令 z = 8 + 3i 和 w = 7 + 2i,求 z/w 和 zw
实数和虚数之和称为复数。这些是可以写成 a+ib 形式的数字,其中 a 和 b 都是实数。它用 z 表示。在复数形式中,值“a”称为实部,用 Re(z) 表示,“b”称为虚部 Im(z)。它也被称为虚数。在复数形式 a +bi 中,“i”是一个称为“iota”的虚数。
The value of i is (√-1) or we can write as i2 = -1.
例如
- 3 + 4i 是复数,其中 3 是实数 (Re),4i 是虚数 (Im)。
- 2 + 5i 是复数,其中 2 是实数 (Re),5i 是虚数 (im)
The Combination of real number and imaginary number is called a Complex number.
对于 z = 8 + 3 i 和 w = 7 + 2 i,求 z/w。即确定 (8 + 3i) (7 + 2i) 并尽可能简化,将结果写成 a + bi 的形式,其中 a 和 b 是实数。
解决方案:
Given: z = 8 + 3 i
w = 7 + 2i
To find z/w,
= (8 + 3i )/ (7 + 2i)
Multiplying the numerator and denominator with the conjugate of denominators.
= {(8 + 3i) / (7 + 2i) } × {(7 – 2i) / (7 – 2i)}
= {(8 + 3i)(7 – 2i)} / {(7 + 2i)(7 – 2i)}
= {56 – 16i + 21i – 6i2} / {(7)2 – (2i)2}
= {56 + 5i – 6(-1)} / {49 – 4(i2 )}
= {56 + 5i +6} / {49 +4}
= (62 +5i) / 53
z/w = 62/53 + 5/53i
Now, (8 + 3i) (7 + 2i)
= {56 + 16i + 21i + 6i2}
= {56 + 37i + 6(-1)}
= {56 + 5i – 6}
= 50 + 5i
类似问题
问题 1:以标准形式表达 (2 – i)/(1 + i)?
解决方案:
Given: (2 – i)/(1 + i)
Multiplying the numerator and denominator with the conjugate of denominators,
= {(2 – i)/(1 + i) × (1 – i)/(1 – i)}
= {(2 – i)(1 – i)} / {(1)2 – (i)2}
= {2 – 2i – i – i2} / (1-i2)
= {2 – 3i – (-1)} / (1+1)
= ( 3 – 3i) / 2
= 3/2 – 3/ 2 i
问题2:化简为a + ib, (-5i)(2/8i)
解决方案:
Given: (-5i)(2/8i)
= (- 10/8 )i2
= (- 10/8 )(-1)
= 10/8 + 0i
= 5/4 + 0i
问题 3:对于 z = 3 + 3 i 和 w = 5 + 2 i,求 z/w。即确定 (3 + 3i) (5 + 2i) 并尽可能简化,将结果写成 a+bi 的形式,其中 a 和 b 为实数。
解决方案:
Given: z = 3 + 3 i
w = 5 + 2i
To find z/w
= (3 + 3i )/ (5 + 2i)
Multiplying the numerator and denominator with the conjugate of denominators.
= {(3 + 3i) / (5 + 2i)} × {(5 – 2i) / (5 – 2i)}
= {(3 + 3i)(5 – 2i)} / {(5 + 2i)(5 – 2i)}
= {15 – 6i +15i – 6i2} / {(5)2 – (2i)2}
= {15 + 9i – 6(-1)} / {49 – 4(i2 )}
= {15 + 9i + 6} / {49 + 4}
= (21 + 9i) / 53
z/w = 21/53 + 9/53i
Now (3 + 3i) (5 + 2i)
= {15 + 6i +15i + 6i2}
= {15 + 21i + 6(-1)}
= {15 + 21i – 6}
= 9 + 21i
问题 4:执行指定的操作并以标准格式写出答案:(2 – 14i)(2 + 14i)
解决方案:
Given: (2 – 14i)(2 + 14i)
= {(4 + 28i – 28i -196i2 )}
= ( 4 + 196)
= 200 + 0i
问题 5:执行指定的操作,并以标准格式写出答案:(7 + 2i) × (5 – 4i)
解决方案:
(7 + 2i) × (5 – 4i)
= (35 – 28i + 10i – 8i2)
= {35 – 18i – 8(-1)}
= 35 – 18i + 8
= 43 – 18i
问题6:简化-3 + 8i?
解决方案:
The multiplicative inverse of a complex number z is simply 1/z.
It is denoted as: 1 / z or z – 1 (Inverse of z)
Here z = -3 + 8i
Therefore z = 1/z
= 1 / (-3 + 8i)
Now rationalizing,
= 1/(-3 + 8i) × (-3 – 8i)/(-3 – 8i)
= (-3 – 8i ) / {(-3)2 – 82i2}
= (-3 – 8i) / {9 + 64}
= (-3 – 8i)/ (73)
= -3/73 – 8i/73
问题 7:以下问题 (-3i)(9i)(-1) 的答案是什么。
解决方案:
Given: (-3i)(9i)(-1)
= -3i × 9i × (-1)
= -27i2 × -1 {i2 = -1}
= -27 (-1) × -1
= 27 × -1
= -27 + 0i