" 10."(-2-(1)/(3)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)

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)/(2)+i(sqrt(3))/(2) , then 8+10z+7z^(2) is equal to a) -(1)/(2)-i(sqrt(3))/(2) b) (1)/(2)+isqrt(3)/(2) c) -(1)/(2)+i(3sqrt(3))/(2) d) (sqrt(3))/(2)i

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

If i=sqrt(-)1, then 4+5(-(1)/(2)+(i sqrt(3))/(2))^(334)+3(-(1)/(2)+(i sqrt(3))/(2))^(365) is equal to (1)1-i sqrt(3)(2)-1+i sqrt(3)(3)i sqrt(3)(4)-i sqrt(3)

If sum_(i=1)^(10) ( (""^10C_(i-1))/(""^10C_i + ""^10C_(i-1) ))^3 = (k)/(11) , then k equals :

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

Let z_1=((1+sqrt(3)i)^2(sqrt(3)-i))/(1-i) and z_2=((sqrt(3)+i)^2(1-sqrt(3)i))/(1+i) then

Find the area of the triangle whose vertices are (i)\ \ \ (2,3)," "(-1,0)," "(2,-4) (i i)(-5, -1)," "(3, -5)," "(5,2)