Sachchidanand Prasad

14-03-2023

Problem: Express the following complex numbers in the form of $a+ b\iota $.

  1. $\frac{1}{3+4\iota }$.
  2. $\frac{(1+\iota )(2+\iota )}{1-\iota }$.
  3. $\left( \frac{\sqrt{3} }{2} + \frac{1}{2}\iota \right) ^4$.
  4. $\iota ^n$, where $n\in \mathbb{Z} $.

Solution:

  1. Note that \begin{align*} \frac{1}{3+4\iota } & = \frac{1}{3+4\iota } \times \frac{3-4\iota }{3-4\iota } \\ & = \frac{3-4\iota }{9+16} \\ & = \frac{3}{25} - \frac{4}{25} \iota. \end{align*} Therefore, \[ \textcolor{blue}{\boxed{\frac{1}{3+4\iota } = \frac{3}{25} - \frac{4}{25} \iota}} \]
  2. With the same as above, we have \begin{align*} \frac{(1+\iota )(2+\iota )}{1-\iota } & = \frac{(1+\iota )(2+\iota )}{1-\iota } \times \frac{1+\iota }{1+\iota } \\ & = \frac{(1+\iota )^2 (2+\iota )}{1+1} \\ & = \frac{(1-1+2\iota )(2+\iota )}{2} \\ & = \frac{2\iota (2+\iota )}{2} \\ & = -1 + 2\iota . \end{align*} Therefore, \[ \textcolor{blue}{\boxed{\frac{(1+\iota )(2+\iota )}{1-\iota } = -1 + 2\iota }} \]
  3. We will use the DeMoivre's Theorem.
    If $z=r \left( \cos \theta + \iota \sin \theta \right) $, then for any integer $n$, \[ z^n = r^n \left( \cos (n \theta ) + \iota \sin (n \theta ) \right). \]
    In the given problem we have \begin{align*} \left( \frac{\sqrt{3} }{2} + \frac{1}{2}\iota \right) ^4 & = \left( \cos \frac{\pi}{6} + \iota \sin \frac{\pi}{6} \right)^4 \\ & = \cos \frac{4\pi}{6} + \iota \sin \frac{4\pi}{6} \\ & = \cos \frac{2\pi}{3} + \iota \sin \frac{2\pi}{3} \\ & = -\frac{1}{2} + \frac{\sqrt{3} }{2}\iota . \end{align*} Therefore, \[ \textcolor{blue}{\boxed{\left( \frac{\sqrt{3} }{2} + \frac{1}{2}\iota \right) ^4 = -\frac{1}{2} + \frac{\sqrt{3} }{2}\iota }} \]
  4. Note that \[ i = \cos \frac{\pi}{2} + \iota \sin \frac{\pi}{2}. \] So, again using the DeMoivre's Theorem, we have \begin{align*} i^n & = \left( \cos \frac{\pi}{2} + \iota \sin \frac{\pi}{2} \right) ^n \\ & = \cos \left( \frac{n\pi}{2} \right) + \iota \sin \left( \frac{n\pi}{2} \right) \\ & = \begin{dcases} 1, &\text{ if } n=4k ;\\ \iota , &\text{ if } n=4k+1 ;\\ -1 , &\text{ if } n=4k+2 ;\\ -\iota , &\text{ if } n=4k+3. \end{dcases} \end{align*} Therefore, \[ \textcolor{blue}{\boxed{ \iota ^n = \begin{dcases} 1, &\text{ if } n=4k ;\\ \iota , &\text{ if } n=4k+1 ;\\ -1 , &\text{ if } n=4k+2 ;\\ -\iota , &\text{ if } n=4k+3. \end{dcases} }} \]