Question

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

Answer 1

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))`.