We have (1 i)^4=(1 i)^2× (1 i)^2 =(1+i^2 2i) × (1+i^2 2i) =( 2i)( 2i) =4i^2 = 4 × ( 1) = 4. [∵ i^2= 1] ...

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Standard VIII

Mathematics

Numbers in General Form

Prove that 1-...

Question

Prove that $(1 - i)^{4} = - 4$ .

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Solution

We have

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

$= (- 2 i) (- 2 i) = 4 i^{2} = 4 \times (- 1) = - 4$ . $[∵ i^{2} = - 1]$

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Similar questions

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

Q. Prove that

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

Prove that:
(i) $i^{n} + i^{n + 1} + i^{n + 2} + i^{n + 3} = 0$
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Q. The value of

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

is
(a) 8
(b) 4
(c) −8
(d) −4

The value of $(1 + i)^{4} + (1 - i)^{4}$ is

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Prove that (1−i)4=−4.

We have (1−i)4=(1−i)2×(1−i)2=(1+i2−2i)×(1+i2−2i) =(−2i)(−2i)=4i2=4×(−1)=−4. [∵i2=−1]

Prove that $(1 - i)^{4} = - 4$ .

We have

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

$= (- 2 i) (- 2 i) = 4 i^{2} = 4 \times (- 1) = - 4$ . $[∵ i^{2} = - 1]$