简化表达式 7/10i

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。 i 的值为 (√-1) 或者我们可以写成 i ² = -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.

实数和虚数

数字表示一个数制，如正数、负数、零、整数、有理数、无理数、分数等，称为实数。这些表示为 Re()。例如：19、-45、0、1/7、2.8、√5等，都是实数。

非实数称为虚数。对一个虚数进行平方后，结果为负。虚数表示为 Im()。示例：√-8、√-9、√-11 都是虚数。这里的“i”是一个名为“iota”的虚数。

简化表达式 7/10i

解决方案：

Given: 7/10i

Standard form of numerator, 7 = 7 +0i

Standard form of denominator, 10i = 0 +10i

Conjugate of denominator, 0 + 10i = 0 – 10i

Multiply the numerator and denominator with the conjugate,

Therefore, {(7 + 0i) / (0 + 10i)} × {(0 – 10i)/(0 – 10i)}

= {7(0 – 10i)} / {0 – (10i)²}

= {0 – 70i} / {0 – (100(-1))}

= {-70i} / 100           

= 0 – 70/100i

= -7/10i

类似问题

问题 1：求解 (1 – 5i) / (-2i)

解决方案：

Given: (1 – 5i) / (-2i)

Standard form of denominator, -2i = 0 – 2i

Conjugate of denominator, 0 – 2i = 0 + 2i

Multiply the numerator and denominator with the conjugate,

Therefore, {(1 – 5i) / (0 – 2i)} × {(0 + 2i)/(0 + 2i)}

= {(1 – 5i)(0 + 2i)} / {0 – (2i)²}

= {0 + 2i – 5i – 10i²} / {0 –  (4(-1))}

= {-3i – 10 (-1)} / 4

= (-3i + 10) / 4

= 10 /4 – 3i/ 4

= 5/2 – (3/4) i

问题 2：化简 (8 + 4i) / (3 + 2i)

解决方案：

Given: (8 + 4i) / (3 + 2i)

Multiplying the numerator and denominator with the conjugate of denominators,

= {(8 + 4i) / (3 + 2i)} × {( 3 – 2i) / (3 – 2i)}

= {(8 + 4i) × (3 – 2i)} / {(3 + 2i) × (3 – 2i)}

= (24 – 16i +12i – 8i²) / {9  -(2i)²}

= (32 – 4i) / (13)

= (32 – 4i) / 13

= 32/13 – 4/13i

问题 3：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/ 2i

问题4：以a + ib, 2(5 + 3i) + i(7 + 7i)的形式表示

解决方案：

Given: 2(5 + 3i) + i(7 + 7i)

= 10 + 6i + 7i + 7i²

= 10 + 13i  + 7(-1)

= 3 + 13i

问题5：进行如下操作，得到a+ib形式的结果？

(2 – √-25) / (1 – √-16)

解决方案：

Given: (2 – √-25 ) / (1 – √-16)                                                  

= {2 – (i)(5)} / {1 – (i)(4)}, {i = √-1}

= (2 – 5i) / (1 – 4i)

= {(2 – 5i) / (1 – 4i)} × {(1 + 4i) / (1 + 4i)}

= {(2 – 5i) (1 + 4i)} / {(1 – 4i) (1 + 4i)}

= {2 + 8i – 5i – (20i²)} /  {(1-16i²)}, {i² = -1}

= {2 + 3i + 20} / {1 – 16(-1)}

= (22 + 3i ) / (1 + 16)

= (22 + 3i)/17

= {(22/17) + (3i/17)}

= 22/17 + 3i/17

问题6：进行如下操作，得到a+ib形式的结果？

(3 – √-16) / (1 – √-9)

解决方案：

Given: (3 – √-16) / (1 – √-9)                                                  

= {3 – (i)(4)} / {1 – (i)(3)}, {i = √-1}

= (3 – 4i ) / (1 – 3i) {Multiplying the numerator and denominator with the conjugate of denominators}

= {(3 – 4i) / (1 – 3i)} × {(1 + 3i) / (1 + 3i)}

= {(3 – 4i) (1 + 3i)} / {(1 – 3i) (1 + 3i)}

= {3 + 9i – 4i – (12i²)} / {(1 – 9i²)}, {i² = -1}

= {3 + 5i + 12} / {1 – 9(-1)}

= (15 + 5i) / (1 + 9)

= (15 + 5i)/10     

= 15/10 + 5i/10

= 3/2 + 1/2i

问题 7：简化给定的表达式 4/10i

解决方案：

Given: 4/10i

Standard form of numerator, 4 = 4 + 0i

Standard form of denominator, 10i = 0 + 10i

Conjugate of denominator, 0 +10i = 0 – 10i

Multiply the numerator and denominator with the conjugate,

Therefore, {(4 + 0i) / (0 + 10i)} × {(0 – 10i)/(0 – 10i)}

= {4(0 – 10i)} / {0 – (10i)²}

= {0 – 40i} / {0 – (100(-1))}

= {-40i} / 100          

= 0 – 40/100i

= -2/5i