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(1+i)2n + (1−i)2n,n ∈ N is
- a)a purely real number
- b)a purely imaginary number
- c)a non-real complex number
- d)none of these
Correct answer is option 'A'. Can you explain this answer?
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(1+i)2n+ (1−i)2n,n ∈ N isa)a purely real numberb)a purely i...
I)2n = (1+i)2n
To simplify this expression, we can use the binomial theorem. According to the binomial theorem, (a+b)n = Σ(nCk)an-kbk, where nCk is the binomial coefficient "n choose k" (which represents the number of ways to choose k items from a set of n items), and Σ represents the sum over all values of k from 0 to n.
In our case, we have a=1 and b=i, so we can write:
(1+i)2n = Σ(nCk)1n-k(i)k
Now we just need to simplify the terms involving i. Note that i2 = -1, so any term with an odd power of i will be equal to i times a term with an even power of i (i.e. a real number). We can use this fact to rewrite the sum as:
(1+i)2n = Σ(nCk)(-1)k/2 + iΣ(nCk)(-1)(k-1)/2
The first sum contains only real numbers, so we can pull it out of the parentheses and simplify:
(1+i)2n = Σ(nCk)(-1)k/2 + iΣ(nCk)(-1)(k-1)/2
= Σ(nCk)(-1)k/2 + iΣ(nCk)(-1)k/2
= Σ(nCk)(-1)k/2 (1+i)
This is our final answer. We've expressed (1+i)2n as a sum of real numbers times (1+i), which is a simpler and more useful form. Note that the sum simplifies further for even values of n (when all terms with odd powers of i cancel out), and for odd values of n (when the real part of the sum is zero).
To simplify this expression, we can use the binomial theorem. According to the binomial theorem, (a+b)n = Σ(nCk)an-kbk, where nCk is the binomial coefficient "n choose k" (which represents the number of ways to choose k items from a set of n items), and Σ represents the sum over all values of k from 0 to n.
In our case, we have a=1 and b=i, so we can write:
(1+i)2n = Σ(nCk)1n-k(i)k
Now we just need to simplify the terms involving i. Note that i2 = -1, so any term with an odd power of i will be equal to i times a term with an even power of i (i.e. a real number). We can use this fact to rewrite the sum as:
(1+i)2n = Σ(nCk)(-1)k/2 + iΣ(nCk)(-1)(k-1)/2
The first sum contains only real numbers, so we can pull it out of the parentheses and simplify:
(1+i)2n = Σ(nCk)(-1)k/2 + iΣ(nCk)(-1)(k-1)/2
= Σ(nCk)(-1)k/2 + iΣ(nCk)(-1)k/2
= Σ(nCk)(-1)k/2 (1+i)
This is our final answer. We've expressed (1+i)2n as a sum of real numbers times (1+i), which is a simpler and more useful form. Note that the sum simplifies further for even values of n (when all terms with odd powers of i cancel out), and for odd values of n (when the real part of the sum is zero).
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(1+i)2n+ (1−i)2n,n ∈ N isa)a purely real numberb)a purely imaginary numberc)a non-real complex numberd)none of theseCorrect answer is option 'A'. Can you explain this answer?
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