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

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.

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

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.

  • $\begingroup$ 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

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: $$ |z|^2 = z \bar z = (2e^{it}-1)(2e^{-it}-1) \\ =4e^{it}e^{-it} - 2(e^{it}+e^{-it}) + 1 \\ = 5 - 4 \cos t $$


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