1.

If ϕ (x, y) and ψ(x, y) are functions with continuous second derivatives, then ϕ (x, y) + iψ(x, y) can be expressed as an analytic function of \(x + iy\left( {i = \sqrt { - 1} } \right)\), when

A. \(\frac{{\partial \phi }}{{\partial x}} = - \frac{{\partial \psi }}{{\partial x}};\;\frac{{\partial \phi }}{{\partial y}} = \frac{{\partial \psi }}{{\partial y}}\)
B. \(\frac{{\partial \phi }}{{\partial y}} = - \frac{{\partial \psi }}{{\partial x}};\;\frac{{\partial \phi }}{{\partial x}} = \frac{{\partial \psi }}{{\partial y}}\)
C. \(\frac{{{\partial ^2}\phi }}{{\partial {x^2}}} + \frac{{{\partial ^2}\phi }}{{\partial {y^2}}} = \frac{{{\partial ^2}\psi }}{{\partial {x^2}}} + \frac{{{\partial ^2}\psi }}{{\partial {y^2}}} = 1\)
D. \(\frac{{\partial \phi }}{{\partial x}} + \frac{{\partial \phi }}{{\partial y}} = \frac{{\partial \psi }}{{\partial x}} + \frac{{\partial \psi }}{{\partial y}} = 0\)
Answer» C. \(\frac{{{\partial ^2}\phi }}{{\partial {x^2}}} + \frac{{{\partial ^2}\phi }}{{\partial {y^2}}} = \frac{{{\partial ^2}\psi }}{{\partial {x^2}}} + \frac{{{\partial ^2}\psi }}{{\partial {y^2}}} = 1\)


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