Question

what is the value of( 1+i)^n+(1-i) ^n​

Answer 1

Answer:

2^(n/2+1)cos(nπ/4)

Step-by-step explanation:

(1+I)^n + (1 - I)^n

multiplying and dividing by 2^(n/2) i.e. (√2)^n

2^(n/2)*{(1/√2 + I/√2)^n + (1/√2 - I/√2)^n)}

= 2^(n/2){(cosπ/4 + isinπ/4)^n + (cosπ/4 - isinπ/4)^n}

= 2^(n/2)(cosnπ/4 + isin nπ/4 + cos nπ/4 - isin nπ/4)

= 2^(n/2)(2cos nπ/4)

=2^(n/2 + 1)cos nπ/4

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