0
$\begingroup$

$$ |2e^{it}-1|^2$$

I don't understand how to work this out, I know if I had for example $|2ti-1|^2$ then I would square the real and imaginary parts and add them to get the modulus squared, but here I have $|2e^{it}-1|^2$ and I don't understand what to do.

Any help would be much appreciated.

$\endgroup$
  • 1
    $\begingroup$ Remember that $e^{it} = \cos t + i \sin t$. $\endgroup$ – Simon S Mar 31 '15 at 12:16
1
$\begingroup$

Considering that $t$ is real, we use Euler's formula $e^{i\theta}=\cos\theta+i\sin\theta$

$$|2e^{it}-1|^2=|2\cos t+2i\sin t-1|^2=(2\cos t-1)^2+(2\sin t)^2$$

This evaluates as $(5-4\cos t)$ and as you can see, the value is dependent on $t$ which is the argument of the complex number $e^{it}$ in polar form.

$\endgroup$
  • $\begingroup$ If you consider $t$ as not purely real, then it becomes more bashy since you need to introduce hyperbolic trigonometric functions into the mix. $\endgroup$ – Prasun Biswas Mar 31 '15 at 12:21
2
$\begingroup$

conceptually, if you are new to complex numbers then it is worth getting accustomed to thinking of the squared modulus of $z$ as also the product $z \bar z$ where $\bar z$ is the conjugate $x-iy$ of $z=x+iy$. this is a natural operation in the context of quadratic field extensions such as $\mathbb{C}=\mathbb{R}[\sqrt{-1}] $ but sometimes may give an advantage in calculation, or offers a variation of method. e.g. in the present case, with $z=2e^{it}-1$ then if $t$ is real you may write: $$ |z|^2 = z \bar z = (2e^{it}-1)(2e^{-it}-1) \\ =4e^{it}e^{-it} - 2(e^{it}+e^{-it}) + 1 \\ = 5 - 4 \cos t $$

$\endgroup$

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.