1 - i1 - i4 - 1 - i1 - i4 -

The correct option is B 2 ( 1+i1 i)^4+ ( 1 i1+i)^4 = ( (1+i)^21 i^2)^4+ ( (1 i)^21 i^2)^4 =12^4 (1+i^2 +2 i)^4+ 12^4 (1+i^2 2i)^4 =12^ ...

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Byju's Answer

Standard XII

Mathematics

Complex Numbers

1 + i1 - i4 +...

Question

${(\frac{1 + i}{1 - i})}^{4} + {(\frac{1 - i}{1 + i})}^{4} =$

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Solution

The correct option is B $2$
${(\frac{1 + i}{1 - i})}^{4} + {(\frac{1 - i}{1 + i})}^{4}$
$= {(\frac{(1 + i)^{2}}{1 - i^{2}})}^{4} + {(\frac{(1 - i)^{2}}{1 - i^{2}})}^{4}$
$= \frac{1}{2^{4}} (1 + i^{2} + 2 i)^{4} + \frac{1}{2^{4}} (1 + i^{2} - 2 i)^{4}$
$= \frac{1}{2^{4}} (1 - 1 + 2 i)^{4} + \frac{1}{2^{4}} (1 - 1 - 2 i)^{4}$
$= \frac{2^{4}}{2^{4}} (i^{4}) + \frac{2^{4}}{2^{4}} (i)^{4}$
$= 1 + 1 = 2$

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Similar questions

Q. The value of

(1 + i)^{4} + (1 - i)^{4}

(1 + i + i^{2} + i^{3} + i^{4} + i^{5}) (1 + i) =

Prove that:
(i) $i^{n} + i^{n + 1} + i^{n + 2} + i^{n + 3} = 0$
(ii) $i^{107} + i^{112} + i^{117} + i^{122} = 0$
(iii) $(1 + i)^{4} \times {(1 + \frac{1}{i})}^{4} = 16$

Q. Solve

(1 + i)^{4} + (1 - i)^{4} =

Q. Prove that

{(1 + i)}^{4} {(1 + \frac{1}{i})}^{4} = 16

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(1+i1−i)4+(1−i1+i)4=

The correct option is B 2(1+i1−i)4+(1−i1+i)4=((1+i)21−i2)4+((1−i)21−i2)4=124(1+i2+2i)4+124(1+i2−2i)4=124(1−1+2i)4+124(1−1−2i)4=2424(i4)+2424(i)4=1+1=2

${(\frac{1 + i}{1 - i})}^{4} + {(\frac{1 - i}{1 + i})}^{4} =$