-
Best of luck to the class of 2024 for their HSC exams. You got this! Let us know your thoughts on the HSC exams here -
YOU can help the next generation of students in the community! Share your trial papers and notes on our Notes & Resources page
Complex no questoin (1 Viewer)
- Thread starter axwe7
- Start date
pikachu975
Premium Member
Do you have an answer, if I got the same I'll post the solution.
Divide (root3 - i)^m to both sides as it can't equal 0.
Change both numerator and denominator to mod arg form.
(2cispi/6)/(2cis(-pi/6)) = 1
(cispi/3)^m = cis0 (De Moivres)
mpi/3 = 0 + 2kpi (k is an integer)
m = 6k and the smallest positive m is when k = 1, therefore
m = 6
Divide (root3 - i)^m to both sides as it can't equal 0.
Change both numerator and denominator to mod arg form.
(2cispi/6)/(2cis(-pi/6)) = 1
(cispi/3)^m = cis0 (De Moivres)
mpi/3 = 0 + 2kpi (k is an integer)
m = 6k and the smallest positive m is when k = 1, therefore
m = 6
Last edited:
How many '*' ?