We have ,

1+2i

1−3i

=

1+2i

1−3i

×

1+3i

1+3i

=

1+3i+2i−6

1+9

=

−5+5i

10

= -

1

2

+

1

2

i
Let -

1

2

=rcosθ ... (i) and

1

2

= r sin θ ... (ii)
Squaring and adding (i) and (ii), we get
r2(cos2θ+sin2θ) =

1

4

+

1

4

=

2

4

=

1

2

⇒ r2 =

1

2

⇒ r =

1

√2

Substituting the value of r in (i) and (ii), we get

1

√2

cos θ = −

1

2

,

1

√2

sin θ =

1

2

⇒ cos θ = −

1

√2

, sin θ =

1

√2

⇒ cos θ = - cos (

π

4

) , sin θ = sin

π

4

Here, cos θ < 0 , sin θ > 0.
∴ θ lies in second quadrant.
θ = π -

π

4

=

3π

4

∴ Modulus is

1

√2

and argument is

3π

4