Question

het
z =
2011
L
1901
L
i 2018 + : 2013 + : 2014
+ i '90.2 + i 1903 + 11904

Answer 1

In both denominator and numerator, we see the same structure on both.

First, we can consider factorization because there are common factors.

Let $\sf{z=\dfrac{i^{2011}+i^{2012}+i^{2013}+i^{2014}}{i^{1901}+i^{1902}+i^{1903}+i^{1904}}$

→ $\sf{z=\dfrac{i^{2011}(1+i+i^2+i^3)}{i^{1901}(1+i+i^2+i^3)}$

But we know that $\sf{1+i+i^2+i^3}$ becomes zero.

$\sf{1+i+(-1)+(-i)=0}$ gives the result.

Therefore, $\sf{z=\dfrac{0}{0} }$ and its value cannot be determined.

In mathematics, dividing by 0 is not accepted.

You get a number that is not determined.

• Let us consider $\sf{0\div0=q+r$.
• Division algorithm shows no remainder.
• → $\sf{0\div0=q}$
• → $\sf{0\times q=0}$ shown by division algorithm
• → Infinitely many solutions, therefore not determined.