Write in polar form - (1+3i)/(1-2i)

z = ( - 1 - 3i ) / ( 1 - 2i ) = [ ( - 1 - 3i )( 1 + 2i ) ]/ [ ( 1 - 2i )( 1 + 2i ) ]

= [ - 1 ( 1 + 2i ) - 3i ( 1 + 2i ) ] / [ 1 - 4 (i^2) ]

= [ - 1 - 2i - 3i - 6 (i^2) ] / [ 1 + 4 ]

= [ 6 - 1 - 5i ] / 5

= ( 5 - 5i ) / 5

= ( 1 - i ).

Let z = r ( cosΘ + i sinΘ ). Then,

r cosΘ = 1 and r sinΘ = - 1

On squaring and adding, we get

r = √2

∴ cosΘ = 1 / √2 and sinΘ = - 1 / √2

⇒ tanΘ = - 1

⇒ tan α = | tanΘ | = 1

rArr: α = (pi) / 4

Now, α = (pi) / 4 and z lies in quad. IV.

∴ Θ = - α = - (pi) / 4.

Hence, { - ( 1 + 3i ) / ( 1 - 2i ) } = √2 [ cos( - pi / 4 ) + i sin( - pi / 4 ) ] is the required polar form.