Complex Numbers
6.0 Euler's formulae
6.0 Euler's formulae
As we derived earlier in polar form, $$z = r\left( {\cos \theta + isin\theta } \right)$$ it is also represented in polar form as $$z = r{e^{i\theta }}$$ where, $\left| z \right| = r{\text{ and }}\arg z = \theta $. Its conjugate is represented as $$\overline z = r{e^{ - i\theta }}$$
It can be derived using the expansion of ${e^x}$ i.e., $${e^x} = 1 + \frac{x}{{1!}} + \frac{{{x^2}}}{{2!}} + \frac{{{x^3}}}{{3!}} + ...$$
Put $x = i\theta $ in above equation, we get $$\begin{equation} \begin{aligned} {e^{i\theta }} = 1 + \frac{{i\theta }}{{1!}} + \frac{{{{(i\theta )}^2}}}{{2!}} + \frac{{{{(i\theta )}^3}}}{{3!}} + ... \\ {\text{ = }}\left( {1 - \frac{{{\theta ^2}}}{{2!}} + \frac{{{\theta ^4}}}{{4!}} - ...} \right) + i\left( {\theta - \frac{{{\theta ^3}}}{{3!}} + \frac{{{\theta ^5}}}{{5!}} - ...} \right) \\ {\text{ = }}\cos \theta + i\sin \theta \\\end{aligned} \end{equation} $$
Note: If $\theta $ is real, then $$\begin{equation} \begin{aligned} \cos \theta = \frac{{{e^{i\theta }} + {e^{ - i\theta }}}}{2} \\ \sin \theta = \frac{{{e^{i\theta }} - {e^{ - i\theta }}}}{{2i}} \\\end{aligned} \end{equation} $$
Question 7. Find the modulus and argument of $$\frac{{i - 1}}{{\cos \frac{\pi }{3} + i\sin \frac{\pi }{3}}}$$
Solution: $$\frac{{i - 1}}{{\cos \frac{\pi }{3} + i\sin \frac{\pi }{3}}} = \frac{{i - 1}}{{\frac{1}{2} + i\frac{{\sqrt 3 }}{2}}} = \frac{{2i - 2}}{{1 + i\sqrt 3 }}$$ Now, multiply by $1 - i\sqrt 3 $ in numerator and denominator, we get $$\begin{equation} \begin{aligned} \Rightarrow \frac{{2i - 2}}{{1 + i\sqrt 3 }} \times \frac{{1 - i\sqrt 3 }}{{1 - i\sqrt 3 }} \\ \Rightarrow \frac{{2i - {i^2}2\sqrt 3 - 2 + i2\sqrt 3 }}{4} \\ \Rightarrow \frac{{2(\sqrt 3 - 1)}}{4} + i\frac{{2(1 + \sqrt 3 )}}{4} \\ \Rightarrow \frac{{\sqrt 3 - 1}}{2} + i\frac{{1 + \sqrt 3 }}{2} \\\end{aligned} \end{equation} $$
Now, modulus $$r = \sqrt {{{\left( {\frac{{\sqrt 3 - 1}}{2}} \right)}^2} + {{\left( {\frac{{1 + \sqrt 3 }}{2}} \right)}^2}} = \sqrt {\frac{{4 - 2\sqrt 3 + 4 + 2\sqrt 3 }}{4}} = \sqrt 2 $$
and
$$\begin{equation} \begin{aligned} \tan \theta = \frac{{\sqrt 3 + 1}}{{\sqrt 3 - 1}} = \frac{{1 + \frac{1}{{\sqrt 3 }}}}{{1 - \frac{1}{{\sqrt 3 }}}} = \frac{{\tan {{45}^ \circ } + \tan {{30}^ \circ }}}{{1 - \tan {{45}^ \circ }\tan {{30}^ \circ }}} \\ \tan \theta = \tan ({45^ \circ } + {30^ \circ }) = \tan {75^ \circ } \\ \Rightarrow {\text{Arg}}(z) = \theta = {75^ \circ } \\\end{aligned} \end{equation} $$
Therefore, in polar form it is written as $$\sqrt 2 (\cos {75^ \circ } + i\sin {75^ \circ })$$
