How do you express the complex number in trigonometric form: -6i?

1 Answer
Apr 16, 2016

Answer:

#-6i = 6(cos((3pi)/2) + isin((3pi)/2))#

Explanation:

Trigonometric form is where

#a + bi = absz(cosx + isinx)#.

#absz# is given by Pythagoras', #sqrt(a^2+b^2)#, using the values from #a + bi#.

#a + bi = 0 - 6i#
#sqrt(a^2+b^2) = sqrt(0^2 + (-6)^2) = sqrt36 = 6#

Now you can divide both sides by #6#, so

#(-6i)/6 = (6(cosx + isinx))/6#
#-i = cosx + isinx#

This gives us, obviously, that

#cosx = 0#
#sinx = -1#

and so, superimposing the graphs and finding points that satisfy the simultaneous equations,

http://www.bbc.co.uk/bitesize/standard/maths_ii/trigonometry/graphs/revision/1/

At the point #x = 270#, the #cosx# graph is #0# and the #sinx# graph is at #-1#, which satisfies the equations. Being repeating patterns, obviously there will be other results, such as #x = -450, -90, 630, 990# etc.

Putting all of this back together,

#-6i = 6(cos270 + isin270)#

or

#-6i = 6(cos((3pi)/2) + isin((3pi)/2))#