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TO FIND THE MODULI AND ARGUMENTS OF THE PRODUCT
AND QUOTIENT OF TWO COMPLEX NUMBERS
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).
z1 = r1(cosθ1+ isinθ1) and z2 = r2(cosθ2+ isinθ2).
- To find the modulus and argument of z1. z2
- 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)
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:
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
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,
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,
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