How can I right this

`-2+i2sqrt(3)`

TO

`re^(itheta)` 

Expert Answers

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You need to convert the given algebraic form of complex number `z = -2 + 2sqrt3*i` to the polar form `z = r*e^(i*theta)` , using Euler's formula `e^(i*theta) = cos theta + i*sin theta` .

You need to make the conversion from cartesian coordinates `(x,y)` of complex number `z = x + i*y` into polar coordinates `(r, theta)` ,  where r represents the magnitude of z and `theta` is the angle between the position vector bar z and x axis, such that:

`r = |z| = sqrt(x^2 + y^2)`

`theta = tan^(-1) (y/x)`

Identifying `x = -2, y = 2sqrt3` , yields:

`r = sqrt(4 + 12) => r = sqrt 16 => r = 4`

`theta = tan^(-1)(2sqrt3/(-2)) => theta = tan^(-1)(-sqrt3)`

`theta = -tan^(-1)(sqrt3) =>tan theta = -sqrt3 =>theta = pi - pi/3 => theta = (2pi)/3`

`e^(i*(2pi)/3) = cos ((2pi)/3) + i*sin ((2pi)/3)`

Hence, converting the given algebraic form of the complex number to the polar form `re^(i*theta)` yields `-2 + 2sqrt3*i = 4*e^(i*(2pi)/3).`

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