Question

express the given complex no. in the form a + ib :

i^9 + i^19​

Answer 1

hey mate ur ans is in the given attachment

Answer 2

The required form is $i^9 + i^{19}=0+0i$.

Step-by-step explanation:

Given : Expression $i^9 + i^{19}$.

To find : Express the given complex no. in the form a + ib ?

Solution :

The expression is re-written as,

$i^9 + i^{19}=(i^3)^3+i^{4\times 5-1}$

$i^9 + i^{19}=(i^2\cdot i)^3+(i^4)^{5}\cdot (i)^{-1}$

$i^9 + i^{19}=(i^2\cdot i)^3+((i^2)^2)^{5}\cdot (i)^{-1}$

We know that, $i^2=-1$

$i^9 + i^{19}=(-i)^3+(1)^{5}\cdot (i)^{-1}$

$i^9 + i^{19}=-(-i)+\frac{1}{i}$

$i^9 + i^{19}=\frac{i^2+1}{i}$

$i^9 + i^{19}=\frac{-1+1}{i}$

$i^9 + i^{19}=0$

The complex number into a+ib form is $i^9 + i^{19}=0+0i$.

Where, a=0 and b=0

#Learn more  

Express (-51)(1/8i) in x+iy form​

brainly.in/question/10427207