# Thread: Vectors + COmplex numbers

1. ## Vectors + COmplex numbers

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!

Thank you!

2. ## Re: Vectors + COmplex numbers

For the answer is that the argument of z+1 is pi/2
Was this a typo or is this what you understood. The argument is theta/2 not pi/2

3. ## Re: Vectors + COmplex numbers

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 + cos2(θ/2) - sin2(θ/2)]

= (2sin(θ/2)cos(θ/2))/(2cos2(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 + sin2(θ)] = √[1 + 2cos(θ) + cos2(θ) + sin2(θ)]

= √(2 + 2cos(θ)) = √(2 + 2cos(θ/2 + θ/2)) = √(2 + 2cos2(θ/2) - 2sin2(θ/2)) = √(4cos2(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.

4. ## Re: Vectors + COmplex numbers

Hi thank you so much!

How did you know that arg(z+1) = arctan(sin(θ)/(1+cos(θ)), why is the one added to the cos and not the sin etc?

Thank you!

5. ## Re: Vectors + COmplex numbers

the "c" in "cis" stands for cosine, which is the real part of the number cos(θ) + i sin(θ). the number 1 is real, so we add it to the real part.

6. ## Re: Vectors + COmplex numbers

Originally Posted by Tutu
[COLOR=#1a1a18]Suppose z = cis A where A is acute. Find the argument of z + 1

There is a very easy and intuitive geometric proof for this.

The figure $OCBA$ is a rhombus, because $z$ is on the unit circle, $|z|=1$. As vectors $z+1$ is a diagonal of that rhombus which bisects the angle that is $\text{arg}(z).$

7. ## Re: Vectors + COmplex numbers

Originally Posted by Plato
There is a very easy and intuitive geometric proof for this.

The figure $OCBA$ is a rhombus, because $z$ is on the unit circle, $|z|=1$. As vectors $z+1$ is a diagonal of that rhombus which bisects the angle that is $\text{arg}(z).$
yes! such a proof is depicted in the attachment of the original post (the blue diagram).