Computers and Technology, 02.03.2020 16:54 highspeed7100

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)

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The provided complex number $(-1 + i\sqrt{3})$ 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 $\theta$ .

The rectangular shape:

$\to x + iy$

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

$\to x = -1 \\\\\to y = \sqrt{3}\\$

Polar form $\to r(\cos(\theta) + i \sin( \theta)\\\\$

If r is the magnitude of the real or complex and $\theta$ denotes the angle, r can be calculated with

$\to r = \sqrt{(x^2+y^2)}\\\\$

$=\sqrt{((-1)^2+ \sqrt{3^2})}\\\\= \sqrt{(1+3)}\\\\= \sqrt{4}\\\\= 2$

When the $\theta$ is found:

$\to \theta= \tan^{-1} \ (\frac{y}{x})\\\\\to \theta= \tan^{-1} (\frac{\sqrt{3}}{-1})\\\\\to \theta= -60^{\circ}\\\\$

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

$\to \theta = -60^{\circ} + 360^{\circ} = 300^{\circ}$

Converting angle from degrees into radian:

$\to \theta = 300^{\circ} (\frac{\pi }{180^{\circ}})\\\\\to \theta = \frac{5pi}{3} \ \ radians\\$

Therefore, r = 2 and $\theta = \frac{5 \pi }{3}\ \ radians$

Polar form:

$\to 2(\cos( \frac{5\pi}{3}) + i \sin(\frac{5pi}{3}))$

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Answer from: kaylahill14211

Question:

The complex number -1 + i√3 can be rewritten in the form r(cos(θ) + isin(θ) where θ is in the range [0, 2π]. What is r and θ?

r = 2 and θ = 5π/3 radians

Polar form: 2(cos(5π/3) + isin(5π/3))

Explanation:

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))

Answer from: Quest

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Answer from: Quest

Hypothesis: volumes of both figures are equal.the volume of figure 1 is: $v_1=a_{basis}\cdot h=\frac{1}{2} \cdot 5\cdot7\sin \alpha\cdot 8=280\sin \alpha$ .the volume of figure 2 is: $v_2=a\cdotb\cdotc=5\cdot7\cdot8=280$ .since $0\le \alpha\le \pi$ , you can conclude that $0\le\sin\alpha\le1$ and $0\le v_1\le 140$ . so, $v_1\le v_2$ .volumes of figures are equal only when an angle in the basis is right (then figures will be the same), in another cases volume of figure 1 is less than volume of figure 2.the hypothesis was not correct in all cases.

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