Convert the following in the polar form : (i) `(1+7i)/((2-i)^2)`(ii) `(1+3i)/(1-2i)`

Convert the following in the polar form : (i) `(1+7i)/((2-i)^2)`(ii) `

1.	Convert the following in the polar form : (i) `(1+7i)/((2-i)^2)`(ii) `(1+3i)/(1-2i)`
Answer» Let `z =((1+7i))/((2-i)^(2))=((1+7i))/((4+i^(2)-4i))=((1+7i))/((3-4i))xx((3+4i))/((3+4i))` `rArr" "z=((1+7i)(3+4i))/((9+16))=(-25+25i)/(25)=(-1+i)`. Let its polar form be `z = r(cos theta + i sin theta)`. Now, `r=\|z\|=sqrt((-1)^(2)+1^(2))=sqrt(2)`. Let `alpha` be the acute angle, given by `tan alpha=\|(Im(z))/(Re(z))\|=\|(1)/(-1)\|= 1 rArr alpha = (pi)/(4)`. Clearly, the point representing `z = (-1+i)` is P`(-1, 1)`, which lies in the second quadrant. `therefore" "arg(z) = theta = (pi-alpha)=(pi-(pi)/(4))=(3pi)/(4)`. Thus, `r=\|z\|=sqrt(2) and theta = (3pi)/(4)`. Hence, the required polar form is `z = sqrt(2)("cos"(3pi)/(4) + "i sin"(3pi)/(4))`.