Question

What is the value of the [i^19+(1÷i)^25]^2

Answer 1

hope this helps.....
if correct...
mark as brainlist plzz

Answer 2

we know that
${i}^{2} = - 1 \\ {i}^{4} =1$
so multiple of 4 iota's value is 1.
${i}^{19} = {i}^{4 + 4 + 4 + 4 + 3} = {i}^{3} = - i$
${( \frac{1}{i} })^{25} = { \frac{1}{ {i}^{6 \times 4 + 1} } }$
$= \frac{1}{i}$
$( { - i + \frac{1}{i} })^{2} \\ = {( \frac{ - {i}^{2} + 1}{i} })^{2} = ( { \frac{2}{i} })^{2}$
$= \frac{4}{ {i}^{2} } \\ = - 4 \: ans$
my answers is correct,please mark it brainiest