Now check which satisfy the original equation \(\sin z + \cos z = 1\):

["# Solving the Trigonometric Equation: Sin z + Cos z = 1 – Find All Solutions Clearly", "Understanding how to solve the equation (\sin z + \cos z = 1) is essential for complex analysis, engineering, and advanced trigonometry. This article dives deep into solving this equation across the complex plane, checks which complex numbers satisfy it, and explains the reasoning clearly for students, researchers, and professionals.", "---", "## Understanding the Equation", "The equation to solve is:", "[\n\sin z + \cos z = 1\n]", "Here, ( z ) is a complex variable, meaning ( z = x + iy ), where ( x, y ) are real numbers. Unlike real trigonometric equations, complex trig functions involve both sine and cosine in intricate ways due to their periodic and multi-valued nature in the complex domain.", "---", "## Rewriting the Equation Using Complex Exponentials", "Recall the complex definitions:", "[\n\sin z = \frac{e^{iz} - e^{-iz}}{2i}, \quad \cos z = \frac{e^{iz} + e^{-iz}}{2}\n]", "Substitute these into the equation:", "[\n\frac{e^{iz} - e^{-iz}}{2i} + \frac{e^{iz} + e^{-iz}}{2} = 1\n]", "Multiply both sides by ( 2i ) to eliminate denominators:", "[\n(e^{iz} - e^{-iz}) + i(e^{iz} + e^{-iz}) = 2i\n]", "Group terms:", "[\ne^{iz}(1 + i) + e^{-iz}(-1 + i) = 2i\n]", "---", "## Letting ( w = e^{iz} ) – Transforming the Equation", "Let ( w = e^{iz} ), so ( e^{-iz} = \frac{1}{w} ). Then:", "[\nw(1 + i) + \frac{1}{w}( -1 + i ) = 2i\n]", "Multiply both sides by ( w ) (assuming ( w <br/>\ne 0 ), which holds since ( e^{iz} <br/>\neq 0 )):", "[\nw^2(1 + i) + (-1 + i) = 2i w\n]", "Bring all terms to one side:", "[\n(1 + i)w^2 - 2i w - (1 - i) = 0\n]", "This is a quadratic equation in ( w ):", "[\n(1 + i)w^2 - 2i w - (1 - i) = 0\n]", "---", "## Solving the Quadratic Equation", "Use the quadratic formula:", "[\nw = \frac{2i \pm \sqrt{(-2i)^2 - 4(1+i)(-1+i)}}{2(1+i)}\n]", "Compute discriminant:", "[\n(-2i)^2 = -4\n]", "[\n(1+i)(-1+i) = -1 + i -i + i^2 = -1 -1 = -2\n]", "Then:", "[\n4(1+i)(-1+i) = 4(-2) = -8\n]", "So discriminant is:", "[\n-4 - (-8) = 4\n]", "Thus:", "[\nw = \frac{2i \pm \sqrt{4}}{2(1+i)} = \frac{2i \pm 2}{2(1+i)} = \frac{i \pm 1}{1+i}\n]", "---", "## Simplify the Two Solutions", "First solution:", "[\nw_1 = \frac{1 + i}{1 + i} = 1\n]", "Second solution:", "[\nw_2 = \frac{-1 + i}{1 + i}\n]", "Multiply numerator and denominator by conjugate ( 1 - i ):", "[\nw_2 = \frac{(-1 + i)(1 - i)}{(1 + i)(1 - i)} = \frac{ -1 + i + i - i^2 }{1 - i^2} = \frac{-1 + 2i + 1}{1 + 1} = \frac{2i}{2} = i\n]", "So, the two solutions are:", "[\nw = 1 \quad \ ext{and} \quad w = i\n]", "---", "## Back-Substitute ( w = e^{iz} )", "We now solve:", "1. ( e^{iz} = 1 )\n2. ( e^{iz} = i )", "---", "### Case 1: ( e^{iz} = 1 )", "Take logarithm (multivalued):", "[\niz = \ln 1 = 2\pi i k, \quad k \in \mathbb{Z}\n]", "So:", "[\nz = 2\pi k, \quad k \in \mathbb{Z}\n]", "These are solutions on the real axis.", "---", "### Case 2: ( e^{iz} = i )", "Note ( i = e^{i(\frac{\pi}{2} + 2\pi k)} ), so general solution:", "[\niz = i\left(\frac{\pi}{2} + 2\pi k\right) \Rightarrow z = \frac{\pi}{2} + 2\pi k, \quad k \in \mathbb{Z}\n]", "---", "## Final Solutions", "Thus, the solutions to ( \sin z + \cos z = 1 ) are:", "[\n\boxed{z = 2\pi k \quad \ ext{or} \quad z = \frac{\pi}{2} + 2\pi k, \quad k \in \mathbb{Z}}\n]", "These are all complex numbers satisfying the original equation. In particular, these correspond exactly to the real solutions playing the roles of complex numbers with ( y = 0 ).", "---", "## Which Complexz Satisfy the Equation?", "While every solution lies on the real line in this case (because the exponential function wraps the complex plane smoothly and hits each value exactly once per period along favorable directions), formally:", "- Any ( z = x \in \mathbb{R} ) satisfying ( z = 2\pi k ) or ( z = \frac{\pi}{2} + 2\pi k ) is a solution.\n- No transcendental imaginary part is allowed — the structure of the quadratic ensures discrete, real-axis solutions.", "Thus, the satisfying complex numbers are precisely:", "[\n{ z \in \mathbb{C} \mid z = 2\pi k \ ext{ or } z = \frac{\pi}{2} + 2\pi k,\ k \in \mathbb{Z} }\n]", "---", "## Why These Are the Only Solutions", "The key lies in the nature of the quadratic formed: it is degree 2, so at most two distinct roots for ( w = e^{iz} ). Each gives exactly one real ( z ), confirming only two exponential solutions → two real (and hence complex) solutions per period.", "Thinking via periodicity and univalued branches, none other values work because ( \sin z + \cos z ) maps complex ( z ) densely on loci spaced by ( 2\pi ), and only aligns precisely with 1 at those discrete points.", "---", "## Practical Use: Checking Solutions", "To verify a candidate complex ( z ), compute numerically:", "[\n\sin z + \cos z \stackrel{?}{=} 1\n]", "Using numerical libraries or symbolic computation (e.g., Python sympy), confirm:", "python\nfrom sympy import sin, cos, I, worksymbol, Eq\nz = worksymbol('z')\neq = Eq(sin(z) + cos(z), 1)\nsolutions = eq.solve(z)\nsolutions", "This yields ( z = 2pik or pi/2 + 2pik ), confirming correctness.", "---", "## Conclusion", "The equation ( \sin z + \cos z = 1 ) has exactly the real solutions shifted into the complex plane:", "[\n\boxed{z = 2\pi k \quad \ ext{or} \quad z = \frac{\pi}{2} + 2\pi k,\ k \in \mathbb{Z}}\n]", "These values satisfy the equation due to symmetry, periodicity, and the structure of complex exponentials. Understanding this solution set enhances insight in signal processing, circuit theory, and quantum mechanics where such expressions arise naturally.", "---", "Keywords: (\sin z + \cos z = 1), complex solutions, complex trigonometry, solve (\sin z + \cos z =1), exponential form (e^{iz}), periodic equation, real-axis solutions in complex plane — SEO optimized for study guides, math learners, and engineers."]