By Dr. Shahid Ali Yousafzai

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Let’s discuss the Euler's formula which is $$ { e }^{ ix }=\cos { x } +i\sin { x } $$ It can also be written as $$ { e }^{ i\theta }=\cos { \theta } +i\sin { \theta } $$ This \(i\) is defined as $$ { i }^{ 2 }=-1 $$ It is an imaginary and unique number and it is unique because $$ 1\times i=-1 $$ No such other identical number exists such that we multiply it and we get -1. We know this series $$ { e }^{ x }=1+\frac { x }{ 1! } +\frac { { x }^{ 2 } }{ 2! } +\frac { { x }^{ 3 } }{ 3! } +...+\frac { { x }^{ n } }{ n! } $$ which is expanded by Taylor's series. Now if I replace this \(x\) with \(ix\), we get $$ { e }^{ ix }=1+\frac { ix }{ 1! } +\frac { { ix }^{ 2 } }{ 2! } +\frac { { ix }^{ 3 } }{ 3! } +\frac { { ix }^{ 4 } }{ 4! } +... $$ $$ { e }^{ ix }=1+ix-\frac { { x }^{ 2 } }{ 2! } -\frac { { ix }^{ 3 } }{ 3! } +\frac { { x }^{ 4 } }{ 4! } +... $$ If we split it now $$ { e }^{ ix }=1-\frac { { x }^{ 2 } }{ 2! } +\frac { { x }^{ 4 } }{ 4! } +...+ix-i\left( \frac { { x }^{ 3 } }{ 3! } +... \right) $$ This gives us the expansion of \(\cos { x } \) and \(\sin { x } \). Thus $$ { e }^{ ix }=\cos { x } +i\sin { x } $$ Or $$ { e }^{ i\theta }=\cos { \theta } +i\sin { \theta } $$ This is called Euler's formula. Now if I consider a circle in cartesian coordinates, then $$ \frac { x }{ r } =\cos { \theta } $$ $$ \frac { y }{ r } =\sin { \theta } $$ $$ x=r\cos { \theta } $$ $$ y=r\sin { \theta } $$ Squaring them $$ { x }^{ 2 }+{ y }^{ 2 }={ r }^{ 2 }\cos ^{ 2 }{ \theta } +{ r }^{ 2 }\sin ^{ 2 }{ \theta } $$ Taking \(r^2\) common $$ { r }^{ 2 }\left( \cos ^{ 2 }{ \theta } +\sin ^{ 2 }{ \theta } \right) ={ x }^{ 2 }+{ y }^{ 2 } $$ $$ r=\sqrt { { x }^{ 2 }+{ y }^{ 2 } } $$ Or, it can also be done as $$ \frac { y }{ x } =\frac { r\sin { \theta } }{ r\cos { \theta } } =\tan { \theta } $$ $$ \theta =\tan ^{ -1 }{ \frac { y }{ x } } $$ Cartesian coordinates \((x,y)\) and polar coordinates \((r, \theta )\) are interconvertible.

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