Fill in the blanks of the following .
(i) For any two complex numbers `z_(1), z_(2)` and any real numbers a, b, `|az_(1) - bz_(2)|^(2) + |bz_(1) + az_(2)|^(2) =* * * `
(ii) The value of `sqrt(-25) xxsqrt(-9) is … `
(iii) The number `(1 - i)^(3)/(1 -i^(3))` is equal to ...
(iv) The sum of the series `i + i^(2) + i^(3)+ * * * ` upto 1000 terms us ...
(v) Multiplicative inverse of 1 + i is ...
(vi) If `z_(1) "and" z_(2)` are complex numbers such that `z_(1) + z_(2)` is a real number, then `z_(1) = * * *`
(vii) `arg(z) + arg barz "where", (barz ne 0) is ...`
(viii) If `|z + 4| le 3, "then the greatest and least values of" |z + 1| are ... and ...`
(ix) If `|(z - 2)/(z + 2)| =(pi)/(6) ,` then the locus of z is ...
`(x) If |z| = 4 "and arg" (z) = (5pi)/(6), then z = * * *`

Explanation

(i) `|az_(1) - bz_(2)|^(2) +|bz_(1) + az_(2)|^(2)`
= `|az_(1)|^(2) + |bz_(2)|^(2) - 2Re (az_(1)* b barz_(2)) + | bz_(1)|^(2) +|az_(2)|^(2)+ 2 Re (az _(1)* barbz_(2))`
`= (a^(2) + b^(2)) |z_(1)|^(2) + (a^(2) + b^(2)) |z_(2)|^(2)`
`= (a^(2) + b^(2)) (|z_(1)|^(2) + |z_(2)|^(2))`
(ii) `=sqrt(-25)xxsqrt(-9) = i sqrt(25)xxisqrt(9) =i (5xx3) = - 15`
(iii) `(1-i)^(3)/(1-i^(3)) = ((1-i^(3)))/((1-i^(3))(1+i+i^(2))) `
`(1-i)^(2)/(i) = ((1+i^(2)-2i))/(i) =(-2i)/(i) = -2 `
(iv) `i + i^(2) + i^(3)+ ...` upto 1000 terms `= i + i^(2) + i^(3)+ i^(4)+... i^(1000) = 0`
`[:' i^(n) + i^(n+1)+i^(n+2)+i^(n+3)= 0, "where n"in N i.e., underset (n=1) overset1000sum i^(n)=0]`
(v) Multiplicative inverse of `1+ i = (1)/(1+i) = (1+i)/(1+i)^(2) = (1)/(2) (1-i)`
(vi) Let `z_(1) = x_(1) + iy_(1) "and" z_(2) = x_(2) + iy_(2)`
`z_(1) + z_(2) = (x_(1) + x_(2))+ i (y_(1) + y_(2))`, which is real
If `z_(1) + z_(2) "is real, then" y_(1) + y_(2) = 0`
`rArr y_(1) = - y_(2)`
`:' z_(2) = x_(2)-iy_(1)`
`rArr z_(2) = barz_(1) ["when" x_(1) = x_(2)]`
(vii) arg`(z) + arg(barz), (barz ne 0)`
`rArr theta + (-theta) = 0`
(viii) Given that, `|z + 4|le 3`
For the greatest value of `|Z + 1|`.
`rArr | z + 1| = | Z + 4 - 3 | le|z + 4 | + -3|`
`|z + 4 - 3| le 3 + 3`
= `|z + 4 - 3 | le 6`
So, greatest value of `|z +1|`, we knoe that the least value of the modulus of a complex number is zero . So. the least value of `|z + 1|` is zero.
(ix) Given that . `|(z - 2)/(z + 2)| = (pi)/(6)`
`rArr (|x + i -2 |)/(|x + i +2|) = (pi)/(6) rArr (|x - 2 +iy |)/(|x + 2 +iy|) = (pi)/(6)`
`rArr 6|x -2 +iy|= pi|x + 2 +iy|`
`rArr 6sqrt((x -2)^(2) +y^(2))= pisqrt((x + 2)^(2) +y)`
`rArr 36[x^(2)+4-4x + y^(2)]=pi^(2)[x^(2) + 4x + 4+ y^(2)]`
`rArr (36-pi^(2))[x^(2)+(36-pi)^(2)y^(2) - (144 + 4pi^(2)) x + 144 + 4pi^(2) = 0 , ` which is a circle.
(x) Given that, `|z| = 4 "and arg" (z) =(5pi)/(6) `
Let `z = x + iy = r (costheta + isintheta)`
`rArr |z| = r = 4 " and" arg(z) = theta`
`:' tentheta = (5pi)/(6)`
`rArr z= 4("cos"(pi)/(6) + "i sin" (pi)/(6)) = 4 [cos (pi-pi//6) +isin(pi-pi//6)]`

`= 4["-cos"(pi)/(6) + "i sin" (pi)/(6)] = 4[(-sqrt(3))/(2)+i(1)/(2)]=-2sqrt(3)+2i`