The value of ((1+i/√2))8+((1-i/√2))8 is equal to

Question

The value of ((1+i/√2))8+((1-i/√2))8 is equal to (A) 4 (B) 6 (C) 8 (D) 2

Tardigrade Admin · Accepted Answer

We have, ((1+i/√2))8+((1-i/√2))8 =[ cos (π/4)+i sin (π/4)]8+[ cos (π/4)-i sin (π/4)]8 = cos 2 π+i sin 2 π+ cos 2 π-i sin 2 π =2 cos 2 π=2(1)=2 [ By De-Moivre's theorem ]

Q. The value of $(\frac{1 + i}{2})^{8} + (\frac{1 - i}{2})^{8}$ is equal to

4

6

8

2

We have, $(\frac{1 + i}{2})^{8} + (\frac{1 - i}{2})^{8}$
$= [cos \frac{π}{4} + i sin \frac{π}{4}]^{8} + [cos \frac{π}{4} - i sin \frac{π}{4}]^{8}$
$= cos 2 π + i sin 2 π + cos 2 π - i sin 2 π$
$= 2 cos 2 π = 2 (1) = 2$
$[$ By De-Moivre's theorem $]$

Q. The value of (2​1+i​)8+(2​1−i​)8 is equal to

4

6

8

2

We have, (2​1+i​)8+(2​1−i​)8 =[cos4π​+isin4π​]8+[cos4π​−isin4π​]8 =cos2π+isin2π+cos2π−isin2π =2cos2π=2(1)=2 [ By De-Moivre's theorem ]

Q. The value of $(\frac{1 + i}{2})^{8} + (\frac{1 - i}{2})^{8}$ is equal to

We have, $(\frac{1 + i}{2})^{8} + (\frac{1 - i}{2})^{8}$
$= [cos \frac{π}{4} + i sin \frac{π}{4}]^{8} + [cos \frac{π}{4} - i sin \frac{π}{4}]^{8}$
$= cos 2 π + i sin 2 π + cos 2 π - i sin 2 π$
$= 2 cos 2 π = 2 (1) = 2$
$[$ By De-Moivre's theorem $]$