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Question 122 of 122

Q.If z1=1+i‾z_1=\overline{1+i} and z2‾=1−i\overline{z_2}=1-i, find the inverse of (z1z2)2026\left(\dfrac{z_1}{z_2}\right)^{2026}

Tamil Nadu DgeTamil Nadu HSC (DGE) Board 2026Subjective· 3mImportance★★★★★
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Recovers z1z_1 and z2z_2 from the given conjugate relations, simplifies their ratio to −i-i, raises it to the 2026th power using periodicity, and takes the multiplicative inverse.

  1. z1=1+i‾z_1=\overline{1+i} means z1z_1 is defined as the conjugate of 1+i1+i: z1=1−iz_1=1-i.
  2. z2‾=1−i\overline{z_2}=1-i means the conjugate of z2z_2 equals 1−i1-i; taking the conjugate of both sides (z2‾‾=z2\overline{\overline{z_2}}=z_2): z2=1−i‾=1+iz_2=\overline{1-i}=1+i.
  3. Compute z1z2=1−i1+i\dfrac{z_1}{z_2}=\dfrac{1-i}{1+i}. Multiply numerator and denominator by the conjugate of the denominator, 1−i1-i: 1−i1+i×1−i1−i=(1−i)2(1+i)(1−i)=1−2i+i21−i2=1−2i−11−(−1)=−2i2=−i\dfrac{1-i}{1+i}\times\dfrac{1-i}{1-i}=\dfrac{(1-i)^2}{(1+i)(1-i)}=\dfrac{1-2i+i^2}{1-i^2}=\dfrac{1-2i-1}{1-(-1)}=\dfrac{-2i}{2}=-i
  4. So z1z2=−i\dfrac{z_1}{z_2}=-i, and we need (−i)2026(-i)^{2026}. …

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