Show for all points in S, where a,b,c∈ℝ, not all 0, show that f(z) is constant?
Let S⊆ℂ be a domain. Let f: S→ℂ be analytic on S with f(z)=u(x,y)+iv(x,y). If au(x,y)+bv(x,y)=c
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Let S⊆ℂ be a domain. Let f: S→ℂ be analytic on S with f(z)=u(x,y)+iv(x,y). If au(x,y)+bv(x,y)=c
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Differentiate au + bv = c with respect to x and y, respectiely:
au_x + bv_x = 0
au_y + bv_y = 0.
Using the Cauchy-Riemann equations u_x = v_y and u_y = -v_x, we can rewrite these in terms of u:
au_x - bu_y = 0
au_y + bu_x = 0.
Multplying the first equation by a, the second by b, and adding yields
(a² + b²) u_x = 0.
==> u_x = 0 in S, since both a and b are not both 0 (else c = 0).
Similarly, u_y = 0 in S.
Hence, u(x,y) is constant in S.
Using the Cauchy-Riemann equations again, we see that v_x = 0 and v_y = 0 in S, and thus v(x,y) is constant in S.
Therefore, f(z) = u(x,y) + iv(x,y) is constant in S.
I hope this helps!