Question

O número complexo $<var>z = 6i^{25} + (2i)^6 + (i)^{-3} </var>$ é igual a quanto?

Answer 1

$6i^{25} + (2i)^6 + (i) ^{-3} = 6i^1 + 64i^2 + (i)^-3 =$




$6i - 64 + ( \frac{1}{i})^3 =$


$6i - 64 + \frac{1}{-i} =$


$\frac{1}{-i} * \frac{i}{i}$

$\frac{i}{-i^2}$

$\frac{i}{1}$

$6i - 64 + i = -64 + 7i$