What is the trigonometric form of # (-6+i) #?

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Answer 1 · 2018-04-10T18:42:14

#color(blue)(sqrt(37)[cos(2.976)+isin(2.976)]#

The trigonometric for of a complex number #( a+bi)# is given by:

#z = r[cos(theta)+isin(theta)]#

Where:

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

#theta=arctan(b/a)#

#r=sqrt((-6)^2+(1)^2)=sqrt(37)#

#theta=arctan(1/-6)~~-0.1651486774#

This is in the IV quadrant.

Remember that #tan(theta)# only has an inverse for the domain:

#-pi/2 < theta < pi/2#

So #arctan(y)# will return angles in this range.

So we need to add #pi# to this result, since #(-6+i)# is in the II quadrant:

#-0.1651486774+pi=2.976# 3 d.p.

So:

#theta=2.976#

Trigonometric form is therefore:

#color(blue)(sqrt(37)[cos(2.976)+isin(2.976)]#