Question

Show that $1+i^{10}+i^{20}+i^{30}$ is real number.

Answer 1

Answer:

0

Step-by-step explanation:

Hi,

Consider 1 + i¹⁰ + i²⁰ + i³⁰

We know that i² = -1

i⁴ = (i²)² = (-1)² = 1

i¹⁰ = i⁸⁺² = i⁸*i² = (i⁴)²*i² = (1)²*(-1) = -1

i²⁰ =  (i⁴)⁵ = (1)⁵ = 1

i³⁰ = i²⁸⁺² = i²⁸*i² = (i⁴)⁷*i² = (1)⁷*(-1) = -1

Hence,1 + i¹⁰ + i²⁰ + i³⁰

= 1 - 1 + 1 - 1

= 0

Hence, 1 + i¹⁰ + i²⁰ + i³⁰ is a real number.

Hope, it helps !

Answer 2

Solution :

$1+i^{10}+i^{20}+i^{30}$

$\textsf { We know that , i^{2} = -1 }$

= $1+\left ( i^{2} \right )^{5} +\left ( i^{2} \right )^{10} + \left ( i^{2} \right )^{15}$

= $1+ \left ( -1 \right )^{5} + \left ( -1 \right )^{10} +\left ( -1 \right )^{15}$

= $ 1 + (-1 ) + 1 + ( -1 ) $

= $ 1 - 1 + 1 - 1 $

= $ 0 $

$\textsf { Real number }$

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