Question

Is ($1+i^{14}+i^{18}+i^{22}$) a real number? Justify your answer.

Answer 1

Answer:

-2

Step-by-step explanation:

Hi,

We know that i² = -1

i⁴ = 1

i¹⁴ can be written as i¹²⁺² = i¹²*i² = (i⁴)³*i²

= (1)³*i² = -1

i¹⁸ can be written as i¹⁶⁺² = i¹⁶*i² = (i⁴)⁴*i²

= (1)⁴*i² = -1

i²² can be written as i²⁰⁺² = i²⁰*i² = (i⁴)⁵*i²

= (1)⁵*i² = -1

So, (1 +  i¹⁴ + i¹⁸ + i²²)

= ( 1 -1 -1 -1)

= -2 which is a real number

Hope, it helps !

Answer 2

$\textsf \green { Solution }$

Given

($1+i^{14}+i^{18}+i^{22}$)

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

= ($1+ \left ( i^{2} \right )^{7} + \left ( i^{2} \right )^{9} + \left ( i^{2} \right )^{11}$)

= ($1+ \left ( -1 \right )^{7} + \left ( -1 \right )^{9} + \left ( -1 \right )^{11}$)

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

= $ 1 - 1 - 1 - 1 $

= $ -2 $

$\textt \blue { Real number }$

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