$\displaystyle Since,\ f(z)=\ u+iv\ be\ analytic\ \\[2ex] \displaystyle By\ Cauchy\ rehmans\ equations \\[2ex] \displaystyle u_x=v_y\ and\ u_y=-v_x\ \ \ ……(A) \\[2ex] \displaystyle Differentiating\ u-v\ =\ e^x\ \ \left(cos\ y\ -\ sin\ y\right)\\[2ex]partially\ w.r.t\ x \\[2ex] \displaystyle u_x-v_x=e^x(\cos{y-\sin y} \ldots{} \ldots{} (B)\\[3ex] \displaystyle Differentiating\ u-v\ =\ e^x\ \ \left(\cos\ y\ -\sin\ y\right)\\[2ex]partially\ w.r.t\ y \\[2ex] \displaystyle u_y-v_y=e^x(-\sin{y-\cos y}) \ldots{} \ldots{} (C)\\[3ex] From \ (A) \ and \ (B) \ we \ get,\\[2ex] \displaystyle v_y-v_x=\ e^x(\cos{y-\sin y}) \\[2ex] From \ (A) \ and \ (C) \ we \ get,\\[2ex] \displaystyle {-v}_x-v_y=e^x(-\sin{y-\cos y}) \\[2ex] Adding \ these \ two \ equations,\\[2ex] \displaystyle -2v_x=e^x(-2\sin{y\ }) \\[2ex] $

$\displaystyle v_x=e^x\sin {y\ }=\ -u_y \\[2ex] From \ (A) \ and \ (B) \ we \ get,\\[2ex] \displaystyle u_x+u_y=\ e^x(\cos{y-\sin y}) \\[2ex] From \ (A) \ and \ (C) \ we \ get,\\[2ex] \displaystyle u_y-u_x=e^x(-\sin{y-\cos y}) \\[2ex] Subtracting \ these \ two \ equations,\\[2ex] \displaystyle 2u_x=e^x\left(2\cos {y\ }\right) \\[2ex] \displaystyle u_x=e^x\cos{y\ }=v_y \\[2ex] \displaystyle f\left(z\right)=u+iv \\[2ex] Partially \ differentiating \ it \ w.r.t x \\[2ex] \displaystyle f^{'}\left(z\right)=u_x+iv_x \\[2ex] \displaystyle f^{'}\left(z\right)=e^x\cos{y\ }+i\ e^x\sin{y\ } \\[2ex] Integrating \ w.r.t x \\[2ex] \displaystyle \int{}f^{'}\left(z\right)dx\ =\int{}{(e}^x\cos{y\ }+i\ e^x\sin{y)\ dx\ } \\[2ex] \displaystyle f\left(z\right)=e^x\cos {y+i\ e^x\sin y+f(y)}\ \ldots{} \ldots{} (D)\\[2ex] \displaystyle f^{'}\left(z\right)=u_y+iv_y\ ={-v}_x+iu_x……from\ (A) \\[2ex] \displaystyle f^{'}\left(z\right)=-e^x\sin y+ie^x\cos y\ \\[3ex] Integrating w.r.t. y \\[3ex] \displaystyle \int{}f^{'}\left(z\right)dy=\int{}{(-e}^x\sin y+ie^x\cos y)\ dy\ \\[2ex] $

$\displaystyle f\left(z\right)=e^x\cos{y\ }+ie^x\sin{y\ }+f(x) \ldots{} \ldots{} (E)\\[2ex] From \ (D) \ and \ (E), f(y) = f(x) = 0\\[2ex] \displaystyle f\left(z\right)=e^x\cos{y\ }+ie^x\sin{y\ }\ \\[3ex] Using \ Euler's \ Identity,\\[3ex] \displaystyle e^z=\ e^x\cos{y\ }+ie^x\sin{y\ }\ \\[2ex] \displaystyle f\left(z\right)=e^z\ \\[2ex] $