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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 $$(5-3i)(1-i)^n=24+yi$$

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newhere
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1 Answers1

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$$1-i=\sqrt2\left(\cos\dfrac\pi4-i\sin\dfrac\pi4\right)$$

$\implies$ $$(1-i)^n=2^{n/2}\left(\cos\dfrac{n\pi}4-i\sin\dfrac{n\pi}4\right)$$

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 $24$

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