Solve the complex no. z which satisfy the following equation:
[3z*/(4z*-1)]= i
Originally posted by rs rs:Solve the complex no. z which satisfy the following equation:
[3z*/(4z*-1)]= i
z= a + bi so z* is a - bi
3a - 3bi = 4(a-bi)i - i
3a - 3bi = 4ai + 4b - i
3a - 3bi = 4b + (4a - 1)i
by comparing coefficients,
3a = 4b -------(1)
-3b = 4a - 1 ---(2)
Solve by simultaneous equations. Yeah?
wow thanks (: haha!