" Buesiót "(-2-(1)/(3),i)^(3)

(-2-i(1)/(3))^(3)

Convert the following in the form of (a+ib) : (i) (1+i)^(4) (ii) (-3+(1)/(2)i)^(3) (iii) (1-i)(3+4i) (iv) (1+i)(1+ 2i)(1+ 3i) (v) (3+5i)/(6-i) (vi) ((2+3i)^(2))/(2+i) (vii) ((1+ i)(2+i))/((3+i)) (viii) (2-i)^(-3)

Convert the following in the form of (a+ib) : (i) (1+i)^(4) (ii) (-3+(1)/(2)i)^(3) (iii) (1-i)(3+4i) (iv) (1+i)(1+ 2i)(1+ 3i) (v) (3+5i)/(6-i) (vi) ((2+3i)^(2))/(2+i) (vii) ((1+ i)(2+i))/((3+i)) (viii) (2-i)^(-3)

If z_(1)=1-2i, z_(2)=1+i and z_(3) =3+4i," then "((1)/(z_(1))+(3)/(z_(2)))(z_(3))/(z_(2))=

((1+i)^(2))/(3-i)

If a_(1),a_(2),a_(3),... are in A.P.and a_(i)>0 for each i,then sum_(i=1)^(n)(n)/(a_(i+1)^((2)/(3))+a_(i+1)^((1)/(3))a_(i)^((1)/(3))+a_(i)^((2)/(3))) is equal to

((1)/(3)+3i)^(3)=((1)/(3))^(3)+(3i)^(3)+3(1)/(3))(( 1)/(3)+3i)

((1-i)^(3))/(1-i^(3))=-2

Reduce ((1)/(1+2i)+(3)/(1-i))((3-2i)/(1+3i)) to the form (a + ib).