Question

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Answer 1

Question:-

If

$\sf u + iv = \frac{2 + i}{z + 3} and \: z = x + iy \: find \: u,v.$

Given:-

$\sf u + iv = \frac{2 + i}{z + 3} and \: z = x + iy \:$

To Find:-

• Value of u,v.

Solution:-

$\sf u + iv = \frac{2 + i}{x + iy + 3} = \frac{2 + i}{(x + 3) + iy}$

$\sf u + iv = \frac{2 + i}{(x + 3) + iy} \times \frac{(x + 3) - iy}{(x + 3) - iy}$

$\sf u + iv = \frac{(2 + 1) \times[(x + 3) - iy] }{(x + 3) {}^{2} + y {}^{2} }$

$\sf{{Z_{1}}}. \sf{{Z_{2}}} \sf = (ac - bd,ad + bc)$

a = 2, b = 1, c = x + 3, d = – y.

$\sf u + iv = \frac{(2x + 6 + y - 2y + x + 3)}{(x + 3) {}^{2} + y {}^{2} }$

$\sf u + iv = \frac{(2x + y + 6) + i(x - 2y + 3)}{(x + 3) {^{2} + y {}^{2} }}$

$\sf u = \frac{2x + y + 6}{(x + 3) {}^{2} + y {}^{2} } ;v = \frac{x - 2y + 3}{(x + 3) {}^{2} + y {}^{2} }$

Answer:-

$\sf \red{ u = \frac{2x + y + 6}{(x + 3) {}^{2} + y {}^{2} } \: ; \: v = \frac{x - 2y + 3}{(x + 3) {}^{2} + y {}^{2} } }$

Hope you have satisfied.⚘

Answer 2

Answer:

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