Trying to fit in 2016

Algebra Level 3

The number u = ( 14 + 3 i ) 2016 + ( 14 i + 3 ) 2016 u = (14 + 3i)^{2016} + (14i + 3)^{2016} is:

None of these choices Real Has modulus 1 Purely imaginary

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2 solutions

u = ( 14 + 3 i ) 2016 + ( 3 + 14 i ) 2016 = ( 14 + 3 i ) 2016 + i 2016 i 2016 × ( 3 + 14 i ) 2016 = ( 14 + 3 i ) 2016 + ( 3 i 14 ) 2016 1 = ( 14 + 3 i ) 2016 + ( 1 ) 2016 ( 14 3 i ) 2016 = ( 14 + 3 i ) 2016 + ( 14 3 i ) 2016 = k = 0 2016 ( 2016 k ) 1 4 k ( 3 i ) 2016 k + k = 0 2016 ( 1 ) k ( 2016 k ) 1 4 k ( 3 i ) 2016 k \begin{aligned} u & = (14 + 3i)^{2016} + (3 + 14i)^{2016} \\ & = (14 + 3i)^{2016} + \frac{i^{2016}}{i^{2016}} \times (3 + 14i)^{2016} \\ & = (14 + 3i)^{2016} + \frac{(3i - 14)^{2016}}{1} \\ & = (14 + 3i)^{2016} + (-1)^{2016}(14-3i)^{2016} \\ & = (14 + 3i)^{2016} + (14-3i)^{2016} \\ & = \sum_{k=0}^{2016} \begin{pmatrix} 2016 \\ k \end{pmatrix} 14^k (3i)^{2016-k} + \sum_{k=0}^{2016} (-1)^k \begin{pmatrix} 2016 \\ k \end{pmatrix} 14^k (3i)^{2016-k} \end{aligned}

We note that all the k = k = odd terms, which are imaginary, cancel out. Therefore, u u is real \boxed{\text{real}} .

Mohammad Hamdar
Jan 1, 2016

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Pi Han Goh - 5 years, 4 months ago

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