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小 發表於 2016-9-18 09:28 PM (第 2748 天)
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25.
(a + 3i)(-2 + 3i)
= -2a + 3ai - 6i + 9i^2
= -2a - 9 + (3a - 6)i
To let (a + 3i)(-2 + 3i) be a real number, the imaginary part must be zero.
i.e. 3a - 6 = 0
=> a = 2
26.
(2x + yi) - (x - 3i)/i = -y + 4i
(2x + yi)i - (x - 3i) = (-y + 4i)i
2xi + yi^2 - x + 3i = -yi + 4i^2
-y - x + 2xi + 3i = -yi - 4
-y - x + (2x + 3)i = -4 - yi
Then by comparing real part and imaginary part,
we have 聯立二元一次方程.
(1): -y - x = -4
(2): 2x + 3 = -y
(1) + (2) gives -y - x + 2x + 3 = -4 - y
x = -7
y = 11
[ 本帖最後由 斧頭 於 2016-9-18 09:38 PM 編輯 ]