Solve for z cosz log e 2 3 Solve for z sinz 5Solution1 cos

Solve for z: cos(z) = log e (2) (3) Solve for z: sin(z) = 5.

1) cos(z) = log_e(2)

cosz = (e^iz +e^-iz)/2

So, (e^iz +e^-iz)/2 = loge(2)

e^2iz + 1 - 2e^izloge(2) =0

solve the quadratic: x^2 -2xloge(2) +1 =0

x = ( 2loge(2) +/- sqrt(4loge(2)^2 -4)/2

= loge(2) +/- sqrt(loge(2)^2 -1)

= 0.69 +/- i*0.52

e^iz = 0.69 +/- i*0.52

taking natural log on both sides:

z = [ln( 0.69 +/- i*0.52)]/i

2) sin(z) = 5

sinz = (e^iz - e^-iz)/2i

(e^iz - e^-iz)/2i = 5

e^2iz - 1= 10ie^z

e^2iz - 10ie^z -1 =0

solve the quadratic:

x^2 -10ix -1 =0

x = ( 10 +/- sqrt(-100 +4)/2

= 5 +/- i*sqrt(-24)

= 5 +/- i*2isqrt6 = 5 +/- 2sqrt6

e^iz = 5 +2sqrt6 ---> z = ln(5 +2sqrt6)/i

e^iz = 5 -2sqr6 ---> z = -2.29