The value of 15-i6-i7-i8-i9-1-i

The correct option is A 1/2(1+i) i^5+i^6+i^7+i^8+i^9/1+i =i 1 i+1+i/1+i [As, i^5=i,i^6= 1,i^7= i,i^8=1,i^9=i] =i/i+1 =i/i+1×i 1/i 1 =i(i 1)/i^2 1 =i^2 i/2 ...

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Byju's Answer

Standard XII

Mathematics

Properties of Iota

The value of ...

Question

$The value of \frac{(i^{5} + i^{6} + i^{7} + i^{8} + i^{9})}{(1 + i)}$

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

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$\frac{1}{2} (1 - i)$

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$\frac{1}{2}$

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Solution

The correct option is A
$\frac{1}{2} (1 + i)$

$\frac{i^{5} + i^{6} + i^{7} + i^{8} + i^{9}}{1 + i} = \frac{i - 1 - i + 1 + i}{1 + i} [As, i^{5} = i, i^{6} = - 1, i^{7} = - i, i^{8} = 1, i^{9} = i] = \frac{i}{i + 1} = \frac{i}{i + 1} \times \frac{i - 1}{i - 1} = \frac{i (i - 1)}{i^{2} - 1} = \frac{i^{2} - i}{2} = \frac{1}{2} (1 + i)$

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Similar questions

Q. Find the value of

\frac{i^{6} + i^{7} + i^{8} + i^{9}}{i^{2} + i^{3}}

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\frac{1}{2} (1 + i)

(b)

\frac{1}{2} (1 - i)

\frac{1}{2}

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Q. The value of

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is :

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The value of (i5+i6+i7+i8+i9)(1+i)

The correct option is A 12(1+i) i5+i6+i7+i8+i91+i=i−1−i+1+i1+i[As, i5=i,i6=−1,i7=−i,i8=1,i9=i]=ii+1=ii+1×i−1i−1=i(i−1)i2−1=i2−i2=12(1+i)

$The value of \frac{(i^{5} + i^{6} + i^{7} + i^{8} + i^{9})}{(1 + i)}$