Converting standard to polar form

  • #1
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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.
 
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Answers and Replies

  • #2
Buzz Bloom
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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
 
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  • #3
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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 270o?
 
  • #4
LCKurtz
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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##.
 
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  • #5
Buzz Bloom
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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
 
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  • #6
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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##.
LCKurtz said:
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##.
 
  • #7
LCKurtz
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@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)##.
 

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