Complex Number Equations

[Carlb]

Hey, I am having trouble solving a couple of equations including complex numbers.

"In Exercise 5-15, find the modulus r = absolute value(z) and the principal argument(theta)=Arg(z) of each given complex number z, and express z in terms of r and theta"

11) z=-3-4i
My work: absolute value(z)= sqrt(a(squared)+b(squared))
=sqrt(9+16)
Abs Value(z)= 5
therefore....
z=5(-3/5-4/5i)
Arg(z)=5(cos-3/5-isin4/5)

Now i am just stuck figuring out the value of Arg(z)
Apparently the answer is theta=-pi+tan^(-1)(4/3) the negative one is the power of tan while 4/3 is the value in the brackets of tan

15) z=3cos4pi/5+3isin4pi/5
I have no clue where to start because I am not sure whether or not I include the cos and sin in the expression absolute value(z)= sqrt(a(squared)+b(squared))

I don't know how to take the cos or sin of 4pi/5 either

"In exercises 18-23, express in the form z=x=yi the complex number z whose modulus and argument are given"

19) Abs Val(z)=5
Arg(z)= tan^(-1)(3/4) the (-1) and (3/4) are separate

The furthest I got is z=5(x+yi)

 

[Ishuda]

You are good down to
z=5(-3/5-4/5i)

To continue, z can also be written as
z=5(cos(\(\displaystyle \theta\)) + i sin(\(\displaystyle \theta\)))
where \(\displaystyle \theta\) is Arg(z). Thus
cos(\(\displaystyle \theta\)) = -3/5
and
sin(\(\displaystyle \theta\)) = -4/5
That means that
tan(\(\displaystyle \theta\)) = 4/3
that is, putting \(\displaystyle \theta\) in the proper quadrent,
\(\displaystyle \theta\) = -\(\displaystyle \pi\) + tan^-1(4/3)

Edit: correct arithmetic - hopefully all of it and to add that hopefully this will help with the other problems

 

[Carlb]

You are good down to
z=5(-3/5-4/5i)

To continue, z can also be written as
z=5(cos(\(\displaystyle \theta\)) + i sin(\(\displaystyle \theta\)))
where \(\displaystyle \theta\) is Arg(z). Thus
cos(\(\displaystyle \theta\)) = -3/5
and
sin(\(\displaystyle \theta\)) = -4/5
That means that
tan(\(\displaystyle \theta\)) = 4/3
that is, putting \(\displaystyle \theta\) in the proper quadrent,
\(\displaystyle \theta\) = -\(\displaystyle \pi\) + tan^-1(4/3)

Edit: correct arithmetic - hopefully all of it and to add that hopefully this will help with the other problems

That makes a lot of sense, I think I focused to much on how to get to the answer instead of worrying about actually solving the problem, thank you!

Couple more questions just for clarification:

Is the negative pi in the answer a way to indicate what quadrant theta is in because the negatives cancel? or is there some arithmetic way that it came to be

If its a correction for the quadrant, would that mean that i could also write the answer as: \(\displaystyle \theta\) = tan^-1(4/3) -\(\displaystyle \pi\)

 

[Ishuda]

That makes a lot of sense, I think I focused to much on how to get to the answer instead of worrying about actually solving the problem, thank you!

Couple more questions just for clarification:

Is the negative pi in the answer a way to indicate what quadrant theta is in because the negatives cancel? or is there some arithmetic way that it came to be

If its a correction for the quadrant, would that mean that i could also write the answer as: \(\displaystyle \theta\) = tan^-1(4/3) -\(\displaystyle \pi\)

It's a correction for quadrant. Because of the answer, I'm assuming the answer should lie in (-\(\displaystyle \pi\),\(\displaystyle \pi\)]. Yes, the answer could also be written as
\(\displaystyle \theta\) = tan^-1(4/3) - \(\displaystyle \pi\).
Also note that I would assume tan^-1(4/3) lies in the first quadrant.