show the function is an entire function

**Dgphru** · March 24th 2010, 09:39 PM

Hi, can someone please help me..
Question: Write the function $f(z) = sin2(z)$ in the form $f = u + iv$ , with u and v real functions. Hence show that f(z) is an entire function.

here's my working:

$<br /> f(z) = sin2(z), where z = x + iy$
$= sin2(x + iy) <br /> = sin 2x cos 2iy + cos 2x sin iy$

since,
$f(z) = u + iv$
thus $u(x,y) = 0, v (x,y) = sin 2x cos 2iy + cos 2x sin iy$

then i do partial derivaties to satisfy cauchy riemann equations..
partial of u wrt x = 0
partial of v wrt y = cos 2x i sin iy - 2i sin 2x cos 2iy
partial of u wrt y = 0
partial of v wrt x = ..

well i know i did it wrong coz the following are meant to be satisfied, but they just don't work out:
partial of u wrt x = partial of v wrt y
and
partial of u wrt y = - partial of v wrt x

so can someone please help me??

**Prove It** · March 24th 2010, 10:46 PM

Originally Posted by Dgphru

Hi, can someone please help me..
Question: Write the function $f(z) = sin2(z)$ in the form $f = u + iv$ , with u and v real functions. Hence show that f(z) is an entire function.

here's my working:

$<br /> f(z) = sin2(z), where z = x + iy$
$= sin2(x + iy) <br /> = sin 2x cos 2iy + cos 2x sin iy$

since,
$f(z) = u + iv$
thus $u(x,y) = 0, v (x,y) = sin 2x cos 2iy + cos 2x sin iy$

then i do partial derivaties to satisfy cauchy riemann equations..
partial of u wrt x = 0
partial of v wrt y = cos 2x i sin iy - 2i sin 2x cos 2iy
partial of u wrt y = 0
partial of v wrt x = ..

well i know i did it wrong coz
partial of u wrt x = partial of v wrt y
and
partial of u wrt y = - partial of v wrt x

so can someone please help me??

$\sin{2z} = \sin{2(x + iy)}$

$= \sin{(2x + 2iy)}$

$= \sin{2x}\cos{2iy} + \cos{2x}\sin{2iy}$

You should know that $\sin{iX} = i\sinh{X}$ and $\cos{iX} = \cosh{X}$ .

So the equation becomes

$= \sin{2x}\cosh{2y} + i\cos{2x}\sinh{2y}$

$= u + iv$ .

For the function to be an entire function, the Cauchy-Riemann Equations need to be satisfied.

i.e. you need $\frac{\partial u}{\partial x} = \frac{\partial v}{\partial y}$ and $\frac{\partial u}{\partial y} = -\frac{\partial v}{\partial x}$ .

$u = \sin{2x}\cosh{2y}$

$\frac{\partial u}{\partial x} = 2\cos{2x}\cosh{2y}$

$\frac{\partial u}{\partial y} = 2\sin{2x}\sinh{2y}$

$v = \cos{2x}\sinh{2y}$

$\frac{\partial v}{\partial x} = -2\sin{2x}\sinh{2y}$

$= -\frac{\partial u}{\partial y}$ as required.

$\frac{\partial v}{\partial y} = 2\cos{2x}\cosh{2y}$

$= \frac{\partial u}{\partial x}$ as required.

Since the function satisfies the Cauchy-Riemann Equations for all $x, y$ , the function is an entire function.

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