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Question 104 of 122

Q.The value of (1+i2)8+(1−i2)8\left(\dfrac{1+i}{\sqrt2}\right)^8+\left(\dfrac{1-i}{\sqrt2}\right)^8 is :

(a) 88
(b) 44
(c) 22
(d) 66
Tamil Nadu DgeTamil Nadu HSC (DGE) Board 2023MCQ· 1mImportance★★★★★
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Writing each fraction in polar form as a unit complex number at angle ±π/4\pm\pi/4 and using De Moivre's theorem for the 8th power.

  1. 1+i2=12+i2=cos⁡π4+isin⁡π4\dfrac{1+i}{\sqrt2}=\dfrac{1}{\sqrt2}+\dfrac{i}{\sqrt2}=\cos\dfrac{\pi}{4}+i\sin\dfrac{\pi}{4} (modulus 1, argument π/4\pi/4).
  2. By De Moivre's theorem, (1+i2)8=cos⁡8π4+isin⁡8π4=cos⁡2π+isin⁡2π=1\left(\dfrac{1+i}{\sqrt2}\right)^8=\cos\dfrac{8\pi}{4}+i\sin\dfrac{8\pi}{4}=\cos2\pi+i\sin2\pi=1. …

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