How do you divide #( -3i+2) / (-i -1 )# in trigonometric form?

1 Answer
Jul 1, 2018

#z= sqrt26/2(cos(78.69^@)+isin(78.69^@))# or
#z=5/2i+1/2#

Explanation:

Let's start by solving this in the form given and see what we get:

Multiply the conjugate of the denominator:
#( -3i+2) / (-1-i)*(-1+i)/(-1+i)=(3i-2-3i^2+2i)/(1-i^2)=(5i+1)/2=5/2i+1/2#

#1/2+5/2i#
We can convert this to trigonometric form by finding #r#:
#r=sqrt(a^2+b^2)#
#theta=arctan(b/a)#

#r=sqrt((1/2)^2+(5/2)^2)#

#r= sqrt26/2#

#theta=arctan(5)#
#thetaapprox78.69^@#

So we can say the solution in trig form is
#z=sqrt26/2(cos(78.69^@)+isin(78.69^@))#

Let's double check by solving in trigonometric form:
#z_1= 2-3i#
#r= sqrt((2)^2+(-3)^2)#
#r= sqrt(13)#

#theta= arctan(-3/2)approx-56.31^@approx303.69^@#

#z_2=-1-i#
#r= sqrt((-1)^2+(-1)^2)#
#r=sqrt2#
#theta= arctan(-1/-1)= 45^@#, however this in the third quadrant so#+180^@= 225^@#

#z_1/z_2= (sqrt13(cos(303.69^@)+isin(303.69^@)))/(sqrt2(cos(225^@)+isin(225^@))#

WHEN DIVIDING IN TRIGONOMETRIC FORM:
Divide the r by the other r
And subtract the angles

#sqrt13/sqrt2*sqrt2/sqrt2= sqrt26/2#

#303.69^@-225^@= 78.69^@#

So same solution:
#z= sqrt26/2(cos(78.69^@)+isin(78.69^@))#