如何求复数的绝对值?
复平面上原点与给定点之间的距离称为复数的绝对值。实数的绝对值是数字本身,用模数表示,即|x|。
因此,任何值的模数都会给出一个正值,这样;
|6| = 6
|-6| = 6
现在,在复数的情况下,求模有不同的方法,
假设 z = a+ib 是一个复数。那么,z 的模数将是:
|z| = √(a 2 +b 2 ),当我们在复平面上应用勾股定理时,就得到了这个表达式。
因此,复数的 mod,z 从 0 扩展到 z,实数 x 和 y 的 mod 分别从 0 扩展到 x 和 0 到 y。现在它们形成一个直角三角形,其中锐角的顶点为0。
So, |z|2 = |a|2+|b|2
|z|2 = a2 + b2
|z| = √(a2+b2)
示例问题
问题1:求下列复数的绝对值。 z = 2-4i
解决方案:
The absolute value of a real number is the number itself and represented by modulus,
To find the absolute value of complex number,
Given : z = 2-4i
we have : |z| = √(a2+b2)
here a = 2, b = -4
|z| = √(a2+b2)
= √(22+(-4)2)
= √(4 +16)
= √20
hence the absolute value of complex number. z = 3-4i is 5
问题2:求下列复数的绝对值。 z = 3-9i
解决方案:
The absolute value of a real number is the number itself and represented by modulus,
To find the absolute value of complex number,
Given : z = 3 – 9i
we have: |z| = √(a2+b2)
here a = 3, b = -9
|z| = √(a2+b2)
= √(32+(-9)2)
= √(9 +81)
= √90
hence the absolute value of complex number. z = 5 – 9i is √90
问题 3:求下列复数的绝对值。 z = 2- 7i
解决方案:
The absolute value of a real number is the number itself and represented by modulus,
To find the absolute value of complex number,
Given: z = 2 – 7i
we have: |z| = √(a2+b2)
here a = 2, b = -7
|z| = √(a2+b2)
= √(22+(-7)2)
= √(4 +49)
= √53
hence the absolute value of complex number. z = 2 – 7i is √53
问题 4:执行指定的操作,并以标准形式写出答案:(2 + 4i) × (3 – 4i)。求其绝对值?
解决方案:
(2 + 4i) × (3 – 4i)
= (6 – 8i + 12i – 16i2)
= 6 + 4i +16
= 22 – 4i
The absolute value of a real number is the number itself and represented by modulus,
To find the absolute value of complex number,
Given : z = 22 – 4i
we have : |z| = √(a2+b2)
here a = 22, b = -4
|z| = √(a2+b2)
= √(22)2+(-4)2)
= √(484+ 16)
= √500
hence the absolute value of complex number. z = 22 – 4i is √500
问题 5:求下列复数的绝对值。 z = 3 – 3i
解决方案:
The absolute value of a real number is the number itself and represented by modulus,
To find the absolute value of complex number,
Given : z = 3 – 3i
we have : |z| = √(a2+b2)
here a = 3, b = -3
|z| = √(a2+b2)
= √(32+(-3)2)
= √(9 +9)
= √18
hence the absolute value of complex number. z = 3 – 3i is √18
问题6:若z 1 、z 2分别为(1 – i)、(-2 + 2i),求Im(z 1 z 2 /z 1 )。
解决方案:
Given: z1 = (1 – i)
z2 = (-2 + 2i)
Now to find Im(z1z2/z1),
Put values of z1 and z2
Im(z1z2/z1) = {(1 – i) (-2 + 2i)} / (1 – i)
= {( -2 +2i +2i -2i2)} / (1-i)
= {(-2 + 4i + 2) / (1 – i)
= {(4i) /(1 – i)}
= {(0+4i) (1 + i)} / {(1 + i)(1- i)}
= {(4i + 4i2) / (1 + 1)
= 4i -4 / 2
=(-4 + 4i) / 2
= -4/2 + 4/2 i
= -2 + 2i
Therefore, Im(z1z2/z1) = 2
问题 7:执行指定的操作并以标准形式写出答案:(2 – 7i)(3 + 7i)
解决方案:
Given: (2 – 7i)(3 + 7i)
= {6+ 14i – 21i – 49i2}
= (-7i +55)
= 55 -7i