Was this a typo or is this what you understood. The argument is theta/2 not pi/2For the answer is that the argument of z+1 is pi/2
Hi can you help me with:
Suppose z = cis A where A is acute. Find the argument of z + 1:
I understand how to draw the vectors and all on the Argand diagram, so the question with the answers and all are in the first image.
However,t o find the argument of z+1, I don't understand, my understanding is that the argument of z+1 is the red angle, as shown in the next picture. Why is it wrong? For the answer is that the argument of z+1 is pi/2 since it is a rhombus, but I don't see how, not from my understanding of what a z+1 argument is!
Please help, thank you so much!
Thank you!
assuming z = cis(θ) is in the first quadrant, we have arg(z) = arctan(sin(θ)/cos(θ)) = arctan(tan(θ)) = θ.
now arg(z+1) = arctan(sin(θ)/(1+cos(θ))
let's look closer at what we're taking the arctangent OF:
sin(θ)/(1+cos(θ)) = sin(θ/2 + θ/2)/[(1 + cos(θ/2 + θ/2)] = (2sin(θ/2)cos(θ/2))/[1 + cos^{2}(θ/2) - sin^{2}(θ/2)]
= (2sin(θ/2)cos(θ/2))/(2cos^{2}(θ/2)) = sin(θ/2)/cos(θ/2) = tan(θ/2).
hence arg(z+1) = arctan(sin(θ)/(1+cos(θ)) = arctan(tan(θ/2)) = θ/2.
the magnitude is also easy to find:
|z+1| = √[(1+cos(θ))^{2} + sin^{2}(θ)] = √[1 + 2cos(θ) + cos^{2}(θ) + sin^{2}(θ)]
= √(2 + 2cos(θ)) = √(2 + 2cos(θ/2 + θ/2)) = √(2 + 2cos^{2}(θ/2) - 2sin^{2}(θ/2)) = √(4cos^{2}(θ/2))
= 2cos(θ/2) (because we are in the first quadrant).
the "geometric" argument is a bit more elegant, and visual, to boot, taking advantage of the symmetry of z and 1 with respect to z+1. this only works because z is the same magnitude as 1.