12.5 Division in polar form

You can probably guess what happens with division in polar form now – divide modulii
and subtract arguments. Again, check the details for yourself:

(^) (θ 1 )
(^) (θ 2 )=
cosθ 1 +jsinθ 1
cosθ 2 +jsinθ 2

cosθ 1 +jsinθ 1
cosθ 2 +jsinθ 2
×
cosθ 2 −jsinθ 2
cosθ 2 −jsinθ 2
=(cosθ 1 cosθ 2 +sinθ 1 sinθ 2 )+j(sinθ 1 cosθ 2 −sinθ 2 cosθ 1 )
(on using cos^2 θ 2 +sin^2 θ 2 = 1 )
=cos(θ 1 −θ 2 )+jsin(θ 1 −θ 2 )=^ (θ 1 −θ 2 )
Hence
r 1 (θ 1 )
r 2 (θ 2 )

r 1
r 2
(^) (θ 1 −θ 2 )
So, to divide in polar form:

• Divide modulii to obtain the modulus of the quotient


• Subtract arguments to obtain the argument of the quotient.

In symbols ∣

∣

∣

∣

z 1

z 2

∣

∣

∣

∣=

|z 1 |

|z 2 |

arg

(

z 1

z 2

)

=arg(z 1 )−arg(z 2 )

Arg

(

z 1

z 2

)

=Argz 1 −Argz 2 + 2 kπ kan integer

Problem 12.9

Work

√

3 −j

1 −

√

3 j

in Cartesian and polar forms and compare the results.

We already know from Problem 12.8 that

√

3 −j= 2

(

−

π

6

)

and 1−

√

3 j= 2

(

−

π

3

)

So in polar form
√
3 −j
1 −

√

3 j

=

2

(

−

π

6

)

2

(

−

π

3

)=^

(

−

π

6

+

π

3

)

=^

(π

6

)

=cos

(π

6

)

+jsin

(π

6

)

=

√

3

2

+

1

2

j