If z = -1 - i, find z10 in polar form?

Question

2291 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-05-09T01:15:57

#(-1 -i)^{10} = 32( cos(pi/2) + i sin(pi/2)) = 32 i #

#z = -1 -i = \sqrt{2} (- 1/sqrt{2} -i 1/sqrt{2}) = sqrt{2}(cos({5pi}/4) + i sin ({5 pi}/4) )#

# z^{10} = ( sqrt{2}(cos({5pi}/4) + i sin ({5 pi}/4)) )^{10}#

# = (\sqrt{2})^{10} (cos({50 pi}/4) + i sin({50 pi}/4) ) #

#= 2^5 (cos ( {25 pi}/2 - 12 pi) + i sin ( {25 pi}/2 - 12 pi) )#

# = 32( cos(pi/2) + i sin(pi/2)) #

That's the answer in polar form, but we take the next step.

#z^{10} = 32 i #