(i+ 3 )¹⁰⁰+(i- 3 )¹⁰⁰+2¹⁰⁰ is equal to

Step-by-step solution

Let z₁ = i + 3 = 2( 6 + i 6 ) = 2e^ i /6 . Then z₁¹⁰⁰ = 2¹⁰⁰ e^ i 100 /6 = 2¹⁰⁰ e^ i 50 /3 = 2¹⁰⁰ e^ i (16 + 2 /3) = 2¹⁰⁰ e^ i 2 /3 . Using Euler's formula, z₁¹⁰⁰ = 2¹⁰⁰ ( 2 3 + i 2 3 ) = 2¹⁰⁰ (- 1 2 + i 3 2 ) . Let z₂ = i - 3 = 2( 5 6 + i 5 6 ) = 2e^ i 5 /6 . Then z₂¹⁰⁰ = 2¹⁰⁰ e^ i 500 /6 = 2¹⁰⁰ e^ i 250 /3 = 2¹⁰⁰ e^ i (82 + 4 /3) = 2¹⁰⁰ e^ i 4 /3 . Using Euler's formula, z₂¹⁰⁰ = 2¹⁰⁰ ( 4 3 + i 4 3 ) = 2¹⁰⁰ (- 1 2 - i 3 2 ) . Summing the terms: z₁¹⁰⁰ + z₂¹⁰⁰ = 2¹⁰⁰ (- 1 2 + i 3 2 - 1 2 - i 3 2 ) = 2¹⁰⁰ (-1) = -2¹⁰