Complex Analysis- Lecture 19 - The exponential function

3.1 The exponential function:-
The exponential function defined as follow

f(Z)=e^Z=e^(x+iy)=e^x e^iy=e^x (cos?y+i sin?y )
??(f(Z)=e^x (cos?y+i sin?y ) )
3.1.1 Characteristics:-
The absolute of exponential function is e^x ?
Proof:-
|e^Z |=|e^(x+iy) |=|e^x e^iy |=|e^x ||e^iy |=e^x |cos?y+i sin?y |
=e^x (cos^2?y+sin^2?y )^(1/2)=e^x (1)^(1/2)=e^x??(|e^Z |=e^x )
The argument of exponential function is
arg(e^Z )=y+2?k k=0,±1,±2,…..
Polar form of exponential function is
?(f(Z)=P(cos??+i sin?? ) ) ,where P=e^x=|e^Z | and ?=y.
e^Z?0 Every where?
Proof:-
By polar form
e^Z=P(cos??+i sin?? )=Pe^i?
Then P=|e^Z |=e^x>0 & ? e?^i?=cos???+i sin?? ? ,0
So that P>0 & ? e?^i?>0
Thus e^Z?0

e^0=1
e^(Z_1 ) e^( Z_2 )=e^(Z_1+Z_2 )
e^(Z_1 )/e^( Z_2 ) =e^(Z_1-Z_2 )
((e^Z ) ) ?=e^Z ?
e^Z=e^(Z+2?i)
EX(1):- Find all value of Z such that e^Z=1+i?3 ?
Sol:-
e^x=|e^Z |=|1+i?3|=?(1+3)=?4=2
?e^x=2?x=log2
tan?y=?3/1?y=tan^(-1)??(?3/1)?=?/3+2k?,k=0,±1,±2,…
Thus Z=x+iy=log2+i(?/3+2k?),k=0,±1,±2,…
EX(2):- Prove that |e^(2Z+i)+e^(?iZ?^2 ) |?e^2x+e^(-2xy)?
Sol:- |e^(2Z+i)+e^(?iZ?^2 ) |^2=|e^(2x+2yi+i)+e^(i[(x^2-y^2 )+2xyi) |^2
=|e^(2x+i(2y+1))+e^((x^2-y^2 )i-2xy) |^2
=|e^2x [cos?(2y+1)+i sin?(2y+1) ]+e^(-2xy) [cos?(x^2-y^2 )+i sin?(x^2-y^2 ) ] |^2
= |e^2x cos?(2y+1)+ie^2x sin??(2y+1)+e^(-2xy) cos?(x^2-y^2 ) ?+ie^(-2xy) sin?(x^2-y^2 ) |^2

=|e^2x cos?(2y+1)+e^(-2xy) cos?(x^2-y^2 )+i[e^2x sin??(2y+1)+e^(-2xy) sin?(x^2-y^2 ) ? ] |^2
= (e^2x cos?(2y+1)+e^(-2xy) cos?(x^2-y^2 ) )^2+(e^2x sin??(2y+1)+e^(-2xy) sin?(x^2-y^2 ) ? )^2
=e^4x cos^2?(2y+1)+2e^x e^(-2xy) cos?(2y+1) cos??(x^2-y^2 )+e^(-4xy) cos^2??(x^2-y^2 )+? ? e^4x sin^2??(2y+1)+2? e^x e^(-2xy) sin??(2y+1) sin??(x^2-y^2 )+? ? e^(-4xy) sin^2 (x^2-y^2 )
?=e?^4x [cos^2 (2y+1)+sin^2 (y+1) ]+2e^x e^(-2xy) [cos??(2y+1) cos??(x^2-y^2 )+? sin?(2y+1)??sin??(x^2-y^2 )]? ? ?+e^(-4xy) [cos^2 (x^2-y^2 )+sin^2 (x^2-y^2 )
=e^4x+e^(-4xy)+2e^(2x(1-y)) cos??[(2y+1)-(x^2-y^2 )]?
=e^4x+e^(-4xy)+2e^(2x(1-y)) cos?(y^2+2y+1-x^2 )
?e^4x+e^(-4xy)+2e^(2x(1-y))
=(e^2x )^2+2e^2x e^(-2xy)+(e^(-2xy) )^2=(e^2x+e^(-2xy) )^2
|e^(2Z+i)+e^(?iZ?^2 ) |^2?(e^2x+e^(-2xy) )^2 by root two sides, we get
|e^(2Z+i)+e^(?iZ?^2 ) |?e^2x+e^(-2xy)