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Complex Analysis: Theory and Applications · Problem 6.6.8
Taylor and Laurent Series — Howell & Matthews
Mathematics — Complex Analysis · harmonic-functions-mean-value
Problem Let u ( x , y ) u(x,y) u ( x , y ) be harmonic for all ( x , y ) (x,y) ( x , y ) . Show that
u ( x 0 , y 0 ) = 1 2 π ∫ 0 2 π u ( x 0 + R cos θ , y 0 + R sin θ ) d θ , u(x_0,y_0) = \frac{1}{2\pi}\int_0^{2\pi} u\!\left(x_0 + R\cos\theta,\; y_0 + R\sin\theta\right)d\theta, u ( x 0 , y 0 ) = 2 π 1 ∫ 0 2 π u ( x 0 + R cos θ , y 0 + R sin θ ) d θ ,
where R > 0 R > 0 R > 0 . Hint: Let f ( z ) = u ( x , y ) + i v ( x , y ) f(z) = u(x,y) + iv(x,y) f ( z ) = u ( x , y ) + i v ( x , y ) , where v v v is a harmonic conjugate of u u u .
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