Lecture 20 Complex Analysis

3.2 The logarithmic function:-
The logarithmic function of Z is defined as follow
?=log?Z= log?r +i org(Z) ,Z?0
Where r=|Z| & org(Z)=?+2?k k=0,±1,±2,…..
3.2.1 Characteristics:-
The logarithmic function is multiple valued function because any value k is unique branch of logarithmic function if k=0 ,then the branch
?_0=log?Z= log?r+i? ,r>0,0???2?.
is called principal value of logarithmic function or principal branch.
Any branch of logarithmic function is inverse function of exponential function e^Z ?
Sol:- e^log?Z =e^log??|Z|+i arg(Z)? =e^log??r+i(?+2k?)? ,k=0,±1,±2
=e^log?r e^(i(?+2k?))=r[cos?(?+2k?)+i sin?(?+2k?) ]
=r(cos???+i sin?? ? )=Z
The principal value (branch) of logarithmic function is discontinuous in all points non-negative real exist. So that it isn’t analytic in this points because of the ?=0.

The principal value of logarithm function satisfy Cauchy-Riemann equations:-
SOL:-
log?Z=u+iv=log?r+i??u=log?r & v=?
?u/?r=1/r ?v/??=1/r , ?v/?r=0=(-1)/r ?u/??=(-1)/r (0)=0
If r>0 & 0<?<2? then the principle value of logarithmic function is analytic function, and differentiation of it is
?_0=log?Z=log?r+i? & ?_0=u(r,?)+iv(r,?)
Where u(r,?)=log?r
v(r,?)=?
(??_0)/?r=(d?_0)/dZ??Z/?r , Z=re^i?
But (d?_0)/dZ=d/dZ (log?Z )
(??_0)/?r=d/dZ (log?Z )?e^i? *
(??_0)/?r=?u/?r+i ?v/?r=1/r+i(0)=1/r *
From * & ** we get
d/dZ (log?Z )?e^i?=1/r?d/dZ (log?Z )=e^(-i?) 1/r
Thus d/dZ (log?Z )=1/(re^i? )=1/Z

log??Z?w?=log?Z+log?w
log??Z/w?=log?Z-log?w
log??e^Z ?=Z+2k?i (k=0,±1,±2,….)
e^log?Z =Z
log??Z^(1/k) ?=1/k log?Z (k=±1,±2,…….)

EX (1):-Prove that log??(i)^2 ??2log?i
SOL:-
log??(i)^2 ?=log?(-1)
=log??|-1|+? i(?+2k?) k=0,±1,±2,…
=log?1+i?(1+2k)=?(1+2k)i
2 log?i=2((log?1 )+i(?/2+2k?) )
=2(log??1+i?(1/2+2k)? )=2i?(1/2+2k)=i?(1+4k)
EX:-Prove log??(1+i)^2 ?=2log?(1+i)
SOL:-
log??(1+i)^2 ?=log?(1+2i-1)=log?2i=log??2+i(?/2)?
2 log?(1+i)=2(log???2+i(?/4)? ) ?=tan^(-1)??y/x=tan^(-1)??1/1?=tan^(-1)??1=?/4? ?
=2log???2+2i(?/4)?=log??(?2)^2+i ?/2?
=log??2+i(?/2)?
EX(3):- log??(-1+i)^4 ??4log?(-1+i)
Sol:-
log??(-1+i)^4 ?=log??((-1+i)^2 )^2 ?=log??(1-2i-1)^2 ?=log??-4?
=log??4+i??
4 log?(-1+i)=4(log???2+i 3?/4? )=log??4+3?i ?

QUZE:-log??(-1)=(2k+1)?i?
QUZE:-log??(1-i)^2?log?(1-i) ?