    # Modulus And Argument of Product, Quotient Of Complex Numbers TO FIND THE MODULI AND ARGUMENTS OF THE PRODUCT
AND QUOTIENT OF TWO COMPLEX NUMBERS

Let z1 and z2 be two complex numbers and |z1 | = r1, |z2 | = r2, arg z1 = θ1and arg z2 = θ2, where -π< θ1≤π and −π< θ 2≤π. Then,

z1 = r1(cosθ1+ isinθ1) and z2 = r2(cosθ2+ isinθ2).

• To find the modulus and argument of z1. z2

• z1. z2 = r1(cosθ1+ isinθ1). r2(cosθ2+ isinθ2)

= r1 r2[(cosθ1cosθ2- sinθ1 sinθ2)+i(sinθ112+ cosθ1sinθ2) ]

= r1 r2[cos(θ1+ θ2 )+ isin(θ1+ θ2 ) ]

= r(cosθ+ isinθ)

Where, r = r1 r2 and θ = θ1+ θ2

Now |z1 z2 | = r = r1 r2 = |z1 |. |z2 |

Hence, the modulus of the product of two complex numbers is equal to the product of their moduli.

Again, -π< θ1≤π and −π< θ2 ≤π

Hence -2π< θ ≤2π, since θ = θ1+ θ2

or -π< θ+m ≤π (where m = 0 or 2π or -2π)

Hence, arg(z1 z2) = θ + m = θ1+ θ2 + m

Or arg(z1 z2) = arg z1 + arg z2 + m               (1)

(where m = 0 or 2π or -2π).

Hence, it is clear that:

• arg(z1 z2) = arg z1 + arg z2 when -π< arg z1 + arg z2 ≤ π;

• arg(z1 z2) = arg z1 + arg z2 + 2π when -2π< arg z1 + arg z2 ≤ -π;

• arg(z1 z2) = arg z1 + arg z2 - 2π when π< arg z1 + arg z2 ≤ 2π.

• To find the modulus and (z2 ≠ 0)

• Hence, the modulus of the quotient of two complex numbers is equal to the quotient of their moduli.

Since −π< θ2≤π hence, −π< -θ2≤ π and −π< θ1≤π

Hence -2π< θ ≤2π, since θ = θ1- θ2

or -π< θ+m ≤ π (where m = 0 or 2π or -2π)

Hence arg z1/z2 = θ + m = θ1- θ2 + m

or = arg z1 - arg z2 + m               (2)

(where m = 0 or 2π or -2π).

Equations (1) and (2) give the principal values of arguments of (z1 z2) and respectively.

Example1:

If z1 = -3 + 4i and z2 = 12 − 5i, then show that, |z1 z2 | = |z1 |. |z2 | and Solution:

z1 = -3 + 4i, hence |z1 |=√(-3)2 + (4)2 = 5

and z2 = 12 − 5i, hence |z2 |=√(12)2 + (-5)2 = 13

z1 z2 = (-3 + 4i)( 12 − 5i)

= -36 + 15i + 48i − 20i2

= - 36 - 20i2 + 63i

= -36 + 20 + 63i (Since i2 = -1)

= -16 + 63i

Now |z1 z2 | = √(-16)2 + (63)2 = √4225 = 65

Or |z1 z2 | = 5. 13 = |z1 |. |z2 | (Proved) Example2:

Find the arguments of z1 = √3 + i and z2 = -1 - i√3 and hence calculate arg(z1 z2) an .

Solution:

It is evident that in the z-plane the point z1 = √3 + i = (√3, 1) lies in the first quadrant; hence, if arg z1 = θ then, Again in the z-plane the point z2 = -1 - i√3 = (-1, -√3) lies in the third quadrant; hence, if arg z1 = φ then, Online Solution Modulus and Argument of Product, Quotient Complex Numbers Help:

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