The problem has two parts. Part a: Calculate the sum $z = 1 + i + i^2 + i^3 + i^4 + \dots + i^{2017}$. Part b: Given $z_k = i^{2k} + i^{2k+1}$, prove that $z_k + z_{k+1} = 0$.

1. Problem Description

The problem has two parts.

Part a: Calculate the sum $z = 1 + i + i^2 + i^3 + i^4 + \dots + i^{2017}$.

Part b: Given $z_k = i^{2k} + i^{2k+1}$, prove that $z_k + z_{k+1} = 0$.

2. Solution Steps

Part a:

We have a geometric series with the first term $a = 1$, common ratio $r = i$, and $n = 2018$ terms. The formula for the sum of a geometric series is

$S_n = \frac{a(1-r^n)}{1-r}$.

In this case,

$z = \frac{1(1-i^{2018})}{1-i} = \frac{1 - (i^2)^{1009}}{1-i} = \frac{1 - (-1)^{1009}}{1-i} = \frac{1 - (-1)}{1-i} = \frac{2}{1-i}$.

To simplify, multiply the numerator and denominator by the conjugate of the denominator:

$z = \frac{2(1+i)}{(1-i)(1+i)} = \frac{2(1+i)}{1 - i^2} = \frac{2(1+i)}{1 - (-1)} = \frac{2(1+i)}{2} = 1+i$.

Part b:

We are given $z_k = i^{2k} + i^{2k+1}$.

We want to show that $z_k + z_{k+1} = 0$.

First, let's find $z_{k+1}$:

$z_{k+1} = i^{2(k+1)} + i^{2(k+1)+1} = i^{2k+2} + i^{2k+3}$.

Now, let's calculate $z_k + z_{k+1}$:

$z_k + z_{k+1} = (i^{2k} + i^{2k+1}) + (i^{2k+2} + i^{2k+3}) = i^{2k} + i^{2k+1} + i^{2k+2} + i^{2k+3}$.

We can factor out $i^{2k}$:

$z_k + z_{k+1} = i^{2k}(1 + i + i^2 + i^3) = i^{2k}(1 + i - 1 - i) = i^{2k}(0) = 0$.

Therefore, $z_k + z_{k+1} = 0$.

3. Final Answer

a. $z = 1+i$

b. $z_k + z_{k+1} = 0$