求 e ^i/2的实部和虚部

复数是实数的超集。或者我们可以说复数是数学中数系的一部分。 1799 年，一位数学家 Caspar Wessel 首次发现了复数。很久以后，欧拉引入了将i命名为 √-1 的概念。复数可以用以下方式表示：

z = a + ib

其中a和b是实数， i是虚数，也称为iota，其值为√-1。例如，考虑数字 2/5。这个数字可以写成 2/5 + i*0，其中 a = 2/5 和 b = 0。关于复数的一个有趣的事情是，以图形方式将i乘以向量将使向量逆时针旋转 90°。

复数分类

复数分为以下几种类型：

1. 零复数：这里 a = 0, b = 0 所以 z = 0 + i 0 。例如，0。

2. 纯实数：这里，a ≠ 0，b = 0 所以 z = a + i 0 。例如，5、7、8。

3. 纯虚数：这里，a = 0 , b ≠ 0 所以 z = 0 + ib。例如，9i、-3i、2i。

4、虚数：这里a≠0，b≠0所以z=a+ ib 。例如，2 + 3i、3 – 13i。

欧拉公式

该公式用于建立函数和指数函数之间的关系。欧拉公式是

 e^ix = cos(x) + i * sin(x)

or 

e^iπ as cos π + i * sin π

或者我们可以说，如果任何复数的形式为 e ^{i x} ，那么它可以写成 cos(x) + i * sin(x)。这称为欧拉公式。这里实部是 cos x，虚部是i sin x。

求 e ^i/2的实部和虚部

解决方案：

Let the expression e^i/2 be y.

Therefore t can be written as exp(i/2)

or, t = exp(i * 1/2)

or, t = cos(1/2) + i sin(1/2)

or, t = 0.87758256189 + i * 0.4794255386

Therefore real part is 0.87758256189 and the imaginary part is 0.4794255386.

示例问题

问题 1：求 e ^{i π}的虚部和实部

解决方案：

From Euler’s formula, we can write e^iπ as cos π + i * sin π

cos π = -1 

sin π = 0

Therefore imaginary part is 0 and the real part is -1

So the equation becomes e^iπ +1 = 0, this beautiful equation is called Euler’s identity.

问题 2：求 5 + i 6.9的虚部和实部

解决方案：

This problem is fairly straightforward.  When we are given a complex number like this, 

it is very easy to write the real and imaginary part of it.

imaginary part of the complex number = 6.9

real part of the complex number= 5

问题 3：求复数 50 的实部和虚部。

解决方案：

If a real number is given as a complex number then it is clear that the complex number does not have an imaginary part.

So the imaginary part of the complex number is 0

And, the real part of the complex number is 50.

问题 4：求复数 9 i 的实部和虚部。

解决方案：

If a complex number is given in the form xi then it doesn’t have a real part.

That is real part of the complex number 9i  is 0

Imaginary par is 9i 

问题 5：求复数的实部和虚部 (2 + 3 i )/(1 + i )

解决方案：

In this type of problem, we need to remove the i from the denominator.

If a complex number is given as the ratio of two different complex numbers, then multiply the numerator and 

denominator with the conjugate

The complex conjugate of a complex number is the number itself but with opposite sign. 

For example, there complex conjugate of a number a + ib is a – ib.

So the complex conjugate of the denominator is 1 – i.  

Multiplying this with numerator and denominator we will get, 

((2 + 3i) * (1 – i)) / (1 + i) * (1 – i)

= ((2 + 3i) * (1 – i)) / (1 – i²)

= ((2 + 3i) * (1 – i)) / (1 – (-1)) 

= ((2 + 3i) * (1 – i)) / 2 

= (2(1 – i) ) / 2 + (3i * (1 – i))/2

= 1 – i + 3i/2 + 3/2

= 5/2 + i/2

 = 2.5 + 0.5i 

Therefore the real part of the complex number is 2.5

and the imaginary part of the complex number is 0.5