If $u - v = \dfrac{\cos x + \sin x - e^{-y}} {2 \cos x - \cosh y} $ and $f\left( \dfrac{\pi}{2} \right) = 0$, determine the analytic function $f(z) = u + iv$.
I don't know how to solve this. I don't know where to start?
If $u - v = \frac{\cos x + \sin x - e^{-y}} {2 \cos x - \cosh y} $ and $f( \pi /2) = 0$ determine the function $f(z)$.
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complex-analysis
analytic-functions
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0Is there any typo? $$\cot \frac{z}{2}=\frac{\sin x}{\cosh y-\cos x}+\frac{i\sinh x}{\cos x-\cosh y}$$ – 2017-02-17
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1If $f(z)=u+iv$ is analytic, then $\Delta u=u_{xx}+u_{yy}=0, \Delta v=v_{xx}+v_{yy}=0$ and hence $\Delta(u-v)=0$. However your $u-v$ does not satisfies $\Delta(u-v)=0$. – 2017-02-17
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Note if $f(z)=u+iv$ is analytic, then $u,v$ satisfy $$ u_x=v_y, u_y=-v_x. $$ Let $u-v=g(x,y)$. Note $$ f'(z)=u_x+iv_x\tag{1} $$ or $$ f'(z)=v_y-iu_y. \tag{2}$$ Then by (1) $$ f'(z)=(u-v)_x+(1+i)v_x=g_x+(1+i)v_x \tag{3}$$ and by (2) $$ f'(z)=(1-i)v_y-i(u-v)_y=(1-i)v_y-ig_y. \tag{4}$$ Adding (3) to (4) gives $$ 2f'(z)=(g_x-ig_y)+(1-i)(v_y+iv_x).$$ Thus we have $$ 2f'(z)=(g_x-ig_y)+(1-i)f'(z)$$ from which we have $$ f'(z)=\frac{1-i}{2}(g_x-ig_y). $$ I will come back.
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0The main problem is integration, Just _The detail is omitted_. – 2017-02-16