Show that the locus in the complex plane of all points satisfying #cosv + isinv # where #v in [0,2pi]# is a unit circle?

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Answer 1 · 2017-10-05T21:35:19

Suppose we have a point #z# in the complex plane such that

# z = cosv + isinv \ \ # where #v in [0,2pi]#

Now let us suppose that #z# has the rectangular form:

# z = x+iy #

Equating real and imaginary components, we have:

# x=cosv#
# y = sinv #

So, the locus of the point #z# satisfies:

# x^2 + y^2 = cos^2v+sin^2v = 1 #

Hence the point #z#, or #cosv + isinv# lies on the unit circle,QED