Find the value of- 1-i1-i40

Now,Let S=(1+i1 i)^40 #160; #160; #160; #160; #160; =((1+i)^2(1 i)^2)^20 #160; #160; #160; #160; #160; =(2i 2i)^20 [ ...

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Byju's Answer

Standard XII

Mathematics

Algebra of Complex Numbers

Find the valu...

Question

Find the value of:
${(\frac{1 + i}{1 - i})}^{40}$

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Solution

Now,

Let $S = {(\frac{1 + i}{1 - i})}^{40}$

$= {(\frac{(1 + i)^{2}}{(1 - i)^{2}})}^{20}$

$= {(\frac{2 i}{- 2 i})}^{20}$ [ Since $(1 + i)^{2} = 1 + 2 i - 1$ and $(1 - i)^{2} = 1 - 2 i - 1$ ]

$= (- 1)^{20}$

$S = 1$ .

$∴ {(\frac{1 + i}{1 - i})}^{40} = 1$

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Find the value of:(1+i1−i)40

Now,Let S=(1+i1−i)40 =((1+i)2(1−i)2)20 =(2i−2i)20 [ Since (1+i)2=1+2i−1 and (1−i)2=1−2i−1] =(−1)20 S=1.∴(1+i1−i)40=1

Find the value of:
${(\frac{1 + i}{1 - i})}^{40}$

Now,

Let $S = {(\frac{1 + i}{1 - i})}^{40}$

$= {(\frac{(1 + i)^{2}}{(1 - i)^{2}})}^{20}$

$= {(\frac{2 i}{- 2 i})}^{20}$ [ Since $(1 + i)^{2} = 1 + 2 i - 1$ and $(1 - i)^{2} = 1 - 2 i - 1$ ]

$= (- 1)^{20}$

$S = 1$ .

$∴ {(\frac{1 + i}{1 - i})}^{40} = 1$