(iz3,0)^401.
find all the roots of the equation z^3 + 1=0. give your answer in exponential and algebraic form.
2007-08-12 23:07:36 · 5 answers · asked by Amrinder J 1 in Science & Mathematics ➔ Mathematics
the roots of z^3+1=0 are
z^3=-1 = 1
If you mean (i Zo)401= i^400*i*Zo^401 = 1<401(pi/3+2kpi/3)+pi/2=1<(5pi/6+pi/2)+4kpi/3
2007-08-13 07:53:40 · answer #1 · answered by santmann2002 7 · 0⤊ 0⤋
z^4 + 4jz^2 = a million z^4 + 4jz^2 + (2j)^2 = a million+(2j)^2, yet (2j)^2 = -4, so (z^2+2j)^2 = -3, subsequently z^2+2j = ±j?3, so z^2 = (-2 ± ?3) j = (a million ± ?3/2)(-2j). by way of De Moivre ?(-2j) = a million-j, that is effortless to envision. additionally, ±?(a million ± ?3/2) = ±(±a million/2+?3/2), which you would be able to verify by way of squaring. subsequently the 4 roots are: z = ±(±a million/2+?3/2)(a million-j).
2016-11-12 04:40:18 · answer #2 · answered by Anonymous · 0⤊ 0⤋
I believe that the symbol ^ is used to indicate an exponential.
Thus z^3 is z to the power 3. In other words, z cubed.
2007-08-12 23:21:21 · answer #3 · answered by nontarzaniccaulkhead 6 · 0⤊ 0⤋
Do your OWN Homework you lazy git!
2007-08-13 02:33:42 · answer #4 · answered by Anonymous · 0⤊ 0⤋
what does the ^ stand for?
2007-08-12 23:16:52 · answer #5 · answered by dappa d 2 · 0⤊ 1⤋