 (?^2 v)/(??^2 )=0
so of f ?(z) exist at z?(0,0),
then the complement functions u and v are laplacian .
example:- consider f(z)=1/z
suppose z=re^i?=r(cos???+isin ?? )
then f(z)=1/z=1/(re^i? )=1/r(cos???+isin ?? )
=1/r [cos???-isin ?? ]
=cos??/r+i ((-sin??)/r)
=u(r,?)+iv(r,?)
where
u(r,?)=cos??/r
v(r,?)=(-sin??)/r
u_r=(-1)/r^2 cos?? , u_?=(-sin??)/r
v_r=1/r^2 sin?? , v_?=?-cos???/r
then
c.r.equation are satisfied
u_r=1/r v_?
v_r=(-1)/r u_?
f ?(z)=e^(-i?) [u_r+?iv?_r ]=e^(-i?) [(-1)/r^2 cos???+i sin??/r^2 ? ]=(-1)/r^2 e^(-i?) [cos???-isin ??]
=(-1)/r^2 e^(-i?).e^(-i?)=(-1)/r^2 e^(-2i?)=(-1)/(r^2 e^2i? )=(-1)/z^2