What is the answer to solvefor x,csiny+x^2y=c ?

The answer to solvefor x,csiny+x^2y=c is

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solvefor x,csiny+x^2y=c

solve for

x, cs in y + x^{2} y = c

x = \frac{c ( n ^{2} s ^{2} y ^{2} + 1 + 1 )}{2 y} - \frac{n c s y}{2 n ^{2} s ^{2} y ^{2} + 1 + 1} i, x = - \frac{c ( n ^{2} s ^{2} y ^{2} + 1 + 1 )}{2 y} + \frac{n c s y}{2 n ^{2} s ^{2} y ^{2} + 1 + 1} i, x = \frac{c ( - n ^{2} s ^{2} y ^{2} + 1 + 1 )}{2 y} - \frac{n c s y}{2 - n ^{2} s ^{2} y ^{2} + 1 + 1} i, x = - \frac{c ( - n ^{2} s ^{2} y ^{2} + 1 + 1 )}{2 y} + \frac{n c s y}{2 - n ^{2} s ^{2} y ^{2} + 1 + 1} i

Solution steps

cs in y + x^{2} y = c

Substitute

x = a + bi

cs in y + (a + bi)^{2} y = c

Expand

cs in y + (a + bi)^{2} y : (y a^{2} - y b^{2}) + i (2 y ab + c n sy)

y a^{2} - y b^{2} + i (2 y ab + c n sy) = c

Rewrite

c

in standard complex form:

c + 0 i

y a^{2} - y b^{2} + i (2 y ab + c n sy) = c + 0 i

Complex numbers can be equal only if their real and imaginary parts are equalRewrite as system of equations:

[y a^{2} - y b^{2} = c 2 y ab + c n sy = 0]

[y a^{2} - y b^{2} = c 2 y ab + c n sy = 0] : a = \frac{c ( n ^{2} s ^{2} y ^{2} + 1 + 1 )}{2 y}, a = - \frac{c ( n ^{2} s ^{2} y ^{2} + 1 + 1 )}{2 y}, a = \frac{c ( - n ^{2} s ^{2} y ^{2} + 1 + 1 )}{2 y}, a = - \frac{c ( - n ^{2} s ^{2} y ^{2} + 1 + 1 )}{2 y}, b = - \frac{n c s y}{2 n ^{2} s ^{2} y ^{2} + 1 + 1} b = \frac{n c s y}{2 n ^{2} s ^{2} y ^{2} + 1 + 1} b = - \frac{n c s y}{2 - n ^{2} s ^{2} y ^{2} + 1 + 1} b = \frac{n c s y}{2 - n ^{2} s ^{2} y ^{2} + 1 + 1}

Substitute back

x = a + bi

x = \frac{c ( n ^{2} s ^{2} y ^{2} + 1 + 1 )}{2 y} - \frac{n c s y}{2 n ^{2} s ^{2} y ^{2} + 1 + 1} i, x = - \frac{c ( n ^{2} s ^{2} y ^{2} + 1 + 1 )}{2 y} + \frac{n c s y}{2 n ^{2} s ^{2} y ^{2} + 1 + 1} i, x = \frac{c ( - n ^{2} s ^{2} y ^{2} + 1 + 1 )}{2 y} - \frac{n c s y}{2 - n ^{2} s ^{2} y ^{2} + 1 + 1} i, x = - \frac{c ( - n ^{2} s ^{2} y ^{2} + 1 + 1 )}{2 y} + \frac{n c s y}{2 - n ^{2} s ^{2} y ^{2} + 1 + 1} i

(x + 2 y + 3) + i (3 x - y - 1) = 0

4 z^{2} + (- 13 + 18 i) z - 5 - 10 i = 0

x^{2} + 4 - 3 i = 0

- 3 ai - (- 1 - i) b = 3 a - 2

i (x) = \frac{1}{x}

What is the answer to solvefor x,csiny+x^2y=c ?
The answer to solvefor x,csiny+x^2y=c is