since cos 82° = sin 8°,
z^5 = 2^5 cis(5*8)
= 32 cis 40°
= 32 (cos40° + i sin40°)
= 32 (sin50° + i cos50°)
32+50 = 82
Let z be a complex number such that
z=2(cos8∘+icos82∘).
Then z^5 can be expressed as r(sinα∘+icosα∘), where r is a real number and 0≤α≤90. What is the value of r+α?
1 answer