0
$\begingroup$

I need to calculate $$\left| \frac{3}{\sqrt{20}} + i\!\cdot\!\frac{1}{\sqrt{20}}\!\cdot\!e^{i\!\cdot\!\frac{\pi}{3}} \right|$$

Is there a way to do it without turning the polar form into cartesian, multiply by $i$ and take magnitude of the resulting cartesian complex number? I've tried changing all the numbers to polar complex form but with different argument I'm not sure how to add.

|| is absolute value sign by the way.

$\endgroup$
  • $\begingroup$ Just change the polar part to cartesian and go about your business as usual. The geometric interpretation of the polar form may help you identify it as well. $\endgroup$ – CyclotomicField Jan 16 at 2:29
  • $\begingroup$ @CyclotomicField The question specifically asked if it were possible to do it without turning the polar into Cartesian form. So the answer is 'no', unless the arguments were the same or separated by $\pi$. $\endgroup$ – Deepak Jan 16 at 2:31
  • $\begingroup$ @CyclotomicField Deepak's interpretation is correct, it feels like "brute force" to have to convert everything back to cartesian and distribute, was wondering if there's a more simpler way. $\endgroup$ – MinYoung Kim Jan 16 at 2:38
  • $\begingroup$ You don't need to. They are asking for the absolute value. So you just need to do that. Multiply by the complement and take the square root. $\endgroup$ – fleablood Jan 16 at 3:07
0
$\begingroup$

It's asking you not for the complex number but its absolute value.

The complement of $x + z$ where $x$ is real is $x + \overline z$.

And the complement of $re^{i\theta}$ is $re^{-i\theta}$.

And $i*e^{i\theta} = e^{i\frac \pi 2}e^{i\theta} = e^{i (\theta +\frac \pi 2)}$.

So $|\frac 3{\sqrt {20}} + i\frac 1{\sqrt{20}}e^{i\frac \pi 3}| =$

$\sqrt{(\frac 3{\sqrt {20}} +\frac 1{\sqrt{20}}e^{i(\frac \pi 3+\frac \pi 2)})(\frac 3{\sqrt {20}} +\frac 1{\sqrt{20}}e^{-i(\frac \pi 3+\frac \pi 2)})}=$

$\sqrt{\frac 9{20} + \frac 3{20}(e^{i\frac {5\pi}6} + e^{-i\frac {5\pi}6}) + \frac 1{20}}=$

$\sqrt{\frac {10}{20} + \frac 3{20}*2\cos \frac {5\pi}6}=$

$\sqrt{\frac 12 - \frac {3\sqrt 3}{20}}$

And I probably made an arithmetic error somewhere....

$\endgroup$
  • $\begingroup$ Nope, that is the correct answer I got by turning everything into cartesian. The question actually asked for square of the magnitude, so without that square root, it makes things even simpler, $\frac{1}{2} - \frac{3\sqrt{3}}{20}$ = .24, which is what I got. Thanks, didn't think about just multiplying its conjugate. $\endgroup$ – MinYoung Kim Jan 16 at 16:34
  • $\begingroup$ I say the $\frac {\sqrt{20}$ and the $3$ and the $1$s becoming nice I was expecting and nice clever negative or pythagorean triple simplifying out. $\endgroup$ – fleablood Jan 16 at 16:50
0
$\begingroup$

In general, it's only trivial to add (or subtract) two complex numbers in polar form if their arguments are the same or separated by $\pi$.

When the arguments are the same, $\ r_1e^{i\theta} \pm r_2{i\theta} = (r_1 \pm r_2)e^{i\theta}$

When the arguments are separated by $\pi$, $\ r_1e^{i\theta} \pm r_2e^{i(\theta \pm \pi)} = r_1e^{i\theta} \mp r_2e^{i\theta} = (r_1 \mp r_2)e^{i\theta}$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.