If ( 3 +i)¹⁰⁰=2⁹⁹(a+i b) , then a²+b² is equal to

Step-by-step solution

Given the expression ( 3 + i)¹⁰⁰ = 2⁹⁹(a + ib) . Express the complex number z = 3 + i in polar form. The modulus is r = | 3 + i| = ( 3 )^2 + 1^2 = 3 + 1 = 2 . The argument is given by = 1 3 , so = 6 . Thus, 3 + i = 2( 6 + i 6 ) . Using De Moivre's Theorem, ( 3 + i)¹⁰⁰ = [2( 6 + i 6 )]¹⁰⁰ = 2¹⁰⁰( 100 6 + i 100 6 ) . Simplifying the angle, 100 6 = 50 3 = 16 + 2 3 . Therefore, ( 3 + i)¹⁰⁰ = 2¹⁰⁰( 2 3 + i 2 3 ) = 2¹⁰⁰(- 1 2 + i 3 2 ) = 2⁹⁹(-1 + i 3 ) . Comparing this with 2⁹⁹(a + ib) , we get a = -1 and b = 3 . Then a^