Prove that 1 - i4 1 - 1i4 - 16

#160; LHS #10; #10; #160; ( 1+i #10;)^4( 1+1i)^4 #10; #10; #160; =( 1+i #10;)^2( 1+1i)^2( 1+i #10;)^^2( 1+1i)^2 #10; # ...

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Byju's Answer

Standard XII

Mathematics

Complex Numbers

Prove that ...

Question

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

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Solution

$L H S$

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

$= {(1 + i)}^{2} {(1 + \frac{1}{i})}^{2} {(1 + i)}^{^{2}} {(1 + \frac{1}{i})}^{2}$

$= {(1 + i)}^{2} {(1 + i)}^{^{2}} {(1 + \frac{1}{i})}^{2} {(1 + \frac{1}{i})}^{2}$

$= (1^{2} + i^{2} + 2 i) (1^{2} + i^{2} + 2 i) (1^{2} + \frac{1}{i^{2}} + \frac{2}{i}) (1^{2} + \frac{1}{i^{2}} + \frac{2}{i})$

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

$= 2 i \times 2 i \times \frac{2}{i} \times \frac{2}{i}$

$= 16$

$R H S$

Hence, proved.

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Similar questions

Q. Prove that

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

Q. Prove that

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

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} =

{(\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

LHS (1+i)4(1+1i)4 =(1+i)2(1+1i)2(1+i)2(1+1i)2 =(1+i)2(1+i)2(1+1i)2(1+1i)2 =(12+i2+2i)(12+i2+2i)(12+1i2+2i)(12+1i2+2i) =(1−1+2i)(1−1+2i)(12+1−1+2i)(12+1−1+2i) =2i×2i×2i×2i =16 RHS Hence, proved.

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