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A complex sequence $(z_n)$ is given by: $$z_n=\frac{e^{in^2}}{1+in^2}$$

I found its absolute value, $|z_n|=\frac{1}{\sqrt{1+n^4}}$, by multiplying the top and bottom by the conjugate of the complex denominator, then expanded out the exponential term into sine and cosine terms using Euler's formula, rearranged to get a real part and an imaginary part of $z_n$ and then used $|z_n|=\sqrt{(Re(z_n))^2+(Im(z_n))^2}$ to find the absolute value of $z_n$. With some rearranging and cancellations, I got my answer (above).

I know it's the right answer, but I was wondering if there's a nicer/quicker way of doing this? It seems as if there surely is, but I somehow can't see it.

Thank you.

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    $\begingroup$ Regarding your side question, yes, $|a+b|=\sqrt(a+b)^2$, since if $a$ and $b$ are both real, then $a+b=c$, some other real number, whilst $a+ib$ is irreducible. $\endgroup$
    – AmpleMimic
    Commented Jan 10, 2016 at 23:33
  • $\begingroup$ Notice, for real numbers $a$ & $b$, it's true that $|a+b|=\sqrt{(a+b)^2}$ $\endgroup$ Commented Jan 10, 2016 at 23:34

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First of all, $$ \left| \frac zw \right| = \frac{|z|}{|w|} $$ so you can compute the absolute value of numerator and denominator separately.

From $|z|^2 = z \overline z$ you get $$ |e^{in^2}|^2 = e^{in^2} e^{-in^2} = 1 $$ and $$ |1+in^2|^2 = (1+in^2)(1-in^2) = 1 + n^4 \, . $$ Alternatively, $$ |e^{in^2}| = \sqrt{\cos^2 (n^2) + \sin^2 (n^2)} = 1 $$ and $$ |1+in^2| = \sqrt{ 1^2 + (n^2)^2} = \sqrt{1 + n^4} \, . $$

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Notice, when $a,b\in\mathbb{R}$:

  • $$\left|\frac{a+bi}{c+di}\right|=\frac{\left|a+bi\right|}{\left|c+di\right|}$$
  • $$|a+bi|=\sqrt{\Re^2(a+bi)+\Im^2(a+bi)}=\sqrt{a^2+b^2}$$
  • $$a+bi=|a+bi|e^{\arg(a+bi)i}=|a+bi|\left(\cos\left(\arg(a+bi)\right)+\sin\left(\arg(a+bi)\right)i\right)$$

Assume $n\in\mathbb{R}$:

$$\left|z_n\right|=\left|\frac{e^{in^2}}{1+in^2}\right|=\frac{\left|e^{in^2}\right|}{\left|1+in^2\right|}=\frac{\left|\cos(n^2)+\sin(n^2)i\right|}{\left|1+in^2\right|}=$$


Now, see that the real part of our numerator is $\cos(n^2)$ and denominator is $1$

and the imaginary part of our numerator is $\sin(n^2)$ and denominator is $n^2$.


$$\frac{\sqrt{\cos^2(n^2)+\sin^2(n^2)}}{\sqrt{1^2+\left(n^2\right)^2}}=$$


Use $\cos^2(x)+\sin^2(x)=1$:


$$\frac{\sqrt{1}}{\sqrt{1^2+\left(n^2\right)^2}}=\frac{1}{\sqrt{1^2+n^4}}=\frac{1}{\sqrt{1+n^4}}$$

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