Posted by **Shin** on Saturday, June 22, 2013 at 4:45am.

Let z be a complex number such that z = 2(cos 8∘ + i cos 82∘).Then z^5 can be expressed as r(sin α∘+ i cos α∘), where r is a real number and 0 ≤ α ≤ 90. What is the value of r+α?

Hint to solve:

Example Question:

Let z be a complex number such that z = 2(cos 5∘ + i cos 85∘). Then z^6 can be expressed as r(sin α∘ + i cos α∘), where r is a real number and 0 ≤ α ≤90. What is the value of r+α?

Solution:

Since cos85∘=cos(90∘−5∘)=sin5∘,z=2(cos5∘+icos85∘)=2(cos5∘+isin5∘).

Applying De Moivre's formula gives

z^6=26(cos(6⋅5∘)+isin(6⋅5∘))=64(cos30∘+isin30∘).

Since cos30∘=cos(90∘−60∘)=sin60∘ and sin30∘=sin(90∘−60∘)=cos60∘,

z^6=64(cos30∘+isin30∘)=64(sin60∘+icos60∘).

Therefore, r=64 and α=60, hence r + α = 64 + 60 = 124.