Find the imaginary part of the complex number(1+i)^10 + (1-i)^10.

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justaguide's profile pic

justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

We have to find the imaginary part of (1+i)^10 + (1-i)^10.

(1+i)^10 + (1-i)^10

=> [(1 + i)^2]^5 + [(1 - i)^2]^5

=> [1 + i^2 + 2i]^5 + [1 + i^2 - 2i]^5

=> [1 - 1 + 2i]^5 + [1 - 1 - 2i]^5

=> (2i)^5 + (-2i)^5

=> (2i)^5 - (2i)^5

=> 0

The imaginary part of (1+i)^10 + (1-i)^10 is 0

giorgiana1976's profile pic

giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

We'll recall the rectangular form of a complex number:

z = a + b*i, where a is the real part and b is the imaginary part.

We notice that we can write (1+i)^10 = [(1+i)^2]^5

We'll expand the binomial:

(1+i)^2 = 1 + 2i + i^2

We'll replace i^2 by -1:

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

We'll eliminate like terms:

(1+i)^2 = 2i

(1+i)^10 = (2i)^5 = 2^5*i^5

We know that i^5 = i^4*i = 1*i = i

(1+i)^10 = 32 i (1)

We notice that we can write (1-i)^10 = [(1-i)^2]^5

We'll expand the binomial:

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

(1-i)^2 = -2i

(1-i)^10 = (-2i)^5 = -32i (2)

We'll add (1) + (2):

(1-i)^2 + (1-i)^10 = 32 i - 32 i

(1-i)^2 + (1-i)^10 = 0

We can write the result as a complex number, as it follows:

(1-i)^2 + (1-i)^10 = 0 + 0i

We notice that the imaginary part of the complex number is b = 0.

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