Is it true that $ |\sin^2z+\cos^2z|=1, \forall z \in\Bbb C$?

Question

We know that equation $ \sin^2z+ \cos^2z=1$ which holds $ \forall z \in\Bbb R$, actually holds $ \forall z \in\Bbb C$.

Thanks in advance.

Actually I wanted to asking something else. Previously I put modulus on wrong place. Please check edited question. — Sara Tancredi, Oct 09 '13 at 00:16

score 11 · Answer 1 · edited Oct 09 '13 at 07:48

11

In general, if an identity holds for a smooth function over real numbers, then the same identity also holds over the complex numbers. (Amazing!)

In this case: $$\cos^2(z)+\sin^2(z)=(\cos(z)+i\sin(z))(\cos(z)-i\sin(z))=e^{iz}e^{-iz}=1$$ where you need the exponential definition of trigonometric functions. There is no need for the absolute value in your question.

edited Oct 09 '13 at 07:48

Dennis Meng

1,397

answered Oct 09 '13 at 03:08

Kieran Cooney

1,437

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While there is some truth to your first sentence (due to the identity theorem for holomorphic functions) it is not true as stated. I'm not even sure what "the same identity" is supposed to mean if you start with a smooth function defined on $\mathbb{R}$: Arguably, $|x|^2 = x^2$ is an identity for a "smooth function over real numbers", but the "same identity", $|z|^2 = z^2$ doesn't hold for $z \in \mathbb{C}$. – mrf Oct 09 '13 at 09:17

Mhenni Benghorbal · Answer 2 · 2013-10-09T02:33:15.063

2

Here is a start

$$ \sin z = \sin(x+iy)=\sin(x)\cos(iy)+\sin(iy)\cos(x),\quad i=\sqrt{-1}. $$

$$ \implies \sin z = \sin(x) \cosh(y)+i\sinh(y)\cos(x). $$

Now, do the same with $\cos z$ and substitute back in your equation

$$ |\sin^2 z +\cos^2 z | $$

and see what you get.

Added: If you work the problem further, then we have

$$ \sin^2 z +\cos^2 z $$

$$= \sin^2 x \cosh^2 y - \cos^2 x \sinh^2 y + \cos^2 x \cosh^2 y - \sin^2 x \sinh^2 y =1. $$

Note:

$$\cosh^2 t - \sinh^2 t =1. $$

edited Oct 09 '13 at 02:33

answered Oct 09 '13 at 01:09

Mhenni Benghorbal

47,431

It gave us some weird result. $|cosh^2y-sinh^2+2isinx.sinhy.coshy(siny+cosx)|$. It doesn't give us any clue. What do you think now? Are we heading in right direction? – Sara Tancredi Oct 09 '13 at 02:02
@SaraTancredi: You are not done yet. You need to evaluate the modulus and make further simplifications. – Mhenni Benghorbal Oct 09 '13 at 02:07
Oh i see. So should it be 1 in the end? Or you don't know about the final answer? – Sara Tancredi Oct 09 '13 at 02:13
@SaraTancredi: I added more material. – Mhenni Benghorbal Oct 09 '13 at 02:17
@SaraTancredi: Note that, when you simplify $\sin^2 z +\cos^2 z$ you end up with one and its modulus is one. Work the problem step by step to see what's going on. – Mhenni Benghorbal Oct 09 '13 at 02:27
1

Thank you so much Mhenni. Let me do it. – Sara Tancredi Oct 09 '13 at 02:54
@SaraTancredi: You are very welcome. – Mhenni Benghorbal Oct 09 '13 at 03:02

score 1 · Answer 3 · answered Oct 09 '13 at 00:12

1

When you take the Absolute Value of $z$ you get a number in $\mathbb R$, and since $\left|z\right|\in \mathbb R$, $\sin^2|z|+\cos^2|z|=1$, so the answer is yes

answered Oct 09 '13 at 00:12

user93089

2,395
1
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I am really sorry. My question was different. I did mistake while writing the question. Please check it now. Thanks – Sara Tancredi Oct 09 '13 at 00:14

Is it true that $ |\sin^2z+\cos^2z|=1, \forall z \in\Bbb C$?

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