D

#### What is wrong with this calculation?

23 viewed last edited 4 years ago
Anonymous
0

Here are 2 ways of computing: \frac{1}{i} leading to different results. Which one is correct and why?

Method 1: \frac{1}{i}=\frac{i^4}{i}=i^3=i^2.i=-i

Method 2: \frac{1}{i}=\frac{1}{\sqrt{-1}}=\sqrt{\frac{1}{-1}}=\sqrt{\frac{\left(-1\right)^2}{-1}}=\sqrt{-1}=i

I cannot figure out which of the methods gives the correct answer and what's wrong with the incorrect method.

Vivekanand Vellanki
0

Clearly, Method 1 is correct. To verify:

\frac{1}{i\ }=-i\ \Rightarrow\ 1=-i.i\ \Rightarrow\ 1\ =\ -i^2\ \Rightarrow\ 1\ =\ -\left(-1\right)\ =\ 1

Method 2 is incorrect. Lets verify and see:

\frac{1}{i}=i\ \Rightarrow\ 1\ =\ i^2\ \Rightarrow\ 1\ =\ -1

Clearly, Method 2 leads us to an invalid conclusion. Hence, Method 2 is wrong.

The question is where did Method 2 go wrong. Please take a look at this video on how to divide by a complex number: https://www.khanacademy.org/math/precalculus/imaginary-and-complex-numbers/complex-conjugates-and-dividing-complex-numbers/v/dividing-complex-numbers

Following the approach in the video: The complex conjugate of the denominator:

0 + i is 0 - i.

Using that approach:

\frac{1}{i}=\frac{1}{i}\cdot\frac{-i}{-i}=\frac{-i}{-i^2}=\frac{-i}{-\left(-1\right)}=-i

For more practice questions and material: http://www.mesacc.edu/~scotz47781/mat120/notes/complex/dividing/dividing_complex.html

Mahesh Godavarti
0

The identify \sqrt{a} \sqrt{b} = \sqrt{ab} is only valid for real numbers when a, b \geq 0 , since \sqrt{a} is undefined for a \lt 0 when we are only dealing with real numbers.

Therefore, \sqrt{(-1)^2} \neq \sqrt{-1} \sqrt{-1} .