實數與複數,要詳細過程.

[匿名]

2010-10-02 10:45 am

1.若a及b為實數且(2i-1)(4+i)=2a+(b+1)i,求a及b的值.

2.若a及b為實數且(a+bi)(1-3i)=20,求a及b的值

3.若a及b為實數且4a+3bi=(b+8)+(15-a)i,求a及b的值

4.若a及b為實數且i(3i+4)=a+bi,求a及b的值

5.簡化2i÷3-i+5÷1+3i,並以標準式表示答案.

6.簡化4+3i÷5-2i,並以標準式表示答案.

7簡化10-15i÷2-i,並求該公式的實部.

8簡化(5+i)(i-5)

回答 (1)

用戶10012

2010-10-02 4:19 pm

✔ 最佳答案

Q1. Principle : If a + bi = c + di, than a = c and b = d.
(2i - 1)(4 + i) = 8i + 2i^2 - 4 - i = - 6 + 7i.
2a + (b + 1)i.
Based on the above principle, 2a = - 6, so a = -3.
b + 1 = 7, so b = 6.
Q2 to Q4 are similar to Q1, you can solve using the principle.
Q5.
Principle : (a + bi)(a - bi) = a^2 + b^2. Use this to rationalize the denominator.
2i/(3 - i) = 2i(3 + i)/[(3 - i)(3 + i)] = (6i + 2i^2)/[3^2 + 1^2] = (-2 + 6i)/10.
5/(3i + 1) = 5(1 - 3i)/[(1 + 3i)(1 - 3i)] = (5 - 15i)/[1^2 + 3^2] = (5 - 15i)/10
So adding together = [(-2 + 6i) + (5 - 15i)]/10 = (3 - 9i)/10 = 3/10 - 9i/10.
Q6 and Q7 are similar, you can try yourself using the above principle.
Q8.
(5 + i)(i - 5) = 5i - 25 + i^2 - 5i = - 25 - 1 = - 26.
Or using the principle : (5 + i)(i - 5) = - (5 + i)(5 - i) = - [5^2 + 1^2] = - 26.

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