Question

express i^-35 in the form of a+ib

Answer 1

i⁻³⁵
=i⁻³⁴⁻¹
=(i²)⁻¹⁷×(i⁻¹)
=(-1)⁻¹⁷(1/i)
=1/(-1)¹⁷(1/i)
=(-1)(1/i)
=-1/i
=0+i(-1/i²)
=0+i(-1/-1)
=0+i.1
where a=0, b=1

Answer 2

Answer:

$i^{-35}=0+i$

Step-by-step explanation:

Given : Expression $i^{-35}$

To find : Express expression in the form of a+ib ?

Solution :

Write expression as,

$i^{-35}=i^{-34-1}$

$i^{-35}=i^{-34}\times i^{-1}$

$i^{-35}=(i^2)^{-17}\times i^{-1}$

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

$i^{-35}=(-1)^{-17}\times i^{-1}$

$i^{-35}=\frac{1}{(-1)^{17}}\times i^{-1}$

$i^{-35}=\frac{1}{-1}\times i^{-1}$

$i^{-35}=-i^{-1}$

$i^{-35}=-\frac{1}{i}$

Multiply and divide by i,

$i^{-35}=-\frac{1}{i}\times \frac{i}{i}$

$i^{-35}=-\frac{i}{i^2}$

$i^{-35}=-\frac{i}{-1}$

$i^{-35}=i$

$i^{-35}=0+i$

i.e. In the form of a+ib $i^{-35}=0+i$

Where, a=0 and b=1.