Question

i^9+i^19 pls anyone answer this question

Answer 1

First we have to be familiar with,

$\bullet\ \ i^{4n+1}\ =\ i\\ \\ \\ \bullet\ \ i^{4n+2}\ =\ i^2\ =\ -1\\ \\ \\ \bullet\ \ i^{4n+3}\ =\ i^3=-i\\ \\ \\ \bullet\ \ i^{4n}\ =\ i^4\ =\ 1$

∀n ∈ Z.

From these, we just say,

→  If the exponent of i leaves remainder 1 on division by 4, then the given power of i values none other than i.

→  If the exponent of i leaves remainder 2 on division by 4, then the given power of i values -1.

→  If the exponent of i leaves remainder 3 on division by 4, then the given power of i values -i.

→  If the exponent of i leaves remainder 2 on division by 4, then the given power of i values 1.

Okay, what are the given powers of i? Those are i⁹ and i¹⁹.  

Consider i⁹.

Here the exponent is 9. It leaves remainder 1 on division by 4.

9 = 4 · 2 + 1

So i⁹ = i.

Consider i¹⁹.

Here the exponent is 19. It leaves remainder 3 on division by 4.

19 = 4 · 4 + 3

So  i¹⁹ = i³ = - i.

Now,  i⁹ + i¹⁹  =  i - i = 0.

Hence the answer is just 0.

Answer 2

Answer:

the answer is ( 9+19)i^ = 29i^