In Exercises 23 to 26 follow the technique outlined in the t

In Exercises 23 to 26, follow the technique outlined in the text to find all solutions of the given equation. Z^5 = I (z + 1)^4 = 1 - I z^8 = -1 z^3 = 8

Solution

23).

z^5 = e^((pi/2 + 2k pi) i ), k=0,1,2,3,4, so

z = e^(pi i /10), e^5pi i /10), e^(9pi i /10), e^(13pi i/10), e^(17pi i/10).

In trigonometric form:

z = cos(pi/10) + i sin(pi/10)
z = cos(5pi/10) + i sin(5pi/10) = cos(pi/2) + i sin(pi/2) = i
z = cos(9pi/10) + i sin(9pi/10) = -cos(pi/10) + i sin(pi/10)
z = cos(13pi/10) + i sin(13pi/10) = -cos(3pi/10) - i sin(3pi/10)
z = cos(17pi/10) + i sin(17pi/10) = cos(3pi/10) - i sin(3pi/10)

Note these can be expressed as radicals. For example, by starting with cos(pi/5) = (1/4)(1+sqrt(5)) and sin(pi/5) = (1/4)sqrt(10-2sqrt(5)) and using half-angle formulas.

Note2: pi/10 = 18 degrees and 3pi/10 = 54 degrees, angles associated with the regular decagon and regular pentagon.

24).(z+1)^4 = 1-i

1-i = 2^1/2 * (cos (-pi/4) + i*sin(-pi/4)) then i found z= [2^(1/2*1/4) * (cos (-pi/4) + i*sin (-pi/4) )^1/4] -1 then using de moivre\'s thrm z= {2^3/4 *(cos [pi/16 +2k*pi/16] - i*sin[pi/16 + 2k*pi/16])}-1 ; k=0,1,2,3

24).

ubtract  1 from both sides of the equation :
                      z⁸ = -1
                     z =  8th root of (-1)

The equation has no real solutions. It has 8 imaginary, or complex solutions.

                      z= 0.9239 + 0.3827 i  
                      z= 0.3827 + 0.9239 i  
                      z= -0.3827 + 0.9239 i  
                      z= -0.9239 + 0.3827 i  
                      z= -0.9239 - 0.3827 i  
                      z= -0.3827 - 0.9239 i  
                      z= 0.3827 - 0.9239 i  
                      z= 0.9239 - 0.3827 i  

26).

z^3=8

Factoring:  z³-8

Theory : A difference of two perfect cubes,  a³ - b³ can be factored into
              (a-b) • (a² +ab +b²)

=> (z - 2)  •  (z² + 2z + 4) =0