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Gnesinka [82]
3 years ago
6

Solve for x and y 1/(x+yi) + 2/(x-yi) = 1+i

Mathematics
1 answer:
aliina [53]3 years ago
4 0

Answer:

Step-by-step explanation:

x+i*y≠0 ==> x≠0 and y≠0

x-i*y≠0 ==> x≠0 and y≠0

We must exclude (0,0) as solution.

\dfrac{1}{x+i*y} +\dfrac{2}{x-i*y} =1+i\\\\\dfrac{1*(x-i*y)}{(x+i*y)(x-i*y)} +\dfrac{2*(x+i*y)}{(x-i*y)(x+i*y)} =1+i\\\\\\\dfrac{x-i*y+2x+2i*y}{x^2+y^2} =1+i\\\\3x+i*y=(1+i)(x^2+y^2)\\\\3x+i*y=(x^2+y^2)+i(x^2+y^2+i*(x^2+y^2)\\\\\left\{\begin{array}{ccc}3x&=&x^2+y^2\\y&=&x^2+y^2\\\end {array}\right.\\\\\\\left\{\begin{array}{ccc}y&=&3x\\3x&=&x^2+9x^2\\\end {array}\right.\\\\\\\left\{\begin{array}{ccc}y&=&3x\\x(10x-3)&=&0\\\end {array}\right.\\\\

\left\{\begin{array}{ccc}x&=&0\\y&=&0\\\end {array}\right.\ (to\ exclude) \ or \ \left\{\begin{array}{ccc}x&=&\dfrac{3}{10}\\\\y&=&\dfrac{9}{10}\\\end {array}\right.\\\\Sol=\{(\dfrac{3}{10},\dfrac{9}{10})\}\\

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