Demoivre's Theorem Question

Question

Use DeMoivre's Theorem to find (sqrt(2)-i)^4

 

Step by Step, this is a practice problem my professor gave me 

Answer 1

DeMoivre's Theorem states

(cos x + i sin x)ⁿ = cos (nx) + i sin(nx)

 

Comparing to your problem

(√(2) - i)⁴

 

Initially, you'd look for an angle that satisfies

cos x = √(2)

sin x = -1

 

No such angle exists, so you have to play around with it first. If the left side were multiplied by some factor A, there might be an angle that works. If it were

A cos x + i A sin x

 

A cos x = √(2)

A sin x = -1

 

cos x = √(2)/A

sin x = -1/A

 

You can draw this triangle in the 4th quadrant with x = √(2), y = -1, and the hypotenuse of A = √(3) by the Pythagorean Theorem.

Therefore,

cos x = √(2)/√(3) = √(2/3)

sin x = -1/√(3)

 

and your problem can be written as

(√(3) cos x + i √(3) sin x)⁴

= (√(3))⁴ (cos x + i sin x)⁴

= 3 (cos x + i sin x)⁴

= 3 (cos 4x + i sin 4x)

 

Now you need to find cos 4x and sin 4x. To do that, we'll first find cos 2x and sin 2x

cos 2x = 2 cos² x - 1

   = 2 (√(2/3)² - 1

   = 2 (2/3) - 1

   = 4/3 - 1

   = 1/3

 

sin 2x = 2 sin x cos x

   = 2 (-1/√(3)) (√(2/3))

   = -2√(2) / 3

 

cos 4x = 2 cos² 2x - 1

   = 2 (1/3)² - 1

   = 2/9 - 1

   = -7/9

 

sin 4x = 2 sin 2x cos 2x

   = 2 (-2√(2)/3) (1/3)

   = -4 √(2) / 9

 

Plugging these in to the earlier result,

3 (cos 4x + i sin 4x)

= 3 (-7/9 + i (-4 √(2) / 9) )

= -7/3 - (4 √(2) / 3) i