Greatest negative integral value of n for which 1- i -1- i n -1 isA- 4B- -4C- -D- Can-t be determined

1+i/1 i can be brought to standard form by multiplying both numerator amp; denominator by 1+i ( 1+i )/ ( 1 i ) ( 1+i )/ ( 1+i ) = (1+i)^2/1+1 = 2i/2 = i ...

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Standard XII

Mathematics

Proof by mathematical induction

Greatest nega...

Question

Greatest negative integral value of n for which ${(\frac{1 + i}{1 - i})}^{n}$ = 1 is

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-4

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Can't be determined

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Solution

The correct option is B
-4

$\frac{1 + i}{1 - i}$ can be brought to standard form by multiplying both numerator & denominator by 1+i

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

⇒ $i^{n}$ = 1 for n = 4k

i.e., n = ......... , -12,-8,-4,0,4,8,12,.........

So,greatest negative integer among above numbers is -4

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Greatest negative integral value of n for which (1+i1−i)n = 1 is

The correct option is B -4 1+i1−i can be brought to standard form by multiplying both numerator & denominator by 1+i (1+i)(1−i)(1+i)(1+i) = (1+i)21+1 = 2i2 = i ⇒ in = 1 for n = 4k i.e., n = ......... , -12,-8,-4,0,4,8,12,......... So,greatest negative integer among above numbers is -4

Greatest negative integral value of n for which ${(\frac{1 + i}{1 - i})}^{n}$ = 1 is