# Modulus And Argument of Product, Quotient Of Complex Numbers

TO FIND THE MODULI AND ARGUMENTS OF THE PRODUCT

AND QUOTIENT OF TWO COMPLEX NUMBERS

AND QUOTIENT OF TWO COMPLEX NUMBERS

Let z

z

_{1}and z_{2}be two complex numbers and |z_{1}| = r_{1}, |z_{2}| = r_{2}, arg z_{1}= θ_{1}and arg z_{2}= θ_{2}, where -π< θ_{1}≤π and −π< θ_{2}≤π. Then,z

_{1}= r_{1}(cosθ_{1}+ isinθ_{1}) and z_{2}= r_{2}(cosθ_{2}+ isinθ_{2}).**To find the modulus and argument of z**_{1}. z_{2}- arg(z
_{1}z_{2}) = arg z_{1}+ arg z_{2}when -π< arg z_{1}+ arg z_{2}≤ π; - arg(z
_{1}z_{2}) = arg z_{1}+ arg z_{2}+ 2π when -2π< arg z_{1}+ arg z_{2}≤ -π; - arg(z
_{1}z_{2}) = arg z_{1}+ arg z_{2}- 2π when π< arg z_{1}+ arg z_{2}≤ 2π. **To find the modulus and (z**_{2}≠ 0)

z

_{1}. z

_{2}= r

_{1}(cosθ

_{1}+ isinθ

_{1}). r

_{2}(cosθ

_{2}+ isinθ

_{2})

= r

_{1}r

_{2}[(cosθ

_{1}cosθ

_{2}- sinθ

_{1}sinθ

_{2})+i(sinθ

_{11}

_{2}+ cosθ

_{1}sinθ

_{2}) ]

= r

_{1}r

_{2}[cos(θ

_{1}+ θ

_{2})+ isin(θ

_{1}+ θ

_{2}) ]

= r(cosθ+ isinθ)

Where, r = r

_{1}r

_{2}and θ = θ

_{1}+ θ

_{2}

Now |z

_{1}z

_{2}| = r = r

_{1}r

_{2}= |z

_{1}|. |z

_{2}|

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

Again, -π< θ

_{1}≤π and −π< θ

_{2}≤π

Hence -2π< θ ≤2π, since θ = θ

_{1}+ θ

_{2}

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

Hence, arg(z

_{1}z

_{2}) = θ + m = θ

_{1}+ θ

_{2}+ m

Or arg(z

_{1}z

_{2}) = arg z

_{1}+ arg z

_{2}+ m (1)

(where m = 0 or 2π or -2π).

Hence, it is clear that:

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

Since −π< θ

_{2}≤π hence, −π< -θ

_{2}≤ π and −π< θ

_{1}≤π

Hence -2π< θ ≤2π, since θ = θ

_{1}- θ

_{2}

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

Hence arg z

_{1}/z

_{2}= θ + m = θ

_{1}- θ

_{2}+ m

or = arg z

_{1}- arg z

_{2}+ m (2)

(where m = 0 or 2π or -2π).

Equations (1) and (2) give the principal values of arguments of (z

_{1}z

_{2}) and respectively.

**Example1:**

If z

_{1}= -3 + 4i and z

_{2}= 12 − 5i, then show that, |z

_{1}z

_{2}| = |z

_{1}|. |z

_{2}| and

**Solution:**

z

_{1}= -3 + 4i, hence |z

_{1}|=√(-3)

^{2}+ (4)

^{2}= 5

and z

_{2}= 12 − 5i, hence |z

_{2}|=√(12)

^{2}+ (-5)

^{2}= 13

z

_{1}z

_{2}= (-3 + 4i)( 12 − 5i)

= -36 + 15i + 48i − 20i

^{2}

= - 36 - 20i2 + 63i

= -36 + 20 + 63i (Since i

^{2}= -1)

= -16 + 63i

Now |z

_{1}z

_{2}| = √(-16)

^{2}+ (63)

^{2}= √4225 = 65

Or |z

_{1}z

_{2}| = 5. 13 = |z

_{1}|. |z

_{2}|

**(Proved)**

**Example2:**

Find the arguments of z

_{1}= √3 + i and z

_{2}= -1 - i√3 and hence calculate arg(z

_{1}z

_{2}) an .

**Solution:**

It is evident that in the z-plane the point z

_{1}= √3 + i = (√3, 1) lies in the first quadrant; hence, if arg z

_{1}= θ then,

Again in the z-plane the point z

_{2}= -1 - i√3 = (-1, -√3) lies in the third quadrant; hence, if arg z

_{1}= φ then,

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