Question

[(1+i)^100 + (1-i)^100]​

Answer 1

Answer:

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Step-by-step explanation:

LHS

(1+I)^100+(1-i)^100

This can be written as

(1+i)^(2*50)+(1-i)^(2*50)

Now i^2=-1 so

Expanding this we get

(1+2i+i^2)^50+(1-2i+i^2)^50

(2i)^50+(-2i)^50

Now we can take out 2^50 common

So

(2^50)(i^50+(-i)^50)

As square of the value is always positive we can write this as

We can write this as

(2^50)(i^(2*25)+(i)^(2*50))

As i^2=-1

So

(2^50)((-1)^25+(-1)^25)=2^50*(-1-1)

=-2^51

LHS=RHS