HINT: Use the Euler formula for $\sin(z)$. Be sure to take care and define $z=x+iy$ where $x,y\in\Bbb R$.

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2019年11月24日 · Let $z = x + i y \in \C$ be a complex number, where $x, y \in \R$. Let $\sin z$ denote the complex sine function. Then: $\cmod {\sin z} = \sqrt {\sin^2 x + \sinh^2 y}$ where: $\cmod z$ denotes the modulus of a complex number $z$ $\sin x$ denotes the real sine function $\sinh$ denotes the hyperbolic sine function. Proof

2018年2月9日 · sin ⁡ (z + 2 ⁢ π) = sin ⁡ z, cos ⁡ (z + 2 ⁢ π) = cos ⁡ z ∀ z The periodicity of the functions causes that their inverse functions , the complex cyclometric functions , are infinitely multivalued; they can be expressed via the complex logarithm and square root (see general power ) …

2022年12月8日 · Let $z = x + i y \in \C$ be a complex number, where $x, y \in \R$. Let $\sin z$ denote the complex sine function. Then: $\map \Re {\sin z} = \sin x \cosh y$ where: $\map \Re z$ denotes the real part of a complex number $z$ $\sin$ denotes the sine function (real and complex) $\cosh$ denotes the hyperbolic cosine function. Proof. From Sine of ...

The sine of a complex number can be found using the formula sin(z) = (e^(iz) - e^(-iz))/2i, where z is a complex number and i is the imaginary unit. This formula can be derived using the exponential form of a complex number, z = x + iy = r(cosθ + isinθ), where r …

To find the complex numbers z satisfying $\sin (z) = \cos (z)$, can I say: $$\sin (z) = \frac { (e^ {iz}-e^ {-iz})} {2i}=\frac { (e^ {iz}+e^ {-iz})} {2}$$ and solve for z? So we then reduce this to $$-e^ {-iz}...

How do I find the zeros of $\sin(z)$, where $z$ is a complex number? I know that along the real line we have zeros along $k\pi$, where $k$ is an integer. But what about the rest of the plane?

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Compute answers using Wolfram's breakthrough technology & knowledgebase, relied on by millions of students & professionals. For math, science, nutrition, history, geography, engineering, mathematics, linguistics, sports, finance, music…