Question

value of iota 9 + iota 10 + iota 11 is equal to​

Answer 1

Answer:

Value of $i^{9} + i^{10} + i^{11} = -1$

Step-by-step explanation:

As we know that

$i = \sqrt{-1}$

When we square of $i$ we got

$i^{2} = -1$

And when we cube it we got

$i^{3} = -\sqrt{-1}$

And when we put power 4 we got

$i^{4} = 1$

Now sum of all numbers,

$i + i^{2} +i^{3} +i^{4} = 0$

And,

$i^{4n} = 1$ always

Now come on question given is,

Sum of $i^{9} + i^{10} + i^{11}$

We can write,

$i^{9} = i^{8} * i^{1}$

$i^{9} = i$

By this,

$i^{10} = i^{2}$

$i^{11} = i^{3}$

Now,

$i^{9} + i^{10} + i^{11} = i + i^{2} +i^{3}$

We know,

$i + i^{2} +i^{3} +i^{4} = 0$

By this we got,

$i^{9} + i^{10} + i^{11} = -i^{4}$

And we know $i^{4} = 1$

So final asnwer,

$i^{9} + i^{10} + i^{11} = -1$