找出下列问题的答案 (1.5 + 3.2i) – (-2.4 – 3.7i)
复数是可以表示为实数和虚数之和的术语。复数可以写成 a+ib 的形式,其中 a 和 b 都是实数。它用 z 表示。
For example:
6+2i is a complex number, where 3 is a real number (Re) and 4i is an imaginary number (Im).
9+5i is a complex number where 2 is a real number (Re) and 5i is an imaginary number (im)
实数和虚数的组合称为复数。
示例: √-5、-8i、-13i 都是虚数。这里的“i”是一个名为“iota”的虚数
找出下列问题的答案 (1.5 + 3.2i) – (-2.4 – 3.7i)
答案:
Given: (1.5 + 3.2i) – (-2.4 – 3.7i)
By open bracket
= 1.5 + 3.2i + 2.4 + 3.7i
= 3.9 + 9.9i
类似问题
问题 1:简化 (-i)(2i)(-1/8i) 3 ?
解决方案:
Given: (-i)(2i)(-1/8i)3
= (-i)(2i) (-1/8i)3
= -2i2 (-1/512i3)
= -2i2 (-1/512 i3)
= -2(-1) {(-1/512)(- i)} {i2 = -1 and i3 = -i}
= 2(1/512i)
= (2/512)i
= 0 + 1/256i
问题 2:表达 (1-i) – (-1+6i)?
解决方案:
Given : (1-i) – (-1+6i)
= 1 – i + 1 – 6i
= 2 – 7i
问题 3:化简 (5 -3i) 3 ?
解决方案:
Given: (5 -3i)3
Here we will use identity (a-b)3 = a3 – b3 – 3a2b + 3ab2
= (5)3 – (3i)3 – 3(5)2(3i) + 3 (5) (3i)2
= 125 – (27i3) – 225i + 15(9i2)
= 125 -[27(-i)] – 225 i + 15(-9)
= 125 +27i -225i – 135
= -10 – 198i
问题 4:表达 {1/5 + 2/5i} – {4 + 5/2i}
解决方案:
Given: {1/5 +2/5 i} – { 4 + 5/2 i}
= {(1+2i)/5} – {(8+5i)/ 2}
= [{2(1+2i) – 5(8+5i)} /10]
= {2+4i-40 -25i} /10
= (-38 -21i)/10
= -38/10 – 21/10 i
= -19/5 – 21/10 i
问题 5:简化 (-6i)(7i)(-2)
解决方案:
Given: (-6i)(7i)(-2)
= -6i x 7i x (-2)
= -42i2 x -2 {i2 = -1}
= -42 (-1) x -2
= 42 x -2
= -84 + 0i
问题 6:证明 {(2+3i) / (3+4i)} {(2-3i)/(3-4i)} 复数是纯实数?
回答:
Given : {(2+3i) / (3+4i) } { (2-3i)/(3-4i) }
= {(2+3i)(2-3i) } / {(3+4i)(3-4i)}
= {4 -6i +6i -9(i2)} / {9 -12i + 12i – 16(i2)}
= {4 +9} / {9 +16}
= 13/25 + 0i
Therefore {(2+3i) / (3+4i)} {(2-3i)/(3-4i)} it is purely real
问题 7:执行指定的操作,并以标准格式 (3-i)/(1+2i) 写出答案?
解决方案:
Given : (3-i)/(1+2i)
Multiply with the conjugate of denominator
= {(3-i)/(1+2i) x (1-2i)/(1-2i)}
= {(3-i)(1-2i)} / {(1)2 -(2i)2} {difference of squares formula . i.e (a+b)(a-b) = a2 – b2}
= {3 -6i -i + 2i2} / {1-4(-1)}
= {3-7i -2} / {1 + 4} { i2 = -1 }
= (1-7i)/5
= 1/5 -7/5 i