Question

Find the value of i^49+i^68+i^89+i^110​

Answer 1

Answer: i^49+i^68+i^89+i^110=(2*i)

Step-by-step explanation:
We know, (i^4)=(√-1)^4=1 So, (i^4)=1
Then, (i^8)=[(i^4)^2]
We got above that, (i^4)=1
So, (i^8)=[(1)^2]=1
Similarly, (i^12)=[(i^4)^3]=[(1)^3]=1
Similarly, (i^16)=[(i^4)^4]=[(1)^4]=1
So we got,
(i^4)=(i^8)=(i^12)=(i^16)=1

This means that,
[i^(multiple of 4)=1]
Hence (i^20)=(i^24)=(i^28)=1 and so on .

i^49 + i^68+ i^89 + i^110=
(i^48)*i + (i^68) + (i^88)*i + (i^108)*(i^2) -equation (1)

{Since, [i^(multiple of 4)=1]
Hence
(i^48)=((i^68)=(i^88)=(i^108)=1}

Solving equation (1) further,
(1)*i + (1) + (1)*i + (1)*(i^2)=
i + 1 + i + (i^2)=
i + 1 + i + (√-1)^2=
i + 1 + i + (-1)=
(2*i)

Answer 2

Answer:

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