Converting standard to polar form

[Arnoldjavs3]

Homework Statement

you are given the standard form z = 3 - 3i

Homework Equations

The Attempt at a Solution

so to convert this to polar form, i know that $r = 3√2$ but how do i find theta here? There are so many mixed answers it seems online that I can't tell... i know that $(3,-3)$ is in the last quadrant and that $tan^-1(-3/3) = -45$.

But how can I do this all without a calculator first of all? I have a final where no calculators are allowed. Some sites are telling me that theta is just -45 or -pi/4 here. Others are telling me that its 360 - (-45) or 360 + -45.
What the heck is the right answer?

Also, just for my understanding here. say I have a different standard form where $z=-8i$ and I want to find its cubed roots. Would theta be 270 here? or $3pi/2$? Because $tan^-1(-8/0)$ is undefined.

 

[Buzz Bloom]

Some sites are telling me that theta is just -45 or -pi/4 here. Others are telling me that its 360 - (-45) or 360 + -45.

Hi Arnoldjavs3:

What is the difference between the two answers: (a) -45, and (b) 360-45=315.

BTW: I don't know what your teacher requires, but in general it is better to include a symbol like "^o" or "deg" for an angle using degrees as a unit rather than omit it.

Regards,
Buzz

 

[Arnoldjavs3]

Hi Arnoldjavs3:

What is the difference between the two answers: (a) -45, and (b) 360-45=315.

BTW: I don't know what your teacher requires, but in general it is better to include a symbol like "^o" or "deg" for an angle using degrees as a unit rather than omit it.

Regards,
Buzz

Oh... right. I didn't know how to add the degree symbol with latex. I feel stupid now.

How about the degree for $z=-8i$? Am I right to think that it is 270^o?

 

[LCKurtz]

Homework Statement

you are given the standard form z = 3 - 3i

Homework Equations

The Attempt at a Solution

so to convert this to polar form, i know that $r = 3√2$ but how do i find theta here? There are so many mixed answers it seems online that I can't tell... i know that $(3,-3)$ is in the last quadrant and that $tan^-1(-3/3) = -45$.

But how can I do this all without a calculator first of all? I have a final where no calculators are allowed. Some sites are telling me that theta is just -45 or -pi/4 here. Others are telling me that its 360 - (-45) or 360 + -45.
What the heck is the right answer?

Draw a line from the origin to $(3,-3)$. Label it $r$. Then draw an arc counterclockwise from the positive $x$ axis to $r$. That arc subtends the angle you want. Don't use any inverse trig formula, just look at it. You should see that it is $180^\circ + 45^\circ$ or $\pi +\frac \pi 4 =\frac{5 \pi} 4$. Just draw a quick picture for this kind of problem.
[Edit, corrected] As Mark44 points out in post #6, I meant
$270^\circ + 45^\circ$ or $\frac{3\pi} 2 +\frac \pi 4 =\frac{7 \pi} 4$.

Also, just for my understanding here. say I have a different standard form where $z=-8i$ and I want to find its cubed roots. Would theta be 270 here? or $3pi/2$? Because $tan^(-1)[-8/0]$ is undefined.

Again, don't use inverse trig functions here. You want$$
r^3e^{i3\theta} = 8e^{\frac {3\pi i} 2}$$ So $r=2$ and $3\theta = \frac {3\pi} 2 + 2n\pi$.

 

[Buzz Bloom]

How about the degree for z=-8i?

Hi Arnoldjavs3:

What do you think the answer is?

BTW: How to represent the value of an angle in the third or fourth quadrant is an arbitrary convention. The two choices are
(a) 180 < θ < 360, or
(b) 0 > θ > - 180.
You might want to notice which convention your teacher usually uses, and do the same.

Another BTW re

I didn't know how to add the degree symbol with latex. I feel stupid now.

There are many useful symbols available by selecting "∑" on the formatting option bar.

Regards,
Buzz

 

[Mark44]

Draw a line from the origin to $(3,-3)$. Label it $r$. Then draw an arc counterclockwise from the positive $x$ axis to $r$. That arc subtends the angle you want. Don't use any inverse trig formula, just look at it. You should see that it is $180^\circ + 45^\circ$ or $\pi +\frac \pi 4 =\frac{5 \pi} 4$. Just draw a quick picture for this kind of problem.

@LCKurtz, I'm sure you really mean $270^\circ + 45^\circ$ or $\frac {3\pi} 2 + \frac \pi 4 = \frac{7\pi} 4$.

Again, don't use inverse trig functions here. You want$$
r^3e^{i3\theta} = 8e^{\frac {3\pi i} 2}$$ So $r=2$ and $3\theta = \frac {3\pi} 2 + 2n\pi$.

 

[LCKurtz]

@LCKurtz, I'm sure you really mean $270^\circ + 45^\circ$ or $\frac {3\pi} 2 + \frac \pi 4 = \frac{7\pi} 4$.

Yes, of course. For some reason I copied his point as $(-3,-3)$.