How do you write #-i# in trigonometric form?

1 Answer
Mar 1, 2018

# \qquad \qquad \qquad \qquad \ \ -i \ = \ cos \pi + i sin \pi. #

Explanation:

# "One way to do this is as follows." #

# "If" \ \ z \ \ "is a complex number, then:" #

# \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad z \ = \ r ( cos theta + i sin theta ); \qquad \quad "where:" #

# \quad r = "magnitude (length) of" \ \ z, #

# \quad \theta = "the angle" \ \ z \ \ "makes, in radians when" \ z \ "is placed in" #
# \qquad \qquad \qquad "standard position in the complex plane." #

# "Now visualize" \ z = -i \ \ "on the complex plane. It lies on the" #
# "negative" \ x "-axis, 1 unit in length, to the left of the origin. The" #
# "angle of" \ -i \ \ "in this, its standard position, is clearly" \ \ \pi \ \"radians." #

# "So now we see, for" \ z = -i : #

# \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad r = 1, #

# \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \theta = \pi. #

# "Thus:" #

# \qquad \qquad \qquad \qquad \qquad \qquad \quad -i \ = \ 1 cdot ( cos \pi + i sin \pi ) #

# \qquad \qquad \qquad \qquad \qquad \qquad \qquad \quad \quad \ \ = \ cos \pi + i sin \pi. #

# "So:" #

# \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad -i \ = \ cos \pi + i sin \pi. #

# "This is our answer." #