Show that -14 n-3- i- where n is a positive integer-

We have ( √( 1))^4n+3 = ( i)^4n+3 [∵√( 1)=i] =( i)^4n× ( i)^3 = [( i)^4]^n× ( i)^3 = (1 × i) =i [∵ ( i)^4 =1 and i^3= ( i)=i] Hence, ( √( 1))^4n+3=i. ...

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Standard VIII

Mathematics

Cube-Root of a Negative Perfect Cube

Show that -√-...

Question

Show that $(- \sqrt{- 1})^{4 n + 3} = i$ , where n is a positive integer.

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Solution

We have

$(- \sqrt{- 1})^{4 n + 3} = (- i)^{4 n + 3}$ $[∵ \sqrt{- 1} = i]$

$= (- i)^{4 n} \times (- i)^{3}$

$= [(- i)^{4}]^{n} \times (- i)^{3} = (1 \times i) = i$ $[∵ (- i)^{4} = 1 and - i^{3} = - (- i) = i]$

Hence, $(- \sqrt{- 1})^{4 n + 3} = i$ .

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Show that (−√−1)4n+3=i, where n is a positive integer.

We have (−√−1)4n+3=(−i)4n+3 [∵√−1=i] =(−i)4n×(−i)3 =[(−i)4]n×(−i)3=(1×i)=i [∵(−i)4=1 and −i3=−(−i)=i] Hence, (−√−1)4n+3=i.

Show that $(- \sqrt{- 1})^{4 n + 3} = i$ , where n is a positive integer.

We have

$(- \sqrt{- 1})^{4 n + 3} = (- i)^{4 n + 3}$ $[∵ \sqrt{- 1} = i]$

$= (- i)^{4 n} \times (- i)^{3}$

$= [(- i)^{4}]^{n} \times (- i)^{3} = (1 \times i) = i$ $[∵ (- i)^{4} = 1 and - i^{3} = - (- i) = i]$

Hence, $(- \sqrt{- 1})^{4 n + 3} = i$ .