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

Answer

N/a

Explanation

`(1+2i)/(1- 3i) = (1+2i)/(1- 3i)xx(1+3i)/(1+3i)`
`" "=(1+2i+3i+6i^(2))/(1-9i^(2))=(-5 +5i)/(10)`
` " "=-(1)/(2)+(1)/(2)i`
Let ` " "-(1)/(2)+(1)/(2)i=r(costheta+isintheta)`
`rArr" "rcostheta=-(1)/(2) and rsintheta=(1)/(2)`
Squaring and adding above equations,
` " "r^(2)cos^(2)theta+r^(2)sin^(2)theta=(1)/(4) +(1)/(4)`
` rArr" "r^(2)=(1)/(2)rArrr=(1)/(sqrt(2))`
and `" "(rsintheta)/( rcostheta)=(1//2)/(-1//2)=-1`
`rArrtantheta=- 1=-tan ""(pi)/( 4)`
`rArr tantheta=tan(pi-(pi)/(4))" "(because` In second quadrant `sin` is positive and `cos` is negative)
`rArr" " theta=(3pi)/(4)`
`therefore ` For given complex number, modulus = `(1)/(sqrt(2))` and argument = `(3pi)/(4)`