How do you write #1-i# in trigonometric form?

1 Answer
Oct 23, 2017

#sqrt(2)*cis ((7pi)/4)#

Explanation:

First, recall the basic conversion formulas to convert from a complex number in #a+bi# format into trigonometric #r*cis theta# format. These come mainly from the rectangular to polar conversion methodology:

#r = sqrt(a^2 + b^2) #
# tan theta_{ref} = |b/a| color(white)("aaaaaa")"Reference Angle"#

In this problem, #a = 1# and #b = -1#. We can find #r# readily:

#r = sqrt(a^2 + b^2) = sqrt(1^2 + (-1)^2) = sqrt(1 + 1) = sqrt(2)#

To find the proper #theta# angle, we begin with finding the reference angle #theta_{ref}# first:

#tan theta_{ref} = |(-1)/(1)|#
#tan theta_{ref} = 1#
#theta_{ref} = pi/4#

(This could have been readily apparent by noticing that both #a# and #b# are the same absolute value (1).)

If we consider that the complex number #1-i# is graphed in Quadrant IV of the complex plane, and we recall how to use reference angles, we can see that the proper #theta# in this case is:

#theta = 2pi - pi/4 = (7pi)/4#

Thus, the complex number #1-i# in trigonometric form is #sqrt(2)*cis (7pi)/4#, also written as #sqrt(2)(cos ((7pi)/4) + i*sin ((7pi)/4))#