Integral power of iota, Algebraic operations and Equality of complex numbers Questions in English

(D) Given expression: $i^n + i^{n+1} + i^{n+2} + i^{n+3}$
Factor out $i^n$: $i^n(1 + i + i^2 + i^3)$
We know that $i^2 = -1$ and $i^3 = -i$.
Substituting these values: $i^n(1 + i - 1 - i)$
Simplifying the expression inside the parentheses: $i^n(0) = 0$
Thus,the sum of any four consecutive powers of $i$ is $0$.

(D) Given the equation: $50 \left(\frac{2x}{1 + 3i} - \frac{y}{1 - 2i}\right) = 31 + 17i$.
Multiply the numerators and denominators by their conjugates: $\frac{2x(1-3i)}{1^2+3^2} = \frac{2x-6xi}{10}$ and $\frac{y(1+2i)}{1^2+2^2} = \frac{y+2yi}{5}$.
Substituting these into the equation: $50 \left(\frac{2x-6xi}{10} - \frac{y+2yi}{5}\right) = 31 + 17i$.
$50 \left(\frac{2x-6xi - 2(y+2yi)}{10}\right) = 31 + 17i$.
$5(2x - 6xi - 2y - 4yi) = 31 + 17i$.
$10x - 30xi - 10y - 20yi = 31 + 17i$.
Grouping real and imaginary parts: $(10x - 10y) + i(-30x - 20y) = 31 + 17i$.
Comparing real and imaginary parts:
$10x - 10y = 31$ (Equation $1$)
$-30x - 20y = 17$ (Equation $2$)
Multiply Equation $1$ by $2$: $20x - 20y = 62$ (Equation $3$).
Add Equation $2$ and Equation $3$: $(-30x - 20y) + (20x - 20y) = 17 + 62 \Rightarrow -10x - 40y = 79$.
Alternatively,from Equation $1$,$x - y = 3.1 \Rightarrow x = y + 3.1$.
Substitute into Equation $2$: $-30(y + 3.1) - 20y = 17 \Rightarrow -30y - 93 - 20y = 17 \Rightarrow -50y = 110 \Rightarrow y = -2.2$.
Then $x = -2.2 + 3.1 = 0.9$.
Calculate $10(x - 3y) = 10(0.9 - 3(-2.2)) = 10(0.9 + 6.6) = 10(7.5) = 75$.