Complex Sine and Cosine Functions

An expert is a person who has made all the mistakes that can be made in a very narrow field, Niels Bohr

The Complex Exponential Function

The complex exponential function, denoted as $e^z$ or $\exp(z)$, is one of the most important functions in all of mathematics. It is the unique entire function that extends the real exponential function $e^x$ to the complex plane. This function is not only entire but also periodic.

To define the complex exponential function, we seek a function $f: \mathbb{C} \to \mathbb{C}$ that satisfies two key properties:

1. Additivity (Functional Equation): $f(z_1 + z_2) = f(z_1) \cdot f(z_2)$ for all $z_1, z_2 \in \mathbb{C}$.
2. Consistency with the Real Exponential: $f(x) = e^x$ for all $x \in \mathbb{R}$.

Let $z = x + iy$ with $x, y \in \mathbb{R}$. Using properties (1) and (2): $f(z) = f(x + iy) \overset{(1)}{=} f(x) \cdot f(iy) \overset{(2)}{=} e^x \cdot f(iy).$

Writing $f(iy) = A(y) + iB(y)$ where $A$ and $B$ are real-valued functions of $y$, $f(z) = e^x A(y) + ie^x B(y),$ so $u(x, y) = e^x A(y)$ and $v(x, y) = e^x B(y)$.

For $f$ to be analytic (differentiable everywhere), its real and imaginary parts must satisfy the Cauchy–Riemann equations.

Computing partial derivatives: $u_x = e^x A(y), \quad u_y = e^x A'(y), \quad v_x = e^x B(y), \quad v_y = e^x B'(y).$

The equation $u_x = v_y$ gives $e^x A(y) = e^x B'(y)$, hence $A(y) = B'(y) (\star)$.

Similarly, the equation $u_y = -v_x$ gives $e^x A'(y) = -e^x B(y)$, hence $A'(y) = -B(y) (\star\star)$.

Differentiating $A(y) = B'(y) (\star)$, we get $B''(y) = A'(y)$ and substituting $A'(y) = -B(y) (\star\star)$: $B''(y) = A'(y) = -B(y) \quad \Longrightarrow \quad B''(y) + B(y) = 0.$

This is a linear, homogeneous, constant-coefficient ODE with characteristic equation $r^2 + 1 = 0$, giving roots $r = \pm i$.

Recall. For complex conjugate roots $a \pm bi$, the general solution is $y(t) = e^{at}(\alpha\cos(bt) + \beta\sin(bt))$, where $\alpha, \beta \in \mathbb{R}$.

With $a = 0$ and $b = 1$ ($r = \pm i$), the general solution is: $B(y) = \alpha\cos y + \beta\sin y, \qquad A(y) =[(\star)] B'(y) = -\alpha\sin y + \beta\cos y.$

From property (2), $f(0) = e^0 = 1$. But $f(0) = e^0 A(0) + ie^0 B(0) = A(0) + iB(0)$, so $A(0) = 1 \quad \text{and} \quad B(0) = 0.$

From $A(0) = 1$: $-\alpha\sin 0 + \beta\cos 0 = \beta = 1$.

From $B(0) = 0$: $\alpha\cos 0 + \beta\sin 0 = \alpha = 0$.

Therefore $A(y) = \cos y$ and $B(y) = \sin y$, and we finally arrive at the definition of the complex exponential function: $\boxed{e^z = e^x(\cos y + i\sin y), \qquad z = x + iy.}$

The exponential function is well-defined for all $z \in \mathbb{C}$, entire, non-zero, consistent with the real exponential, periodic with period $2\pi i$ ($e^{z+2\pi i}=e^z$), satisfies $\frac{d}{dz}e^z = e^z$, and can be represented as a power series $\boxed{e^z = \sum_{n=0}^∞ \frac{z^n}{n!}}$ with infinite radius of convergence.

Sine and Cosine functions

From Euler’s Formula to Complex Definitions

For a complex number $z = x + iy$ (where $x, y \in \mathbb{R}$), the complex exponential is $exp(e) = e^{x+iy} = e^x(cos(y) + isin(y))$

Euler’s formula $e^{iy} = cos(y) + isin(y)$ leads to elegant expressions for the trigonometric functions. If we also write its complex conjugate $e^{-i y} = \cos(y) - i\sin(y),$ then by adding and subtracting these two expressions, we could isolate the real trigonometric functions and get: $\boxed{cos(y) = \frac{eⁱʸ+e⁻ⁱʸ}{2}, sin(y) = \frac{eⁱʸ-e⁻ⁱʸ}{2i}}$.

Complex Definitions of Trigonometric Functions

These relationships naturally extend to complex arguments, giving us the standard complex-analytic definitions: $\boxed{\cos z = \frac{e^{iz} + e^{-iz}}{2}, \qquad \sin z = \frac{e^{iz} - e^{-iz}}{2i}}$.valid for all $z \in \mathbb{C}$.

From these, all other trigonometric functions follow: $\tan z = \frac{\sin z}{\cos z}, \quad \cot z = \frac{\cos z}{\sin z}, \quad \sec z = \frac{1}{\cos z}, \quad \csc z = \frac{1}{\sin z}.$

Definition. The hyperbolic sine and cosine are defined for all $z \in \mathbb{C}$ by $\boxed{\sinh z = \frac{e^z - e^{-z}}{2}}, \qquad \boxed{\cosh z = \frac{e^z + e^{-z}}{2}}.$

These are entire functions (it follows from the fact that they are build from the entire exponential function) built from the same exponential building blocks as sine and cosine, but without the factor of $i$ in the exponent. These definitions are valid for all complex numbers $z \in \mathbb{C}$, and use the exponential function to build smooth, entire functions.

Bridge identities. Evaluating the circular functions at imaginary arguments (or vice versa) reveals that trigonometric and hyperbolic functions are essentially the same functions, rotated in the complex plane: $\sin(iz) = i\sinh z, \qquad \cos(iz) = \cosh z$, $\sinh(iz) = i\sin z, \qquad \cosh(iz) = \cos z$

Proof: $\sin(iz) = \dfrac{e^{i(iz)} - e^{-i(iz)}}{2i} = \dfrac{e^{-z} - e^{z}}{2i} = \dfrac{-(e^z - e^{-z})}{2i} = \dfrac{i(e^z - e^{-z})}{2} = i\sinh z.$

Key Properties

• Differentiation: $\frac{d}{dz} \sinh z = \cosh z,\frac{d}{dz} \cosh z = \sinh z$
• Fundamental identity (hyperbolic version of the Pythagorean identity): $\boxed{\cosh^2 z - \sinh^2 z = 1}$.
• Even/odd symmetry: $\cosh(-z) = \cosh z, \quad \sinh(-z) = -\sinh z$
• Separation into Real and Imaginary Parts. $\boxed{\cos(x+iy) = \cos x\,\cosh y \;-\; i\sin x\,\sinh y}$, $\boxed{\sin(x+iy) = \sin x\,\cosh y \;+\; i\cos x\,\sinh y}$.

Let’s write $\sin z$ and $\cos z$ in terms of their reasl and imaginary parts $x = \operatorname{Re} z$ and $y = \operatorname{Im} z$.

$$ \begin{aligned} \cos(x+iy) = \frac{e^{i(x+iy)} + e^{-i(x+iy)}}{2} &=\frac{e^{-y+ix} + e^{y-ix}}{2} \\[2pt] &= \frac{e^{-y}(\cos x + i\sin x) + e^{y}(\cos x - i\sin x)}{2}\\[2pt] &=\cos x \cdot \frac{e^y + e^{-y}}{2} - i\sin x \cdot \frac{e^y - e^{-y}}{2}\\[2pt] &=\cos x\cosh y - i\sin x \sinh y. \end{aligned} $$$$ \begin{aligned} \sin(x+iy) = \frac{e^{i(x+iy)} - e^{-i(x+iy)}}{2i} &=\frac{e^{-y+ix} - e^{y-ix}}{2i} \\[2pt] &= \frac{e^{-y}(\cos x + i\sin x) - e^{y}(\cos x - i\sin x)}{2i} \\[2pt] &= \sin x \cdot \frac{e^y + e^{-y}}{2} + i\cos x \cdot \frac{e^y - e^{-y}}{2} \\[2pt] &= \sin x \cosh y + i\cos x \sinh y \end{aligned} $$

Properties

• Reduction to Real Functions. When restricted to the real line ($y = 0$), since $\cosh 0 = 1$ and $\sinh 0 = 0$: $\cos(x + i\cdot 0) = \cos x \cdot 1 - i\sin x \cdot 0 = \cos x$ (the familiar real cosine), $\sin(x + i\cdot 0) = \sin x \cdot 1 + i\cos x \cdot 0 = \sin x$ (the familiar real sine).

This confirms that the complex definitions are consistent extensions of the real trigonometric functions — they agree on $\mathbb{R}$ and extend holomorphically to all of $\mathbb{C}$.

• Entire Functions. Both $\sin z$ and $\cos z$ are entire functions — holomorphic on the entire complex plane $\mathbb{C}$, with no singularities of any kind.

  Since the exponential function $e^z$ is entire, and $\sin z$ and $\cos z$ are defined through linear combinations of the entire functions $e^{iz}$ and $e^{-iz}$, they inherit this property. No discontinuities, no branch cuts, no singularities — just smooth, “well complex-analytic behavior” everywhere.

Their reciprocals and quotients, however, introduce isolated singularities (poles) where their denominators vanish. All these poles are simple (order 1. This is the case because the zeroes of sin(z) and cos(z) are simple, their derivatives are non-zero at the zeroes), and the functions are meromorphic on $\mathbb{C}$.

A meromorphic function is almost analytic everywhere, except for countably many singularities (poles) where it “blows up” in a controlled way, e.g., $\frac{1}{z}, \frac{1}{z^2}$, etc.

Function	Poles at
$\tan z = \sin z / \cos z$	$z = \frac{\pi}{2} + n\pi, \quad n \in \mathbb{Z}$
$\sec z = 1/\cos z$	$z = \frac{\pi}{2} + n\pi, \quad n \in \mathbb{Z}$
$\cot z = \cos z / \sin z$	$z = n\pi, \quad n \in \mathbb{Z}$
$\csc z = 1/\sin z$	$z = n\pi, \quad n \in \mathbb{Z}$

• Derivatives of Sine and Cosine. The derivatives of the complex sine and cosine functions can be found by differentiating their exponential definitions. Using the chain rule and the fact that $\frac{d}{dz}e^z = e^z$, we have $\frac{d}{dz}cos(z) = \frac{d}{dz}\frac{eⁱᶻ+e⁻ⁱᶻ}{2} = \frac{ieⁱᶻ-ie⁻ⁱᶻ}{2} = i\frac{eⁱᶻ-e⁻ⁱᶻ}{2} = -\frac{eⁱᶻ-e⁻ⁱᶻ}{2i} = -sin(z)$. Similarly, for the sine function: $\frac{d}{dz}sin(z) = \frac{d}{dz}\frac{eⁱᶻ-e⁻ⁱᶻ}{2i} = \frac{ieⁱᶻ+ie⁻ⁱᶻ}{2i} = \frac{eⁱᶻ+e⁻ⁱᶻ}{2} = cos(z)$.

These results are identical to the familiar formulas from real calculus and extend naturally to the complex domain, $\boxed{\frac{d}{dz}\sin z = \cos z, \qquad \frac{d}{dz}\cos z = -\sin z}$

• Higher-order derivatives: $\frac{d^n}{dz^n}\cos z = -\sin z$ (n = 1), $-\cos z$ (n = 2), $\sin z$ (n =3), $\cos z$ (n = 4). $\frac{d^n}{dz^n}\sin z = -\cos z$ (n = 1), $-\sin z$ (n = 2), $-\cos z$ (n =3), $\sin z$ (n = 4).

Compact formula using phase shift: $\frac{d^n}{dz^n}\cos z = \cos(z+\frac{n\pi}{2}), \frac{d^n}{dz^n}\sin z = \sin(z+\frac{n\pi}{2})$

The pattern repeats with period 4, reflecting the identity $\left(\frac{d}{dz}\right)^4 = \text{id}$ on the span of $\{\sin z, \cos z\}$. The derivative generates a 4-cycle isomorphic to $\mathbb{Z}_4$.

$\frac{d^2}{dz^2}\sin z = -\sin z, \frac{d^2}{dz^2}\cos z = -\cos z$. Both satisfy the harmonic oscillator equation (second order) $\boxed{f''(z) + f(z) = 0}$.

Equivalently, both $\cos(z)$ and $\sin(z)$ are fundamental solutions to the second-order harmonic oscillator equation: $\boxed{f''(z) + f(z) = 0}$. By the existence and uniqueness theorem for linear ODEs, we can completely and uniquely define these functions across the entire complex plane simply by specifying their initial conditions at $z = 0$:

• cos(z) : f(0) =$[\cos z = \frac{e^{iz} + e^{-iz}}{2}|_{z=0}] = \frac{1+1}{2}=1$, f′(0) = 0.
• sin(z) : f(0) = 0, f′(0) = 1.

This ODE with specified initial conditions has a unique solution. In other words, it uniquely characterizes cos(z) and sin(z) among entire functions.

1. Since the solutions are entire (analytic everywhere), we assume a power series centered at zero:, $f(z) = \sum_{n=0}^{\infty} a_n z^n$
2. Differentiate the series term-by-term. The constant term vanishes, so the index for the first derivative shifts to start at $n=1$, and for the second derivative, it starts at $n=2$: $f'(z) = \sum_{n=1}^{\infty} n a_n z^{n-1} = \sum_{n=0}^{\infty} (n+1) a_{n+1} z^n$ and $f''(z) = \sum_{n=2}^{\infty} n(n-1) a_n z^{n-2} = \sum_{n=0}^{\infty} (n+2)(n+1) a_{n+2} z^n$.
   To add $f''(z)$ and $f(z)$ together, we need their summations to match powers of $z$. We re-index $f''(z)$ by shifting $n \to n+2$. As a result, the starting index drops back to $n=0$.
3. Substitute into $f'' + f = 0$: $\sum_{n=0}^{\infty} (n+2)(n+1) a_{n+2} z^n + \sum_{n=0}^{\infty} a_n z^n = 0$.
   This relation tells us that every coefficient is determined by the one two steps before it, meaning the even indices and odd indices operate completely independently.
4. Combine: $\sum_{n=0}^{\infty} \left[(n+2)(n+1) a_{n+2} + a_n\right] z^n = 0$. For this to hold for all $z$, each coefficient must vanish: $(n+2)(n+1) a_{n+2} + a_n = 0 \implies \boxed{a_{n+2} = -\frac{a_n}{(n+2)(n+1)}}$
5. Using the initial conditions $f(0) = 1 \implies a_0 = 1$ and $f'(0) = 0 \implies a_1 = 0$. Because $a_1 = 0$, all subsequent odd coefficients ($a_3, a_5, \dots$) must also be 0. Let’s unfold the even terms using our recurrence relation:$a_2 = -\frac{a_0}{2 \cdot 1} = -\frac{1}{2!}, a_4 = -\frac{a_2}{4 \cdot 3} = -\frac{-1/2!}{4 \cdot 3} = \frac{1}{4!},a_6 = -\frac{a_4}{6 \cdot 5} = -\frac{1/4!}{6 \cdot 5} = -\frac{1}{6!}$ Following this alternating factorial pattern, $a_{2k} = \frac{(-1)^k}{(2k)!}$. Mutatis mutandis for $\sin(z)$.
6. Solution 1: Set $a_0 = 1$, $a_1 = 0$ (even function, $f(0)=1$, $f'(0)=0$). $C(z) = \sum_{k=0}^{\infty} \frac{(-1)^k}{(2k)!} z^{2k} = 1 - \frac{z^2}{2!} + \frac{z^4}{4!} - \cdots$
   Solution 2: Set $a_0 = 0$, $a_1 = 1$ (odd function, $f(0)=0$, $f'(0)=1$). $S(z) = \sum_{k=0}^{\infty} \frac{(-1)^k}{(2k+1)!} z^{2k+1} = z - \frac{z^3}{3!} + \frac{z^5}{5!} - \cdots$
7. By direct comparison with Taylor series, $C(z) = \sum_{k=0}^{\infty} \frac{(-1)^k}{(2k)!} z^{2k} = \cos(z)$, $\sum_{k=0}^{\infty} \frac{(-1)^k}{(2k+1)!} z^{2k+1} = \sin(z)$.

Evaluating the derivatives at the origin (the Maclaurin series coefficients): $\cos^{(2n)}(0) = (-1)^n, \quad \cos^{(2n+1)}(0) = 0$, $\sin^{(2n)}(0) = 0, \quad \sin^{(2n+1)}(0) = (-1)^n$

• Global Power-Series Expansions. From the above derivatives: $\cos z = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n)!}\,z^{2n} = 1 - \frac{z^2}{2!} + \frac{z^4}{4!} - \frac{z^6}{6!} + \cdots, \sin z = \sum_{n=0}^{\infty} \frac{(-1)^n}{(2n+1)!}\,z^{2n+1} = z - \frac{z^3}{3!} + \frac{z^5}{5!} - \frac{z^7}{7!} + \cdots$. Both series converge for every $z \in \mathbb{C}$ (radius of convergence $R = \infty$), providing an independent proof that sine and cosine are entire.

The ratio test confirms this: $\left|\frac{a_{n+1}}{a_n}\right| = \frac{|z|^2}{(2n+2)(2n+1)} \to 0 \quad \text{as } n \to \infty,$ regardless of $|z|$, so both series have radius of convergence $R = \infty$.

These power series can serve as alternative definitions of sine and cosine, from which all properties (including Euler’s formula itself) can be derived.

• Integration. Since differentiation and integration are inverse operations: $\int \sin z dz = -\cos z + C, \qquad \int \cos zdz = \sin z + C$ where $C$ is an arbitrary constant.

These hold for contour integrals in $\mathbb{C}$ as well, since $\sin z$ and $\cos z$ are entire and hence have antiderivatives on all of $\mathbb{C}$.

• Trigonometric Identities in the Complex Plane.All standard trigonometric identities originally established for real arguments, also hold for complex arguments.

This follows from the identity principle: if two entire functions agree on $\mathbb{R}$, they agree on all of $\mathbb{C}$. For example, the Pythagorean identity, $cos^2(z) + sin^2(z) = (\frac{eⁱᶻ+e⁻ⁱᶻ}{2})^2 + (\frac{eⁱᶻ-e⁻ⁱᶻ}{2i})^2 = \frac{e²ⁱᶻ+ 2 + e⁻²ⁱᶻ}{4} + \frac{e²ⁱᶻ -2 +e⁻²ⁱᶻ}{-4} = \frac{e²ⁱᶻ+ 2 + e⁻²ⁱᶻ -e²ⁱᶻ + 2 - e⁻²ⁱ}{4} = \frac{4}{4} = 1$.

Identity Principle. If $f$ and $g$ are entire (holomorphic on all of $\mathbb{C}$) and $f(z) = g(z)$ for all $z$ in a set $S \subseteq \mathbb{C}$ that has a limit point in $\mathbb{C}$, then $f \equiv g$ on all of $\mathbb{C}$. Key corollary: If two entire functions agree on $\mathbb{R}$ (or any interval, no matter how small), they agree everywhere on $\mathbb{C}$.

Related identities: $1 + \tan^2 z = \sec^2 z, \qquad 1 + \cot^2 z = \csc^2 z$

Angle Addition Formulas: $\sin(z_1 + z_2) = \sin z_1 \cos z_2 + \cos z_1 \sin z_2$, $\cos(z_1 + z_2) = \cos z_1 \cos z_2 - \sin z_1 \sin z_2$, $\sin(z_1 - z_2) = \sin z_1 \cos z_2 - \cos z_1 \sin z_2$, $\cos(z_1 - z_2) = \cos z_1 \cos z_2 + \sin z_1 \sin z_2$

Double Angle Formulas. Setting $z_1 = z_2 = z$ in the addition formulas: $\sin(2z) = 2\sin z \cos z, \cos(2z) = \cos^2 z - \sin^2 z = 2\cos^2 z - 1 = 1 - 2\sin^2 z$

Half-Angle Formulas. Rearranging the double-angle formulas (with $z$ replaced by $z/2$): $\cos^2(z/2) = \frac{1 + \cos z}{2}, \sin^2(z/2) = \frac{1 - \cos z}{2}$

Product-to-Sum Formulas: $\cos z_1\cos z_2 = \frac{1}{2}\bigl[\cos(z_1 - z_2) + \cos(z_1 + z_2)\bigr]$, $\sin z_1\sin z_2 = \frac{1}{2}\bigl[\cos(z_1 - z_2) - \cos(z_1 + z_2)\bigr]$, $\sin z_1\cos z_2 = \frac{1}{2}\bigl[\sin(z_1 + z_2) + \sin(z_1 - z_2)\bigr]$.

• Periodicity and Symmetry Properties.

Periodicity. Both sine and cosine are $2\pi$-periodic: $\sin(z + 2\pi) = \sin z, \qquad \cos(z + 2\pi) = \cos z \qquad \text{for all } z \in \mathbb{C}.$

More generally, $\sin(z + 2n\pi) = \sin z$ and $\cos(z + 2n\pi) = \cos z$ for all $n \in \mathbb{Z}$.

This follows directly from the periodicity of the exponential function: $e^{i(z+2nπ)}=e^{iz}e^{i2nπ}=e^{iz}$. Besides, the period $2\pi$ is the minimal period: there is no smaller $T > 0$ with $\sin(z + T) = \sin z$ for all $z$.

The tangent and cotangent have the shorter minimal period $\pi$: $\tan(z + \pi) = \tan z, \qquad \cot(z + \pi) = \cot z.$

Phase Relationships: $\sin\left(\frac{\pi}{2} + z\right) = \cos z, \cos\left(\frac{\pi}{2} + z\right) = -\sin z, \sin\left(\frac{\pi}{2} - z\right) = \cos z, \cos\left(\frac{\pi}{2} - z\right) = \sin z$. Sine and cosine are the same function, shifted by $\pi/2$.

Symmetry Properties: $\sin(-z) = -\sin z$ (odd function), $\cos(-z) = \cos z$ (even function).

• Exponential growth in the Imaginary Direction. On the real line, sine and cosine are bounded: $|\sin x| \leq 1$ and $|\cos x| \leq 1$ for all $x \in \mathbb{R}$. In the complex plane, this boundedness breaks down dramatically.

$|\sin(x + iy)|^2 = \sin^2 x\cosh^2 y + \cos^2 x\\sinh^2 y$. Next, simplify using $\cosh^2 y = 1 + \sinh^2 y$, and we get, $= \sin^2 x\,(1 + \sinh^2 y) + \cos^2 x\,\sinh^2 y = \sin^2 x + (\sin^2 x + \cos^2 x)\sinh^2 y$. Conclusion: $\boxed{|\sin(x+iy)|^2 = \sin^2 x + \sinh^2 y}$

An analogous computation yields: $\boxed{|\cos(x+iy)|^2 = \cos^2 x + \sinh^2 y}$

Asymptotic growth. The hyperbolic sine function, sinh(y), is defined as: sinh(y) = $\frac{e^y - e^{-y}}{2}$. Now, let’s analyze the behavior as $|y| \to \infty$, the term $e^{-|y|}$ approaches 0, so $\sinh(∣y∣) \approx \frac{e^{|y|}}{2}$. Therefore, as $|y| \to \infty, ∣\sin(x+iy)∣^2 = \sin^2(x) + \sinh^2(y) \approx \sin^2(x) + (\frac{e^{|y|}}{2})^2 ≈ sin²(x) + \frac{e^{2|y|}}{4}$. Since the exponential term $e^{2|y|}$ grows much faster than the bounded term sin²(x) (which is negligible between 0 and 1), the asymptotic behavior is dominated by the hyperbolic sine term, $∣\sin(x+iy)∣^2 \approx \frac{e^{2|y|}}{4}$. Taking the square root of both sides, we find the magnitude: $∣\sin(x+iy)∣ \approx \sqrt{\frac{e^{2|y|}}{4}} = \frac{e^{|y|}}{2}$. Similarly, $∣\cos(x+iy)∣ \approx \frac{e^{|y|}}{2}$

Conclusion: Complex sine and cosine grow exponentially in the imaginary direction — even though on the real line they remain bounded between $-1$ and $1$.

This behavior is not paradoxical but inevitable: by Liouville’s theorem, a bounded entire function must be constant. Since $\sin z$ and $\cos z$ are entire (analytic everywhere) and non-constant, they must be unbounded.

Remark. Along the real axis ($y = 0$), the formulas correctly reduce to $|\sin x|^2 = \sin^2 x$ and $|\cos x|^2 = \cos^2 x$, both $\leq 1$.

So complex sine and cosine grow exponentially in the imaginary direction! —even though on the real line they stay bounded between -1 and 1.

• Zeros of Sine and Cosine. $\sin z = 0$ when $z = n\pi$ for $n \in \mathbb{Z}$. $\cos z = 0$ when $z = \frac{\pi}{2} + n\pi$ for $n \in \mathbb{Z}$

Zeros of sine. We solve $\sin z = 0$: $\frac{e^{iz} - e^{-iz}}{2i} = 0 \quad \Longrightarrow \quad e^{iz} = e^{-iz} \quad \Longrightarrow \quad e^{2iz} = 1 \quad \Longrightarrow \quad 2iz = 2n\pi i \quad \Longrightarrow \quad z = n\pi$ for $n \in \mathbb{Z}$.

Alternatively, using the modulus formula: $|\sin(x+iy)|^2 = \sin^2 x + \sinh^2 y = 0$ requires both $\sin x = 0$ and $\sinh y = 0$, giving $x = n\pi$ and $y = 0$ (So zeros must lie on the real axis). Conclusion: $\boxed{\sin z = 0 \quad \Longleftrightarrow \quad z = n\pi, \quad n \in \mathbb{Z}}$

Zeros of cosine. Similarly, $\cos z = 0$ gives $e^{2iz} = -1 = e^{i(\pi + 2n\pi)}$, so $z = \frac{\pi}{2} + n\pi$, $\boxed{\cos z = 0 \quad \Longleftrightarrow \quad z = \frac{\pi}{2} + n\pi, \quad n \in \mathbb{Z}}$

All zeros are simple (the derivative is nonzero there): at $z_0 = n\pi$, $\sin'(z_0) = \cos(n\pi) = (-1)^n \neq 0$, and similarly for cosine.

The fact that all zeros are real is a special feature of sine and cosine. Many entire functions (like $e^z - 1$) have complex zeros.