回答本题时不得使用计算器。 写出你的解答过程,求满足联立方程组的复数 zz 和 ww:
2iz+w=5+7i,(4−i)w∗=z−3i.\begin{aligned} 2\text{i}z + w &= 5 + 7\text{i}, \\ (4 - \text{i})w^* &= z - 3\text{i}. \end{aligned}

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完整解法

2

1. Rearrange the second equation to make $z$ the subject.

z=(4−i)w∗+3iz = (4 - \text{i})w^* + 3\text{i}

2. Substitute $z$ into the first equation and simplify to obtain an equation in terms of $w$ and $w^*$.

2i[(4−i)w∗+3i]+w=5+7i2\text{i}[(4 - \text{i})w^* + 3\text{i}] + w = 5 + 7\text{i}
(8i−2i2)w∗+6i2+w=5+7i(8\text{i} - 2\text{i}^2)w^* + 6\text{i}^2 + w = 5 + 7\text{i}
(2+8i)w∗−6+w=5+7i(2 + 8\text{i})w^* - 6 + w = 5 + 7\text{i}
w+(2+8i)w∗=11+7iw + (2 + 8\text{i})w^* = 11 + 7\text{i}

3. Let $w = x + y\text{i}$, substitute $w$ and $w^*$ into the equation, and expand.

Let w=x+yiw = x + y\text{i} and w∗=x−yiw^* = x - y\text{i}, where x,y∈Rx, y \in \mathbb{R}.
(x+yi)+(2+8i)(x−yi)=11+7i(x + y\text{i}) + (2 + 8\text{i})(x - y\text{i}) = 11 + 7\text{i}
x+yi+2x−2yi+8xi−8yi2=11+7ix + y\text{i} + 2x - 2y\text{i} + 8x\text{i} - 8y\text{i}^2 = 11 + 7\text{i}
x+yi+2x−2yi+8xi+8y=11+7ix + y\text{i} + 2x - 2y\text{i} + 8x\text{i} + 8y = 11 + 7\text{i}
(3x+8y)+(8x−y)i=11+7i(3x + 8y) + (8x - y)\text{i} = 11 + 7\text{i}

4. Equate the real and imaginary parts to form a pair of simultaneous equations.

Equating real parts:
3x+8y=11— (1)3x + 8y = 11 \quad \text{--- (1)}

Equating imaginary parts:
8x−y=7— (2)8x - y = 7 \quad \text{--- (2)}

5. Solve the simultaneous equations to find $x$ and $y$, thus obtaining $w$.

From (2), y=8x−7y = 8x - 7.
Substitute into (1):
3x+8(8x−7)=113x + 8(8x - 7) = 11
3x+64x−56=113x + 64x - 56 = 11
67x=67  ⟹  x=167x = 67 \implies x = 1
y=8(1)−7=1y = 8(1) - 7 = 1
Therefore, w=1+iw = 1 + \text{i}.

6. Substitute $w^*$ back into the expression for $z$ to find its value.

Using w∗=1−iw^* = 1 - \text{i}:
z=(4−i)(1−i)+3iz = (4 - \text{i})(1 - \text{i}) + 3\text{i}
z=4−4i−i+i2+3iz = 4 - 4\text{i} - \text{i} + \text{i}^2 + 3\text{i}
z=4−5i−1+3iz = 4 - 5\text{i} - 1 + 3\text{i}
z=3−2iz = 3 - 2\text{i}

答案

z=3−2iz = 3 - 2\text{i}, w=1+iw = 1 + \text{i}

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