Question

ls (1+i^14+i^18+i^22)a real number?justify your answer
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Answer 1

It is purely real number...

Answer is -2

1+(i^4)^3.i^2+(i^4)^4.i^2+(i^4)^5.i^2

Here all i^4=1& i^2=-1

....hope this is helpful

Answer 2

Answer:

-2∈R

Step-by-step explanation:

i^14=[i^2]^7=[-1]^7=-1..........................[-1]^odd is always negative

i^18=[i^2]^9=-1     similarly

i^22=[i^2]^11=-1

so,1+i^14+i^18++i^22=1+[-1]+[-1]+[-1]=-2     and -2 is rale number