if we knew that :cos and sin are bounded function $\mathbb{R}$ for any real
number $x$ .
let $z $ be a complex variable , Is there a proof show that :
$\cos(z)$ and $\sin(z)$ are images of unbounded functions ?
Any kind of help is appreciated.
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5non constant + entire implies unbounded (Liouville) – b00n heT Jun 02 '15 at 09:54
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ok, how to show that cosz is enitre ? – zeraoulia rafik Jun 02 '15 at 09:57
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2What do you mean by "image of" ? – Jun 02 '15 at 09:58
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for example if am true cos x is image not function – zeraoulia rafik Jun 02 '15 at 09:59
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2$\cos(it)-i\sin(it)=e^t$, is there a need to say more ? – Jun 02 '15 at 09:59
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so coz is image of cos for any complex variable z – zeraoulia rafik Jun 02 '15 at 09:59
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many students and many people until now didn't know that cosz and sinz are unbounded ,so i'm interest to have a proof which show that – zeraoulia rafik Jun 02 '15 at 10:01
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@boon het , but there is a case where coz is real for some angles , it does means that non constant is not enough to proof the above claim – zeraoulia rafik Jun 02 '15 at 10:08
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1$\sin(z)=\frac{e^{iz}-e^{-iz}}{2i}$. Now let $z$ move along the positive imaginary axis. – Arthur Jun 02 '15 at 10:19
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sin(0)=0 , z=0+0i so it's real (according to ur formula) . constant in this case – zeraoulia rafik Jun 02 '15 at 10:21
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@zeraouliarafik it's not real because $(0,0)\neq 0$ – user42912 Jun 02 '15 at 10:44
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@zeraouliarafik it is definitely enough. Use Arthur's method to show $\cos, \sin$ are unbounded, if you need that – b00n heT Jun 02 '15 at 10:45
1 Answers
1
In the spirit of giving an explicit computation accessible to good high-school students:
If one accepts that $$ \left. \begin{gathered} \exp(iz) = \cos z + i\sin z, \\ \exp(z + w) = \exp(z) \exp(w), \end{gathered}\right\}\quad\text{for all complex $z$ and $w$,} \tag{1} $$ then unboundedness of both $\cos$ and $\sin$ along the imaginary axis follows from Yves Daoust's comment.
More generally, the real and imaginary parts of the circular functions are given by: $$ \left. \begin{aligned} \cos(x + iy) &= \cos x \cosh y - i\sin x \sinh y, \\ \sin(x + iy) &= \sin x \cosh y + i\cos x \sinh y. \end{aligned}\right\}\quad\text{for all real $x$, $y$.} $$ Since the hyperbolic functions $\cosh$ and $\sinh$ are unbounded, $\cos$ and $\sin$ are also unbounded.
Proof: By Euler's formula, $$ \left. \begin{aligned} \exp(iz) &= \cos z + i\sin z, \\ \exp(-iz) &= \cos z - i\sin z, \end{aligned}\right\}\quad\text{for all complex $z$.} \tag{2} $$ Adding and subtracting, we find $$ \left. \begin{aligned} \cos z &= \frac{\exp(iz) + \exp(-iz)}{2}, \\ \sin z &= \frac{\exp(iz) - \exp(-iz)}{2i}, \end{aligned}\right\}\quad\quad\text{for all complex $z$.} \tag{3} $$
Setting $z = x + iy$ with $x$ and $y$ real in (3), so that $iz = ix - y$ and $-iz = -ix + y$, the law of exponents and Euler's formula give \begin{align*} \cos(x + iy) &= \tfrac{1}{2} \bigl[\exp(ix - y) + \exp(-ix + y)\bigr] \\ &= \tfrac{1}{2}\bigl[e^{-y}(\cos x + i\sin x) + e^{y}(\cos x - i\sin x)\bigr] \\ &= \cos x \bigl[\tfrac{1}{2}(e^{y} + e^{-y})\bigr] - i\sin x \bigl[\tfrac{1}{2}(e^{y} - e^{-y})\bigr] \\ &= \cos x \cosh y - i\sin x \sinh y, \end{align*} and \begin{align*} \sin(x + iy) &= \tfrac{1}{2i} \bigl[\exp(ix - y) - \exp(-ix + y)] \\ &= -\tfrac{i}{2} \bigl[e^{-y}(\cos x + i\sin x) - e^{y}(\cos x - i\sin x)\bigr] \\ &= \sin x \bigl[\tfrac{1}{2}(e^{y} + e^{-y})\bigr] + i\cos x \bigl[\tfrac{1}{2}(e^{y} - e^{-y})\bigr] \\ &= \sin x \cosh y + i\cos x \sinh y. \end{align*}
Andrew D. Hwang
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