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Let  y in the form of $a + bi$, where $a$ and $b$ are real numbers, be the cubic roots of complex number $z^{20},$ where $z=\frac{2}{4 + 3i}.$ Find $a + b.$

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$z = \frac{2}{4+3i} = \frac{2(4 – 3i)}{(4+3i)(4-3i)} = \frac{8-6i}{25}=\frac{8}{25} – \frac{6i}{25}$

Now, try to convert it into polar form i.e. $z = a+bi= re^{i\theta} = r (\cos \theta + i \sin \theta)$

where $r= |z| = \sqrt{a^2 + b^2}$ and $\theta = \tan^{-1} (\frac{y}{x})$ for $0 \leq \theta < 2\pi$

So, here, $r = |z| = \frac{2}{5}$ and $\theta = -\tan^{-1}(\frac{3}{4})$

Hence, $z = \frac{2}{5} \left(\cos(-\tan^{-1}\left(\frac{3}{4}\right))\ + i \sin(-\tan^{-1}\left(\frac{3}{4}\right)) \right)$

and so, $z^{20} = \left(\frac{2}{5}\right)^{20} \left(\cos(-20\tan^{-1}\left(\frac{3}{4}\right))\ + i \sin(-20\tan^{-1}\left(\frac{3}{4}\right)) \right)$

It is from De Moivre's Theorem.

For a complex number $z,$ $n^{th}$ root is $z^{1/n}$ which is $r^{1/n}\left(\cos \left(\frac{\theta}{n} +\frac{2k\pi}{n}\right) + i \sin \left(\frac{\theta}{n} +\frac{2k\pi}{n}\right) \right)$ for $k=0,1,2,..n-1$

So, here, say, $w_k$ is the cubic root of $z^{20},$ it means $w_k = (z^{20})^{1/3}$ for $k=0,1,2.$ Hence,

$w_k = \left(\frac{2}{5}\right)^{20/3} \left(\cos \left(\frac{-20\tan^{-1}\left(\frac{3}{4}\right)}{3} +\frac{2k\pi}{3}\right) + i \sin \left(\frac{-20\tan^{-1}\left(\frac{3}{4}\right)}{3} +\frac{2 k\pi}{3}\right) \right)$ for $k=0,1,2.$

Now, you have $w_0,w_1,w_2$ in the form of $a+bi$ and so, you can find $a+b$ for each $w_0,w_1,w_2.$