01. Show that \(w=z+e^z\) is analytic and find \(\frac{dw}{dz}\).
Ans: $$ w=f(z)= z+e^z\Longrightarrow u+iv=(x+iy)+e^{(x+iy)} =(x+iy)+e^x\cdot e^{iy}=(x+iy)+e^x (\cos y+i\sin y)$$ $$\Longrightarrow u=x+e^x \cos y \,\,\,\,\,;\,\,\,\,\, v=y+e^x \sin y$$ $$\Longrightarrow u_x=1+e^x \cos y \,\,\,\,\,;\,\,\,\,\, v_x=e^x \sin y$$ $$\Longrightarrow u_y=-e^x \sin y \,\,\,\,\,;\,\,\,\,\, v_y=1+e^x \cos y$$ From the above set of equations we have \(u_x=v_y; v_x=-u_y\), hence \(u,v\) satisfies the CR-equations, Hence \(f\) is analytic
To find \(\frac{dw}{dz}\): Differentiate \(w=f(z)=u+iv\) with respect to \(x\) (derivative horizontally)
$$\Longrightarrow \frac{dw}{dz}=f'(z)=u_x+iv_x=1+e^x \cos y+i e^x \sin y=1+e^x (\cos y+i \sin y)=1+e^x e^{iy}=1+e^{x+iy}=1+e^z$$
02. Show that \(w=\sin z\) is analytic and hence find \(\frac{dw}{dz}\).
Ans: $$ w=f(z)= \sin z\Longrightarrow u+iv=\sin(x+iy)=\sin x\, \cos iy +\cos x\, \sin iy$$ (we know that \( \sin ix=i\sin hx, \cos ix=cos hx\)) $$\Longrightarrow u+iv=\sin x\, \cosh y +i\cos x\, \sin hy$$ $$\Longrightarrow u=\sin x\,\cosh y \,\,\,\,\,;\,\,\,\,\, v=\cos x\,\sinh y$$ $$\Longrightarrow u_x=\cos x\,\cosh y \,\,\,\,\,;\,\,\,\,\, v_x=-\sin x\,\sinh y$$ $$\Longrightarrow u_y=\sin x\,\sinh y \,\,\,\,\,;\,\,\,\,\, v_y=\cos x\,\cosh y$$ From the above set of equations we have \(u_x=v_y; v_x=-u_y\), hence \(u,v\) satisfies the CR-equations, Hence \(f\) is analytic
03. Construct the analytic function \(f(z)=u+iv\) given that \(u=\log{\sqrt{(x^2+y^2)}}\)
Ans:
Let \(f(z)=u+iv\), since \(f\) is analytic, \(u,v\) satisfies CR-equations and we have \(u\) only, hence we have $$f'(z)=u_x+i v_x=u_x-iu_y$$ Now \(u=\log{\sqrt{(x^2+y^2)}}=\log{(x^2+y^2)}^{1/2}=\frac{1}{2}\log{(x^2+y^2)}\) $$\Longrightarrow u_x=\frac{1}{2}\frac{2x}{x^2+y^2}\,\,\;\,\,\,\,\,\,u_y=\frac{1}{2}\frac{2y}{x^2+y^2}$$ $$\Longrightarrow f'(z)=\frac{1}{2}\frac{2x}{x^2+y^2}-i\frac{1}{2}\frac{2y}{x^2+y^2}$$ Now substitute \(x=z\) and \(y=0\) in the above equation we get $$f'(z)=1/z\Longrightarrow f(z)=\log{(z)}$$ Verified using Maple
04. Construct the analytic function \(f(z)=u+iv\) given that \(u=e^{2x}(x\cos{2y}-y\sin{2y})\)
Ans:
Let \(f(z)=u+iv\), since \(f\) is analytic, \(u,v\) satisfies CR-equations and we have \(u\) only, hence we have $$f'(z)=u_x+i v_x=u_x-iu_y$$ $$\Longrightarrow u_x=2e^{2x}(x\cos{2y}-y\sin{2y})+e^{2x}(\cos{2y})\,\,\;\,\,\,\,\,\,u_y=e^{2x}(-2x\sin{2y}-\sin{2y}-2y\cos{2y})$$ $$\Longrightarrow f'(z)=(2e^{2x}(x\cos{2y}-y\sin{2y})+e^{2x}(\cos{2y}))+i(e^{2x}(-2x\sin{2y}-\sin{2y}-2y\cos{2y}))$$ Now substitute \(x=z\) and \(y=0\) in the above equation we get $$f'(z)=2e^{2z}z+e^{2z}+i(0)=e^{2z}(1+2z)\Longrightarrow f(z)=(1+2z)\frac{e^{2z}}{2}-\frac{e^{2z}}{2}+c=z e^{2z}+c$$
05. Construct the analytic function \(f(z)=u+iv\) given that \(v=e^{x}(x\sin{y}+y\cos{y})\)
Ans:
Let \(f(z)=u+iv\), since \(f\) is analytic, \(u,v\) satisfies CR-equations and we have \(v\) only, hence we have $$f'(z)=u_x+i v_x=v_y+iv_x$$ $$\Longrightarrow v_x=e^x(x\sin{y}+y\cos{y})+e^x(\sin(y))\,\,\;\,\,\,\,\,\,v_y=e^x(x\cos{y}+\cos{y}-y\sin{y})$$ $$\Longrightarrow f'(z)=(e^x(x\cos{y}+\cos{y}-y\sin{y}))+i(e^x(x\sin{y}+y\cos{y})+e^x(\sin(y)))$$ Now substitute \(x=z\) and \(y=0\) in the above equation we get $$f'(z)=e^z(z+1)\Longrightarrow f(z)=(z+1)e^{z}-e^{z}+c=z e^{z}+c$$
06. Construct the analytic function \(f(z)=u+iv\) given that \(u=\frac{x^4-y^4-2x}{x^2+y^2}\,\,\,\)-(OQP,MQP)
Ans:
Let \(f(z)=u+iv\), since \(f\) is analytic, \(u,v\) satisfies CR-equations and we have \(u\) only, hence we have $$f'(z)=u_x+i v_x=u_x-iu_y$$ $$\Longrightarrow u_x=\frac{(x^2+y^2)(4x^3-2)-(x^4-y^4-2x)2x}{(x^2+y^2)^2}\,\,\;\,\,\,\,\,\,u_y=\frac{(x^2+y^2)(-4y^3)-(x^4-y^4-2x)2y}{(x^2+y^2)^2}$$ $$\Longrightarrow f'(z)=\left(\frac{(x^2+y^2)(4x^3-2)-(x^4-y^4-2x)2x}{(x^2+y^2)^2}\right)-i\left(\frac{(x^2+y^2)(-4y^3)-(x^4-y^4-2x)2y}{(x^2+y^2)^2}\right)$$ Now substitute \(x=z\) and \(y=0\) in the above equation we get $$f'(z)=\frac{2z^5+2z^2}{z^4}-i(0)=2z+\frac{1}{z^2}\Longrightarrow f(z)=z^2-\frac{2}{z}+c$$
07. Construct the analytic function \(f(z)=u+iv\) given that \(v=\left(r-\frac{k^2}{r}\right)\sin \theta\,\,\,\)-(MQP)
Ans:
Let \(f(z)=u+iv\), since \(f\) is analytic, \(u,v\) satisfies CR-equations and we have \(u\) only, hence we have $$f'(z)=f'\left(re^{i\theta}\right)e^{-i\theta}=(u_r+iv_r)\Longrightarrow f'(z)=e^{-i\theta} (u_r+iv_r)=e^{-i\theta} \left(\frac{v_\theta}{r}+iv_r\right)$$ $$\Longrightarrow v_r=\left(1+\frac{k^2}{r^2}\right)\sin\theta\,\,\;\,\,\,\,\,\,v_\theta=\left(r-\frac{k^2}{r}\right)\cos\theta$$ $$\Longrightarrow f'(z)=e^{-i\theta} \left[\frac{1}{r}\left(r-\frac{k^2}{r}\right)\cos\theta+i\left(1+\frac{k^2}{r^2}\right)\sin\theta\right]$$ Now substitute \(r=z\) and \(\theta=0\) in the above equation we get $$f'(z)=1-\frac{k^2}{z}\Longrightarrow f(z)=z+\frac{k^2}{z}+c$$
08. Construct the analytic function \(f(z)=u+iv\) given that \(u-v=e^x (\cos y-\sin y)\)
Ans:
$$u-v=e^x (\cos y-\sin y),$$ Differentiate with respect to \(x\) $$\Longrightarrow u_x-v_x=e^x (\cos y-\sin y)\,\,\,\,\,\,\,-(1)$$ Differentiate with respect to \(y\) $$\Longrightarrow u_y-v_y=e^x (-\sin y-\cos y)$$ Using CR-equations, we can write the same as $$\Longrightarrow -v_x-u_x=e^x (-\sin y-\cos y) \,\,\,\,\,\,\,-(2)$$ $$(1)+(2) \Longrightarrow -2v_x=-2e^x\sin y,\,\,\,\,\Longrightarrow v_x=e^x\sin y$$ $$(1)-(2) \Longrightarrow 2u_x=2e^x\cos y,\,\,\,\,\Longrightarrow u_x=e^x\cos y$$
Do it yourself
01. Construct the analytic function \(f(z)=u+iv\) given that \(u-v=(x-y)(x^2+4xy+y^2)\).
02. Construct the analytic function \(f(z)=u+iv\) given that \(u+v=\frac{1}{r^2}(\cos 2\theta -\sin 2\theta)\).
03. Construct the analytic function \(f(z)=u+iv\) given that sum of its real and imaginary parts is \(x^3-y^3+3xy(x-y)\).
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