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Converting from Cartesian to polar form

nacho

Active member
Sep 10, 2013
156
another question:

convert $|\frac{1-i}{3}|$ to polar form

i am getting $\frac{\sqrt{2}}{3} e^{\frac{i\pi}{4}}$

but the solutions say:
$e^{\frac{-i\pi}{4}}$

i did
$ x = r\cos(\theta)$ and $y=r\sin(\theta)$
so

$\frac{1}{3} = {\frac{\sqrt{2}}{3}}\cos(\theta)$
$\frac{1}{3} = \cos(\theta)$
And thus $\theta = \frac{\pi}{4}$
similarly, the same was determined for $\sin(\theta)$

What did I do wrong, why didn't i obtain a negative?
Also, is there a shorter method to find the ans? I would love if someone could give me a fully worked solution with the most efficient way to get the answer.

THanks.
 

ZaidAlyafey

Well-known member
MHB Math Helper
Jan 17, 2013
1,667
another question:

convert $|\frac{1-i}{3}|$ to polar form
The absolute value in complex analysis represents the modulus so the result will be a positive real number.
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,708
another question:

convert $|\frac{1-i}{3}|$ to polar form

i am getting $\frac{\sqrt{2}}{3} e^{\frac{i\pi}{4}}$

but the solutions say:
$e^{\frac{-i\pi}{4}}$

i did
$ x = r\cos(\theta)$ and $y=r\sin(\theta)$
so

$\frac{1}{3} = {\frac{\sqrt{2}}{3}}\cos(\theta)$
$\frac{1}{3} = \cos(\theta)$ This is wrong: $\color{red}{\cos\theta}$ should be $\color{red}{1/\sqrt2}$.
And thus $\theta = \color{red}{\pm}\frac{\pi}{4}$
similarly, the same was determined for $\sin(\theta)$

What did I do wrong, why didn't i obtain a negative?
Also, is there a shorter method to find the ans? I would love if someone could give me a fully worked solution with the most efficient way to get the answer.

THanks.
[I assume the mod signs should not be there, otherwise as ZaidAlyafey says the result should be a positive real number and $\theta$ will be zero.]

When converting to polar form, you should always check which quadrant the number lies in. In this case, $1-i$ is in the fourth quadrant, so $\cos\theta$ will be positive but $\sin\theta$ will be negative. This means that you should choose the negative value for $\theta$.
 

nacho

Active member
Sep 10, 2013
156
Sorry, both typos

The absolute value shouldn't have been there, and I had the correct workings but transcribed them onto here incorrectly!

Thanks for the note about the $1+i$ lying in the 4th quadrant,
I never looked at it the correct way from the beginning! This has been my flaw up until now.

Thank you !!
 

Petrus

Well-known member
Feb 21, 2013
739
Thanks for the note about the $1+i$ lying in the 4th quadrant,
just to make sure you mean that \(\displaystyle 1-i\) lying in the 4th quadrant cause \(\displaystyle 1+i\) lying in the first quadrant:p

Regards,
\(\displaystyle |\pi\rangle\)
 

nacho

Active member
Sep 10, 2013
156
just to make sure you mean that \(\displaystyle 1-i\) lying in the 4th quadrant cause \(\displaystyle 1+i\) lying in the first quadrant:p

Regards,
\(\displaystyle |\pi\rangle\)
thanks for the concern
I seem to be making typos all over the place! :p that's what I meant.

I didn't understand at first you could look at the points these way.

for instance, $1-i$ is referring to a coordinate, correct?

Yet whenever I think of coordinates, I think only in terms of $(1,-i)$ notation