Question

Solve $i^{57}+\frac{1}{i^{125}}$

• A. $0$
• B. $2i$
• C. $-2i$
• D. $2$

Answer 1

Answer: (A)

$0$

We know that $i^2=-1$

$i^3=i^2.i=-i$

$i^4=i^2\times i^2=-1\times -1=1$

Thus, $i^{57}={\left(i^4\right)}^{14}\times i=1^{14}\times i=i$

$i^{125}={\left(i^4\right)}^{31}\times i=1^4\times i=i$

$\frac{1}{i^{125}}=\frac{1}{i}$

$\frac{1}{i}=\frac{1}{i}\times \frac{i}{i}=\frac{i}{i^2}=-i$

Now $i^{57}+\frac{1}{i^{125}}=i+(-i)=i-i=0$