Finding the polar form of a complex number

[javii]

Homework Statement

Homework Equations

r=sqrt(a^2+b^2)
θ=arg(z)
tan(θ)=b/a

The Attempt at a Solution

for a)[/B]

finding the polar form:
r=sqrt(-3^2+(-4)^2)=sqrt(7)
θ=arg(z)
tan(θ)=-4/-3 = 53.13 °
300-53.13=306.87°

-3-j4=sqrt(7)*(cos(306.87+j306.87)

I don't know if my answer is correct because it is given that -π<arg(z)<=π

 

[BvU]

tan(θ)=-4/-3 = 53.13 °

Is indeed the wrong answer. The second = sign makes no sense. You probably mean to say $$ \tan\theta = {-4\over -3}\ \ \Rightarrow \ \ \theta = 53.13^\circ$$but that is not correct. To satisfy

-π<arg(z)<=π

and to not fold everything back to the range $(-\pi/2 , \pi/2]$ the atan2 function was 'invented'.
Your 300-53.13=306.87° (an alternative answer ?) doesn't satisfy "-π<arg(z)<=π, not even if converted to radians.

And to top it all up, $3^2+4^2 \ne 7$ !

So it's back to the drawing board, I'm afraid...

Not much help (except the reference to atan2, perhaps), but with your drawing at hand I'm pretty confident you can manage on your own and that's much better.

 

[javii]

tanθ=−4−3 ⇒ θ=53.13∘tan⁡θ=−4−3 ⇒ θ=53.13∘​

Yes, that was what I meant.

32+42≠732+42≠73^2+4^2 \ne 7 !

my bad its equal 25, meaning r = 5.

I will try to read about atan2, to be honest I didn't knew about it. Thank you

 

[Ray Vickson]

Homework Statement

View attachment 114125

Homework Equations

r=sqrt(a^2+b^2)
θ=arg(z)
tan(θ)=b/a

The Attempt at a Solution

for a)[/B]
View attachment 114128
finding the polar form:
r=sqrt(-3^2+(-4)^2)=sqrt(7)
θ=arg(z)
tan(θ)=-4/-3 = 53.13 °
300-53.13=306.87°

-3-j4=sqrt(7)*(cos(306.87+j306.87)

I don't know if my answer is correct because it is given that -π<arg(z)<=π

(1) Why are you using degrees when the question expresses angles in radians?
(2) In your computation 300-53.13, where does the 300 come from?

 

[Mark44]

(2) In your computation 300-53.13, where does the 300 come from?

Pretty clearly, it's a typo, where the OP typed 300, but meant 360.

300-53.13=306.87°

 

[Ray Vickson]

Pretty clearly, it's a typo, where the OP typed 300, but meant 360.

OK, then: but where does the 360 come from?