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Dvinal [7]
3 years ago
9

The quotient of w decreased by 5 and 6 write it as an expression

Mathematics
1 answer:
mr_godi [17]3 years ago
5 0

Answer:

W/5-6

Step-by-step explanation:

W/5-6

W/1

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How to prove tan z is analytic using cauchy-riemann conditions
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A function f(z)=f(x+iy)=u(x,y)+i v(x,y) is analytic if the C-R conditions are satisfied:

\begin{cases}u_x=v_y\\u_y=-v_x\end{cases}

We have

f(z)=\tan z=\tan(x+iy)

Recall the angle sum formula for tangent:

\tan(x+iy)=\dfrac{\tan x+\tan iy}{1-\tan x\tan iy}

Now recall that

\tan iy=\dfrac{\sin iy}{\cos iy}=\dfrac{-\frac{e^y-e^{-y}}{2i}}{\frac{e^y+e^{-y}}2}=\dfrac{i\sinh y}{\cosh y}=i\tanh y

So we have

\dfrac{\tan x+\tan iy}{1-\tan x\tan iy}=\dfrac{\tan x+i\tanh y}{1-i\tan x\tanh y}
=\dfrac{(\tan x+i\tanh y)(1+i\tan x\tanh y)}{(1-i\tan x\tanh y)(1+i\tan x\tanh y)}
=\dfrac{\tan x+i\tanh y+i\tan^2x-\tan x\tanh^2y}{1+\tan^2x\tanh^2y}
=\dfrac{\tan x(1-\tanh^2y)}{1+\tan^2x\tanh^2y}+i\dfrac{\tanh y(1+\tan^2x)}{1+\tan^2x\tanh^2y}
=\underbrace{\dfrac{\tan x\sech^2y}{1+\tan^2x\tanh^2y}}_{u(x,y)}+i\underbrace{\dfrac{\tanh y\sec^2x}{1+\tan^2x\tanh^2y}}_{v(x,y)}

We could stop here, but taking derivatives may be messy and would be easier to do if we can write this in terms of sines and cosines.

u(x,y)=\dfrac{\tan x\sech^2y}{1+\tan^2x\tanh^2y}=\dfrac{\frac{\sin x}{\cos x\cosh^2x}}{1+\frac{\sin^2x\sinh^2y}{\cos^2x\cosh^2y}}=\dfrac{\sin x\cos x}{\cos^2x\cosh^2y+\sin^2x\sinh^2y}
u(x,y)=\dfrac{\sin2x}{2\cos^2x\cosh^2y+2(1-\cos^2x)(\cosh^2y-1)}=\dfrac{\sin2x}{2\cosh^2y-1+2\cos^2x-1}
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v(x,y)=\dfrac{\sinh2y}{\cos2x+\cosh2y}

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u_x=\dfrac{2\cos2x(\cos2x+\cosh2y)-(-2\sin2x)\sin2x}{(\cos2x+\cosh2y)^2}=\dfrac{2+2\cos2x\cosh2y}{(\cos2x+\cosh2y)^2}
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Similarly, you can check that u_y=-v_x, hence the C-R conditions are satisfied, and so \tan z is analytic.
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