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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