I want to express (1+i)^11 in algebric form by using the trig functions within the process...

Question

For some reason 1+i = root 2 * e ^ pi / 4 

 

which i dont understand how so I would like an explanation for this...

Answer 1

Taking the 11th power of a binomial will be very tedious.  Try converting the complex number to polar form, and work with that.

 

1 + i = re^iφ

 

r = √[(1)² + (1)²] = √2

 

φ = tan^-1(1/1) = π/4

 

1 + i = (√2)e^iπ/4

 

(1 + i)¹¹ = [(√2)e^iπ/4]¹¹ = 2^11/2 e^iπ(11/4) = 2^11/2 cos(11π/4) + i2^11/2 sin(11π/4)

 

You can subtract multiples of 2π from 11π/4 to work with a simpler argument.

 

11π/4 = 8π/4 + 3π/4 = 2π + 3π/4

 

(1 + i)¹¹ = 2^11/2 cos(3π/4) + i2^11/2 sin(3π/4)

 

This gives the same result as De Moivre's Theorem (see Mark's comment).