"i" ( convert the complex number in the polar form)
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From Pythagoras, we have: \displaystyle{r}^{2}={x}^{2}+{y}^{2}r2=x2+y2 and basic trigonometry gives us:
\displaystyle \tan{\ }\theta=\frac{y}{{x}}tan θ=xy\displaystyle{x}={r}\ \cos{\theta}x=r cosθ \displaystyle{y}={r}\ \sin{\theta}y=r sinθ
Multiplying the last expression throughout by \displaystyle{j}jgives us:
\displaystyle{y}{j}={j}{r}\ \sin{\theta}yj=jr sinθ
So we can write the polar form of a complex number as:
\displaystyle{x}+{y}{j}={r}{\left( \cos{\theta}+{j}\ \sin{\theta}\right)}x+yj=r(cosθ+j sinθ)
r is the absolute value (or modulus) of the complex number
θ is the argument of the complex number.
There are two other ways of writing the polar form of a complex number:
\displaystyle{r}\ \text{cis}\ \thetar cis θ [This is just a shorthand for \displaystyle{r}{\left( \cos{\theta}+{j}\ \sin{\theta}\right)}r(cosθ+j sinθ)]
\displaystyle{r}\ \angle\ \thetar ∠ θ [means once again, \displaystyle{r}{\left( \cos{\theta}+{j}\ \sin{\theta}\right)}r(cosθ+j sinθ)]
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\displaystyle \tan{\ }\theta=\frac{y}{{x}}tan θ=xy\displaystyle{x}={r}\ \cos{\theta}x=r cosθ \displaystyle{y}={r}\ \sin{\theta}y=r sinθ
Multiplying the last expression throughout by \displaystyle{j}jgives us:
\displaystyle{y}{j}={j}{r}\ \sin{\theta}yj=jr sinθ
So we can write the polar form of a complex number as:
\displaystyle{x}+{y}{j}={r}{\left( \cos{\theta}+{j}\ \sin{\theta}\right)}x+yj=r(cosθ+j sinθ)
r is the absolute value (or modulus) of the complex number
θ is the argument of the complex number.
There are two other ways of writing the polar form of a complex number:
\displaystyle{r}\ \text{cis}\ \thetar cis θ [This is just a shorthand for \displaystyle{r}{\left( \cos{\theta}+{j}\ \sin{\theta}\right)}r(cosθ+j sinθ)]
\displaystyle{r}\ \angle\ \thetar ∠ θ [means once again, \displaystyle{r}{\left( \cos{\theta}+{j}\ \sin{\theta}\right)}r(cosθ+j sinθ)]
hope you got it
thank you
mark it as brainest answer
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