`(1-2i)^(-3)=?`

(1-i)(1+2i)(1-3i)=

1+(1+i)+(1+i)^(2)+(1+i)^(3)=

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

Represent (1+2i)/(1-3i) in the polar form.

Find the modulus and argument of (1+2i)/(1-3i) .

Convert the following in the form of (a+ib) : (i) (1+i)^(4) (ii) (-3+(1)/(2)i)^(3) (iii) (1-i)(3+4i) (iv) (1+i)(1+ 2i)(1+ 3i) (v) (3+5i)/(6-i) (vi) ((2+3i)^(2))/(2+i) (vii) ((1+ i)(2+i))/((3+i)) (viii) (2-i)^(-3)

Convert the following in the form of (a+ib) : (i) (1+i)^(4) (ii) (-3+(1)/(2)i)^(3) (iii) (1-i)(3+4i) (iv) (1+i)(1+ 2i)(1+ 3i) (v) (3+5i)/(6-i) (vi) ((2+3i)^(2))/(2+i) (vii) ((1+ i)(2+i))/((3+i)) (viii) (2-i)^(-3)

" If "z=((1+i)(1+2i)(1+3i))/((1-i)(2-i)(3-i))" then the principal argument of "z"

(1+2i)/(1+3i) is equal to