(1 + i)6 + (1 – i)3
Let us simplify we get,
(1 + i)6 + (1 – i)3 = {(1 + i)2 }3 + (1 – i)2 (1 – i)
= {1 + i2 + 2i}3 + (1 + i2 – 2i)(1 – i)
= {1 – 1 + 2i}3 + (1 – 1 – 2i)(1 – i)
= (2i)3 + (– 2i)(1 – i)
= 8i3 + (– 2i) + 2i2
= – 8i – 2i – 2 [since i3 = – i, i2 = – 1]
= – 10 i – 2
= – 2(1 + 5i)
= – 2 – 10i
(1 + i)6 + (1 – i)3 = – 2 – 10i
Answered by Aaryan | 1 year agoShow that 1 + i10 + i20 + i30 is a real number?
Solve the quadratic equations by factorization method only 6x2 – 17ix – 12 = 0
Solve the quadratic equations by factorization method only x2 + (1 – 2i)x – 2i = 0
Solve the quadratic equations by factorization method only x2 + 10ix – 21 = 0
Solve the quadratic equations by factorization method only 17x2 – 8x + 1 = 0