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受欢迎的代数 >

solvefor y,six^2+y^2=z^2

解答

求解

y, s i x^{2} + y^{2} = z^{2}

解答

y = - \frac{s x ^{2}}{2 - \frac{z ^{2} + s ^{2} x ^{4} + z ^{4}}{2}} + - \frac{z ^{2} + s ^{2} x ^{4} + z ^{4}}{2} i, y = \frac{s x ^{2}}{2 - \frac{z ^{2} + s ^{2} x ^{4} + z ^{4}}{2}} - - \frac{z ^{2} + s ^{2} x ^{4} + z ^{4}}{2} i, y = - \frac{s x ^{2}}{2 - \frac{z ^{2} - s ^{2} x ^{4} + z ^{4}}{2}} + - \frac{z ^{2} - s ^{2} x ^{4} + z ^{4}}{2} i, y = \frac{s x ^{2}}{2 - \frac{z ^{2} - s ^{2} x ^{4} + z ^{4}}{2}} - - \frac{z ^{2} - s ^{2} x ^{4} + z ^{4}}{2} i

求解步骤

s i x^{2} + y^{2} = z^{2}

替代

y = x + v i

s i x^{2} + (x + v i)^{2} = z^{2}

展开

s i x^{2} + (x + v i)^{2} : (x^{2} - v^{2}) + i (s x^{2} + 2 xv)

x^{2} - v^{2} + i (s x^{2} + 2 xv) = z^{2}

将

z^{2}

改写成标准复数形式:

z^{2} + 0 i

x^{2} - v^{2} + i (s x^{2} + 2 xv) = z^{2} + 0 i

复数仅在实部和虚部均相等时才相等改写为方程组:

[x^{2} - v^{2} = z^{2} s x^{2} + 2 xv = 0]

[x^{2} - v^{2} = z^{2} s x^{2} + 2 xv = 0] : x = - \frac{s x ^{2}}{2 - \frac{z ^{2} + s ^{2} x ^{4} + z ^{4}}{2}}, x = \frac{s x ^{2}}{2 - \frac{z ^{2} + s ^{2} x ^{4} + z ^{4}}{2}}, x = - \frac{s x ^{2}}{2 - \frac{z ^{2} - s ^{2} x ^{4} + z ^{4}}{2}}, x = \frac{s x ^{2}}{2 - \frac{z ^{2} - s ^{2} x ^{4} + z ^{4}}{2}}, v = - \frac{z ^{2} + s ^{2} x ^{4} + z ^{4}}{2} v = - - \frac{z ^{2} + s ^{2} x ^{4} + z ^{4}}{2} v = - \frac{z ^{2} - s ^{2} x ^{4} + z ^{4}}{2} v = - - \frac{z ^{2} - s ^{2} x ^{4} + z ^{4}}{2}

y = x + v i

代回

y = - \frac{s x ^{2}}{2 - \frac{z ^{2} + s ^{2} x ^{4} + z ^{4}}{2}} + - \frac{z ^{2} + s ^{2} x ^{4} + z ^{4}}{2} i, y = \frac{s x ^{2}}{2 - \frac{z ^{2} + s ^{2} x ^{4} + z ^{4}}{2}} - - \frac{z ^{2} + s ^{2} x ^{4} + z ^{4}}{2} i, y = - \frac{s x ^{2}}{2 - \frac{z ^{2} - s ^{2} x ^{4} + z ^{4}}{2}} + - \frac{z ^{2} - s ^{2} x ^{4} + z ^{4}}{2} i, y = \frac{s x ^{2}}{2 - \frac{z ^{2} - s ^{2} x ^{4} + z ^{4}}{2}} - - \frac{z ^{2} - s ^{2} x ^{4} + z ^{4}}{2} i

流行的例子

((2-3i))/(2x+yi)=3+2i

\frac{( 2 - 3 i )}{2 x + y i} = 3 + 2 i

(a+ib)^2=16+30i

(a + ib)^{2} = 16 + 30 i

z^{2} - i \cdot z + 2 = 0

(3-a-2b)/5-3i+a+bi=4-i

\frac{3 - a - 2 b}{5} - 3 i + a + bi = 4 - i

x^2=-1+sqrt(3)i

x^{2} = - 1 + 3 i