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Complex Number

Properties of Complex Number Modulus

Properties of Modulus Operations

Let z1z_1 and z2z_2 be complex numbers.

Modulus of a Number, its Negative, and its Conjugate

The modulus of a complex number is equal to the modulus of its negative, and also equal to the modulus of its conjugate.

z1=z1=z1ˉ|z_1| = |-z_1| = |\bar{z_1}|

Explanation:

Recall that if z1=x+iyz_1 = x + iy, then z1=xiy-z_1 = -x - iy and z1ˉ=xiy\bar{z_1} = x - iy.

  • z1=x2+y2|z_1| = \sqrt{x^2 + y^2}
  • z1=(x)2+(y)2=x2+y2|-z_1| = \sqrt{(-x)^2 + (-y)^2} = \sqrt{x^2 + y^2}
  • z1ˉ=x2+(y)2=x2+y2|\bar{z_1}| = \sqrt{x^2 + (-y)^2} = \sqrt{x^2 + y^2}

All three yield the same value.

Modulus of Difference

The modulus of the difference of two complex numbers is the same if the order is reversed.

z1z2=z2z1|z_1 - z_2| = |z_2 - z_1|

Explanation:

This is a direct consequence of the first property. We know z1z2=(z2z1)z_1 - z_2 = -(z_2 - z_1). Then:

z1z2=(z2z1)=z2z1|z_1 - z_2| = |-(z_2 - z_1)| = |z_2 - z_1|

Square of Modulus

The square of the modulus of a complex number is equal to the complex number multiplied by its conjugate.

z12=z1×z1ˉ|z_1|^2 = z_1 \times \bar{z_1}

Explanation:

If z1=x+iyz_1 = x + iy, then z1ˉ=xiy\bar{z_1} = x - iy.

z1×z1ˉ=(x+iy)(xiy)=x2(iy)2=x2i2y2=x2(1)y2=x2+y2z_1 \times \bar{z_1} = (x + iy)(x - iy) = x^2 - (iy)^2 = x^2 - i^2y^2 = x^2 - (-1)y^2 = x^2 + y^2

We also know that z1=x2+y2|z_1| = \sqrt{x^2 + y^2}, so z12=(x2+y2)2=x2+y2|z_1|^2 = (\sqrt{x^2 + y^2})^2 = x^2 + y^2.

Thus, both sides are equal.

Modulus of Product

The modulus of the product of two complex numbers is equal to the product of their individual moduli.

z1×z2=z1×z2|z_1 \times z_2| = |z_1| \times |z_2|

Modulus of Quotient

The modulus of the quotient of two complex numbers is equal to the quotient of their individual moduli (provided the denominator is non-zero).

z1z2=z1z2,for z20\left| \frac{z_1}{z_2} \right| = \frac{|z_1|}{|z_2|}, \quad \text{for } z_2 \neq 0

Triangle Inequality

The modulus of the sum of two complex numbers is less than or equal to the sum of their individual moduli.

z1+z2z1+z2|z_1 + z_2| \leq |z_1| + |z_2|

Explanation:

Geometrically, if we consider z1z_1, z2z_2, and z1+z2z_1 + z_2 as sides of a triangle on the complex plane, this property states that the length of one side (z1+z2|z_1 + z_2|) cannot be greater than the sum of the lengths of the other two sides (z1+z2|z_1| + |z_2|).

Using Modulus Properties

Suppose we are given the complex number z=12i3+4iz = \frac{1 - 2i}{3 + 4i}. Find z|z|!

Solution:

We can view z=z1z2z = \frac{z_1}{z_2} with z1=12iz_1 = 1 - 2i and z2=3+4iz_2 = 3 + 4i.

Using the Modulus of Quotient property:

z=12i3+4i=12i3+4i|z| = \left| \frac{1 - 2i}{3 + 4i} \right| = \frac{|1 - 2i|}{|3 + 4i|}

Now we calculate the moduli of z1z_1 and z2z_2:

z1=12i=12+(2)2=1+4=5|z_1| = |1 - 2i| = \sqrt{1^2 + (-2)^2} = \sqrt{1 + 4} = \sqrt{5}
z2=3+4i=32+42=9+16=25=5|z_2| = |3 + 4i| = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5

Therefore,

z=55|z| = \frac{\sqrt{5}}{5}

This is much easier than first multiplying by the conjugate of the denominator and then calculating the modulus.

Exercise

  1. If z1=6+8iz_1 = 6 + 8i and z2=3iz_2 = 3 - i, calculate z1×z2|z_1 \times z_2| using the modulus properties.
  2. If z=512iz = 5 - 12i, prove that z2=z×zˉ|z|^2 = z \times \bar{z}.

Answer Key

  1. We use the property z1×z2=z1×z2|z_1 \times z_2| = |z_1| \times |z_2|.

    Calculate each modulus:

    z1=6+8i=62+82=36+64=100=10|z_1| = |6 + 8i| = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10
    z2=3i=32+(1)2=9+1=10|z_2| = |3 - i| = \sqrt{3^2 + (-1)^2} = \sqrt{9 + 1} = \sqrt{10}

    Then:

    z1×z2=z1×z2=10×10=1010|z_1 \times z_2| = |z_1| \times |z_2| = 10 \times \sqrt{10} = 10\sqrt{10}
  2. Given z=512iz = 5 - 12i.

    Calculate the left side (z2|z|^2):

    z=512i=52+(12)2=25+144=169=13|z| = |5 - 12i| = \sqrt{5^2 + (-12)^2} = \sqrt{25 + 144} = \sqrt{169} = 13
    z2=132=169|z|^2 = 13^2 = 169

    Calculate the right side (z×zˉz \times \bar{z}):

    The conjugate of zz is zˉ=5+12i\bar{z} = 5 + 12i.

    z×zˉ=(512i)(5+12i)=52(12i)2=25144i2=25144(1)=25+144=169z \times \bar{z} = (5 - 12i)(5 + 12i) = 5^2 - (12i)^2 = 25 - 144i^2 = 25 - 144(-1) = 25 + 144 = 169

    Since the left side (169) equals the right side (169), the statement z2=z×zˉ|z|^2 = z \times \bar{z} is proven.