如何除以复数？

复数是实数和虚数之和。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)。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.

虚数

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：√-3、√-7、√-11 都是虚数。这里的“i”是一个名为“iota”的虚数。

如何除复数？

解决方案：

There are few steps to divide the complex number:

The process of dividing  two complex numbers is slightly different from that of the division process of two real numbers. It is the concept of rationalizing the denominator in the case of fractions involving irrational numbers as their denominators.  

The following steps are involved:

Step 1: Assure that both the numerator and denominator are in the standard form of complex numbers, i.e., z = a + ib.

Step 2: Calculate the conjugate of the complex number that is at the denominator of the fraction. Say, if the denominator is a + bi, then its conjugate is a − bi.

Step 3: Multiply the numerator and denominator with the conjugate or with both the terms of the fraction.

Step 4: Simplify the numerator with distributive property

Step 5: Simplify the denominator with the use of difference of squares formula . i.e (a+b)(a-b) = a² – b²

Step 6: In last of complex number separate  its real and imaginary parts.

Suppose there are two complex numbers z₁ = a+bi, z₂ = m+ni

Divide: a+bi / m+ni = {(a+bi)(m-ni)} / {(m+ni)(m-ni)}

                                = {am – ani + mbi – bni2 } / (m² + n²)

                                = {am – ani + mbi – bn (-1)} / (m² + n²)

                                = {am – ani + mbi +bn} / (m² + n²)

                                = {(am + bn) + (mb – an)i} / (m²+n²)

                                = {(am + bn) / (m² +n²)} + {(mb-an)/ (m² +n²)} i

here  {(am + bn) / (m² +n²)} is real part and {(mb-an)/ (m² +n²)} i is imaginary part

示例问题

问题 1：简化 {(-3 – 5i) / (2 +2i)}？

解决方案：

Given {(-3 – 5i) / (2 +2i) }

conjugate of denominator 2+2i is 2-2i

Multiply the numerator and denominator with the conjugate 

Therefore {(-3 – 5i) / (2 +2i) } x {(2-2i) / (2-2i) }

              = {-6 -6i -10i +10i²  } /  { 2² – (2i)² }                  {difference of squares formula . i.e (a+b)(a-b) = a² – b² }

              =  { -6 -6i -10i + 10 (-1) } /  { 4 – 4(-1) }         { i² = -1 }

              = {  -6 -6i -10i -10 } / { 4 + 4 } 

              = (-16 – 16i ) / 8 

              = -16 /8  – 16i /8

              = -2  -2i

问题 2：简化 {(-3 + 5i) / (2 – 2i)}？

解决方案：

Given {(-3 + 5i) / (2 – 2i)}

conjugate of denominator 2-2i is 2+ 2i

Multiply the numerator and denominator with the conjugate

Therefore {(-3 + 5i) / (2 -2i) } x {(2+2i) / (2+2i) }

             = {-6 + 6i +10i -10i²  } /  { 2² – (2i)² }                  {difference of squares formula . i.e (a+b)(a-b) = a² – b² }

             = {-6 + 6i +10i – 10 (-1) } /  { 4 – 4(-1)}               { i² = -1 }

             = {-6 + 6i +10i +10 } / { 4 + 4 }

             = (4 +16i ) / 8

             = 4 /8  +16i /8

             = 1/2  + 2i

问题 3：求解 (1-5i) / (-3i)？

解决方案：

Given : (1-5i) / (-3i)

standard form of denominator -3i = 0 – 3i

conjugate of denominator     0-3i = 0 +3i

Multiply the numerator and denominator with the conjugate

therefore, {(1-5i) / (0 -3i) } x { (0+3i )/( 0 +3i )}

              = { (1-5i)(0+3i )  } / { 0 – (3i)² }

              = { 3i – 15i² } / { 0 –  (9(-1) ) }

              = { 3i – 15 (-1) } / 9

              = ( 3i +15 ) / 9

              = 15 / 9 + 3i/ 9

              = 5/3  + (1/9) i

问题4：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 – √-25 ) / (1 – √-16) =  {2 – (i)(5)} / { 1 – (i)(4) }                 { i = √-1 }

=   (2-5i ) / (1-4i)

= { (2-5i ) / (1-4i)} x {(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5：若z ₁ 、z ₂分别为(1-i)、(-2 +4i)，求Im(z ₁ z ₂ /z ₁ )

解决方案：

Given : z₁ = (1-i) 

                            z₂ = (-2 +4i)

now to find Im(z₁z₂/z₁) 

put values of z₁ and z₂

        Im(z₁z₂/z₁)     = {(1-i) (-2 +4i) } / (1-i)

                               = {( -2 +4) +i(2+4)} / (1+i)

                               = {( 2+6i) /(1+i)}

                               = {( 2+6i) /(1+i)} x { (1+i) / (1- i) }

                               = {( 2+6i) (1+i)} / {(1+i)(1-i)}

                               = {(2+6) +i(6-2) } / (1+1)

                               = 4+2i

therefore Im(z₁z₂/z₁) = 2

问题 6：用标准形式表达 (1-i)/(1+i) 并找到它们的共轭？

解决方案：

Given : (1-i)/(1+i)

= {(1-i)/(1+i) x (1-i)/(1-i)}

= { (1-i)²} / {(1)² -(i)²}

= {1 -2i -1 } / (1-i²)

= -2i / 2

= 0 – 2i/2

= 0 -i

now conjugate =  0+i

问题7：以a+ib的形式表示，3(7+7i) + i(7+7i)

解决方案：

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

                        = 21 + 21i + 7i + 7i²

                        = 21 + 28i  + 7(-1)

                        = 21 +28i -7

                        = 14 +28i