Let $a_1=5-3i$ and let $a_2=2-8i$ be a geometric series.
a. Show $\arg(a_{n+8})=\arg(a_n)$
b. let there be an element with a real part of $24$ find its imaginary part and index
c. how much elements do we need to get to a sum of $-515-133i$
a. $$q=\frac{a_2}{a_1}=\frac{a_1\cdot q}{a_1}=1-i$$
now $a_n=a_1\cdot q^n$ and $a_{n+8}=a_1\cdot q^{n+8}=a_1\cdot q^n\cdot q^8$
but $q^8=16$ so $\arctan(\frac{y_n}{x_n})=\arctan(\frac{16y_n}{16x_n})$
b. how can I find both $n$ and $y$ in $$(5-3i)(1-i)^n=24+yi$$
$$1-i=\sqrt2\left(\cos\dfrac\pi4-i\sin\dfrac\pi4\right)$$
$\implies$ $$(1-i)^n=2^{n/2}\left(\cos\dfrac{n\pi}4-i\sin\dfrac{n\pi}4\right)$$
Real part of $(5-3i)(1-i)^n=2^{n/2}(5-3i)\left(\cos\dfrac{n\pi}4-i\sin\dfrac{n\pi}4\right)$
will be $$2^{n/2}\left(5\cos\dfrac{n\pi}4-3\sin\dfrac{n\pi}4\right)$$
We need to eliminate $5$ as $\cos\dfrac{n\pi}4=\pm\dfrac1{\sqrt2},0,\pm1\left(\implies\cos\dfrac{n\pi}4=0\implies n=2(2m+1)\right)$ and $n$ must be even to make the product $24$