Question

Express i9+i19 in the form of a+ib​

Answer 1

Step-by-step explanation:

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

9

+i

19

=0+0i .

Step-by-step explanation:

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

3

)

3

+i

4×5−1

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

9

+i

19

=(i

2

⋅i)

3

+(i

4

)

5

⋅(i)

−1

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

9

+i

19

=(i

2

⋅i)

3

+((i

2

)

2

)

5

⋅(i)

−1

We know that, i^2=-1i

2

=−1

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

9

+i

19

=(−i)

3

+(1)

5

⋅(i)

−1

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

9

+i

19

=−(−i)+

i

1

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

9

+i

19

=

i

i

2

+1

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

9

+i

19

=

i

−1+1

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

9

+i

19

=0

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

9

+i

19

=0+0i .

Where, a=0 and b=0