The complex number $-1 i\sqrt 3$ can be rewritten in the form $r(\cos\theta i\sin\theta),$ where $r \ge 0$ and $\theta\in[0,2\pi).$ What is $(r,\theta)

The given complex number -1 + i√3 is written in the rectangular form and we are asked to write it in the polar form so that we can get the values of r and θ.

The rectangular form is given by

x + iy

Where x is the real part and y is the imaginary part (i and j both are used denote the imaginary part of the complex number)

Comparing the given complex number to the standard rectangular form

x = -1

y = √3

The polar form is given by

r(cos(θ) + isin(θ)

Where r is the magnitude of the complex number and θ is the angle

r can be found by using

r = √(x²+y²)

r = √((-1)²+(√3²)

r = √(1+3)

r = √4

r = 2

θ can be found by using

θ = tan⁻¹(y/x)

θ = tan⁻¹(√3/-1)

θ = -60°

Since x is negative an y is positive so the argument of complex number lies in the 2nd quadrant so we have to add π or 360°

θ = -60° + 360° = 300°

We can convert the angle from the degrees into radians

θ = 300°(π/180°)

θ = 5π/3 radians

Therefore, r = 2 and θ = 5π/3 radians

Polar form: 2(cos(5π/3) + isin(5π/3))

codiepienagoya codiepienagoya · Answer 2 · 2021-12-21T15:02:11+01:00

The provided complex number [tex](-1 + i\sqrt{3})[/tex] is written in rectangular form, and that we are asked to represent it in polar form so that we can corresponding values of r and [tex]\theta[/tex].

The rectangular shape:

[tex]\to x + iy[/tex]

If x represents the real part and y denotes the exponent (i and j both are used to denote the imaginary part of the complex number) Taking the provided complex number & comparing this to the usual rectangular shape

[tex]\to x = -1 \\\\\to y = \sqrt{3}\\[/tex]

Polar form [tex]\to r(\cos(\theta) + i \sin( \theta)\\\\[/tex]

If r is the magnitude of the real or complex and [tex]\theta[/tex] denotes the angle, r can be calculated with

[tex]\to r = \sqrt{(x^2+y^2)}\\\\[/tex]

[tex]=\sqrt{((-1)^2+ \sqrt{3^2})}\\\\= \sqrt{(1+3)}\\\\= \sqrt{4}\\\\= 2[/tex]

When the [tex]\theta[/tex] is found:

[tex]\to \theta= \tan^{-1} \ (\frac{y}{x})\\\\\to \theta= \tan^{-1} (\frac{\sqrt{3}}{-1})\\\\\to \theta= -60^{\circ}\\\\[/tex]

If x is negative and y is positive, the real or complex arguments is in the second quadrant, so we must add [tex]\pi \ \ or\ \ 360^{\circ}[/tex]

[tex]\to \theta = -60^{\circ} + 360^{\circ} = 300^{\circ}[/tex]

Converting angle from degrees into radian:

[tex]\to \theta = 300^{\circ} (\frac{\pi }{180^{\circ}})\\\\\to \theta = \frac{5pi}{3} \ \ radians\\[/tex]

Therefore, [tex]r = 2[/tex] and [tex]\theta = \frac{5 \pi }{3}\ \ radians[/tex]

Polar form:

[tex]\to 2(\cos( \frac{5\pi}{3}) + i \sin(\frac{5pi}{3}))[/tex]

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