Please show work Write 1 i99 in the form a bi where a and

Please show work.

Write (1 + i)^99 in the form a + bi, where a and b are real numbers. 2. The complex number 1 - squareroot 3i has two squareroots. Find them.

Solution

1) (1 +i)^99

use De-movire\'s theorem:

(a+ib)^n = r^n(cosn*x + isinn*x ) where r = sqrt(a^2 +b^2)

x = tan^-1(b/a) = tan^-1(1) = pi/4 ; r = 1

(1 +i)^99 = 1^99( cos(99*pi/4) +i sin(99*pi/4) )

= -0.707 +i*0.707