First, rationalize the denominator to remove #i# from it:
#(i+9)/(2i+2)color(green)(*((2i-2)/(2i-2)))#
#=(2i^2-2i+18i-18)/(4i^2cancel(-4i)cancel(+4i)-4)#
#=(-2+16i-18)/(-4-4)#
#=(-20+16i)/-8#
#=5/2-2i#
So this point can be written in #x+yi# form as #5/2-2i#. Note: since #y<0# and #x>0#, this point is in the bottom-right quadrant, #Q_"IV"#.
Next, find #r#, the length of the vector whose tip is at this point:
#r=sqrt(x^2+y^2)#
#r=sqrt((5/2)^2+(-2)^2)#
#r=sqrt(25/4+4)#
#r=sqrt(41/4)=sqrt(41)/2#
Finally, we need to find the angle #theta# this vector makes with the positive #x#-axis. This can be done using #costheta=x/r# or #sintheta=y/r#.
#sintheta=y/r#
#=>theta=sin^-1(y/r)#
#theta=sin^-1((-2)/(sqrt(41)//2))#
#theta=sin^-1((-4)/sqrt(41))#
#thetaapprox-0.6747approx-38.66°#
Side note: Because #sin^-1# can only return an angle between #-pi/2# and #+pi/2# (that is, an angle in #Q_"I"# or #Q_"IV"#), the #theta# we get by using #sin^-1# should always be considered as the reference angle for our true angle. We then use the reference angle and the vector's quadrant to determine the true angle. In this case, the vector lies in #Q_"IV"#, so our reference angle happens to be the true angle.
To write the vector in trigonometric form, we note that #x=rcostheta# and #y=r*sintheta#. So our vector #z# goes from
#z=x+iy#
to
#z=rcostheta+i*rsintheta#
#=r(costheta+isintheta)#
And since we now know #r# and #theta#, we can write
#z=sqrt(41)/2[cos(-0.6747)+isin(-0.6747)]#.
