An algebra problem by Md Zuhair

Lemma:

The sum of consecutive powers of $i$ in cycles of $4$ is $0$ .

Proof:

Since we know that

$i^n := \begin{cases} i & ; & n = 4k+1 \\ -1 & ; & n = 4k+2 \\ -i & ; & n = 4k+3 \\ 1 & ; & n = 4k \end{cases}$

So over any set of consecutive cycles of $4$ in the exponent of $i$ (eg - starting with $4k+1$ ), we have

$\displaystyle \sum_{j=4n+1}^{4m} i^j = (m-n) \sum_{j=1}^{4} i^j = (m-n) (0) = 0$

Thus, the lemma is true.

Similarly we can write

$\displaystyle \sum_{r=1}^{2013} i^r = \sum_{r=1}^{2012} i^r + i^{2013} = i^{2013} = i \\ \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \color{#3D99F6} \left( \because 2012 \equiv 0 \pmod{4} \implies \sum_{r=1}^{2012} i^r = 0 \right)$

Hence $\arg (i) = \dfrac{\pi}{2} = 90^{\circ}$ .

So $90-45 = \boxed{45}$ .

$\begin{aligned} \arg \left(\sum_{r=1}^{2013} i^r \right) & = \arg \left(i \times \frac {i^{2013}-1}{i-1} \right) \\ & = \arg \left(i \times \frac {i^{4\times503+1}-1}{i-1} \right) \\ & = \arg \left(i \times \frac {i-1}{i-1} \right) \\ & = \arg i = 90^\circ \end{aligned}$

$\implies a = \boxed{90}$

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Alternative solution

Note that $i^n = \begin{cases} 1 & \text{if } n \text{ mod }4 = 0 \\ i & \text{if } n \text{ mod }4 = 1 \\ -1 & \text{if } n \text{ mod }4 = 2 \\ -i & \text{if } n \text{ mod }4 = 3 \end{cases}$

$\begin{aligned} \implies \sum_{\color{#D61F06}r=0}^n i^r & = \begin{cases} 1 & \text{if } n \text{ mod }4 = 0 \\ 1+ i & \text{if } n \text{ mod }4 = 1 \\ i & \text{if } n \text{ mod }4 = 2 \\ 0 & \text{if } n \text{ mod }4 = 3 \end{cases} \\ \sum_{\color{#3D99F6}r=1}^n i^r & = \begin{cases} 1 & \text{if } n \text{ mod }4 = 0 \\ 1+ i & \text{if } n \text{ mod }4 = 1 \\ i & \text{if } n \text{ mod }4 = 2 \\ 0 & \text{if } n \text{ mod }4 = 3 \end{cases} - 1 \\ \implies \sum_{\color{#3D99F6}r=1}^{2013} i^r & = {\color{#3D99F6} 1+i } - 1 = i & \small \color{#3D99F6} \text{As } 2013 \text{ mod }4 = 1 \\ \arg \left( \sum_{r=1}^{2013} i^r \right) & = \arg i = 90^\circ \end{aligned}$

$\implies a = \boxed{90}$