Prove that 1-i4-1-1i4-16

( 1+i ) #160; ^ 4 ×( 1+ 1 i #160; ) #160; ^ 4 (1+i)^4×= i) =(1^2 i^2)^4 (1+i)^4=2^4=16 ...

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Byju's Answer

Standard VIII

Mathematics

Expansion of (x^2 - y^2)

Prove that ...

Question

Prove that ${(1 + i)}^{4} \times {(1 + \frac{1}{i})}^{4} = 16$

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Solution

${(1 + i)}^{4} \times {(1 + \frac{1}{i})}^{4}$
${(1 + i)}^{4} \times = - i)$
$= {(1^{2} - i^{2})}^{4}$
${(1 + i)}^{4} = 2^{4} = 16$

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Similar questions

Q. Prove that

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

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. Prove that

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

Q. Solve

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

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

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Prove that (1+i)4×(1+1i)4=16

(1+i)4×(1+1i)4(1+i)4×=−i)=(12−i2)4(1+i)4=24=16

Prove that ${(1 + i)}^{4} \times {(1 + \frac{1}{i})}^{4} = 16$

${(1 + i)}^{4} \times {(1 + \frac{1}{i})}^{4}$
${(1 + i)}^{4} \times = - i)$
$= {(1^{2} - i^{2})}^{4}$
${(1 + i)}^{4} = 2^{4} = 16$