How to solve this equation #e^(6z)+(1-i)e^(3z)-i=0#, for z where z is a complex number?

1 Answer
Mar 13, 2017

Answer:

See below.

Explanation:

Calling #w=e^(3z)# we have

#w^2+(1-i)w-i=0# solving for #w# we have

#w=-1 = e^(i (pi+2kpi))# and #w=i = e^(i(pi/2+2kpi))# or

1) #e^(3z)=e^(i (pi+2kpi))# or

#3z=i (pi+2kpi) = 3(x+iy)->{(x=0),(y=pi/3+2/3kpi):}#

2) #e^(3z)=e^(i(pi/2+2kpi))# or

#3z = i(pi/2+2kpi)=3(x+iy)->{(x=0),(y=pi/6+2/3kpi):}#