Show that 23-1-i292-4-

We have i^23=i^(4 × 5 +3)=(i^4)^5× i^3= 1 × ( i) = i. [∵ i^4=1 and i^3= i] ( 1/i )^29 = 1/i^29 = 1/i^29×i^3/i^3 = i^3/i^32 = i/1= i. [∵ i^3= i and i^32=1 ...

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Byju's Answer

Standard XI

Mathematics

Binomial Theorem

Show that 23+...

Question

Show that ${i^{23} + {(\frac{1}{i})}^{29}}^{2} = - 4$ .

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Solution

We have

$i^{23} = i^{(4 \times 5 + 3)} = (i^{4})^{5} \times i^{3} = 1 \times (- i) = - i$ . $[∵ i^{4} = 1 and i^{3} = - i]$

${(\frac{1}{i})}^{29} = \frac{1}{i^{29}} = \frac{1}{i^{29}} \times \frac{i^{3}}{i^{3}} = \frac{i^{3}}{i^{32}} = \frac{- i}{1} = - i$ . $[∵ i^{3} = - i and i^{32} = 1]$

$∴ {i^{23} + {(\frac{1}{i})}^{29}}^{2} = (- i - i)^{2} = (- 2 i)^{2} = 4 i^{2} = 4 \times (- 1) = - 4$

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Show that {i23+(1i)29}2=−4.

We have i23=i(4×5+3)=(i4)5×i3=1×(−i)=−i. [∵i4=1 and i3=−i] (1i)29=1i29=1i29×i3i3=i3i32=−i1=−i. [∵i3=−i and i32=1] ∴{i23+(1i)29}2=(−i−i)2=(−2i)2=4i2=4×(−1)=−4

Show that ${i^{23} + {(\frac{1}{i})}^{29}}^{2} = - 4$ .