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

Answer

N/a

Explanation

(i) `(1+7i)/((2-i)^(2))=(1+7i)/(4+i^(2)-4i)=(1+7i)/( 3-4i)`
`" "=(1+7i)/(3-4i)xx(3+4i)/(3+4i)`
` " =(3+ 4i+21i+28 i^(2))/(9-16i^(2))`
`" "=(-25+25i)/(25 )=-1+i `
Then, let given expression
` " "(1+7i)/((2-i)^(2))or-1+i=r(costheta+isintheta)`
`rArr" "rcostheta=-1 and rsintheta=1`
` rArrr^(2)cos^(2) theta+ r^(2)sin^(2)theta=(-1 )^(2)+(1)^(2) =1+ 1`
`rArr" "r^(2) =2 " "rArr" "r=sqrt(2)`
and `" "(rsintheta)/(rcostheta)=(1)/(-1)`
`rArr" "tantheta=- 1=-tan""(pi)/(4)`
`" "=tan(pi- (pi)/(4))`
(`because` In second quadrant `cos` and `tan` are negative.)
`rArr" "theta=(3pi)/(4)`
Therefore, `(1+7i)/((2-i)^(2))=sqrt(2)[cos""(3pi )/(4 )+isin""(3pi)/(4))]`
(ii) Given expression
`" "(1+ 3i)/( 1-2i) =(1+3i\)/(1- 2i)xx(1+ 2i)/(1+ 2i)`
`" "=(1+2i+3i+6i^(2))/(1^(2)-4i^(2))`
`" "=(1+5i+6xx(-1))/(1 - 4(-1))`
`= (-5+5i)/(5)=-1+i`
Then, let given expression ` (1+3i)/( 1-2i)or (-1+i)=r(costheta+isintheta)`
`therefore" "rcostheta=-1" "...(1)`
`" " rsintheta=1" "...(2)`
Squring and adding equations (1) and (2)
`r^(2)cos^(2)theta+ r ^(2 )sin^(2 )theta= (-1)^(2)+(1)^(2)`
`rArr" "r^(2)=2`
`rArr" "r=sqrt(2)`
Divide equation (2) by (1)
`" "(rsintheta)/(rcostheta)=(1)/(-1)`
`rArr" "tantheta=-1=- tan""(pi)/(4)= (tanpi-( pi)/(4))`
`rArr" "theta=(pi-(pi)/(4))(because` In second quadrant `cos` is negative and `sin` is positive. `)`
` rArr" "theta =(3pi)/(4)`
`therefore` Polar form of given complex number is
` " "=sqrt(2)(cos""(3pi)/(4)+isin""(3pi)/(4))`