Mathematics: The lengths of two line segments are shown. Use the ruler provided to measure the length of a third line segment to the nearest 12 inches. Which statement is true about these e three-line segments?

The lengths of two line segments are shown. Use the

Ответ:

ritiguda

28.10.2021

Mathematics

Part I: The complex number √3-i in the polar form will be 2(cos210^0+isin210^0)

The complex number -2-2i in the polar form will be $2\sqrt{2}(cos45^0+isin45^0)$

Part II: The expression of the product of wz in polar form is $4\sqrt{2-\sqrt{3} } (cos45^0+isin45^0)$

Part III: The expression of z² value in polar form is $z^2 = 8(cos\frac{\pi}{8 }+isin \frac{\pi}{8} )\\$

Part IV: The expression of w^4 in polar form is expressed as 16(cos52.5^0+isin52.5^0)

Complex numbers are the square roots of any negative numbers. They have both real and imaginary axis.

The rectangular form of expressing complex numbers is expressed as:

z = x + iy

The polar representation is expressed as:

r(cosθ - isinθ)

Given the complex numbers

z=√3-i

w=-2-2i

Part I: To express in polar form, we need to first get the modulus and argument of each of the complex numbers.

z= √3-i

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

$|z| = \sqrt{3+1} \\|z|=\sqrt{4}\\|z|=2$

For the argument:

$\theta = tan^{-1}\frac{y}{x}\\ \theta = tan^{-1}\frac{-1}{\sqrt{3} }\\\theta =-30^0\\$

Since tan is negative in the 3rd quadrant

$\theta = 180 + 30\\\theta = 210^0$

The complex number √3-i in the polar form will be 2(cos210^0+isin210^0)

For the complex number:

z= -2-2i

|z| = $\sqrt{(-2 )^2+(-2)^2} \\$

$|z| = \sqrt{4+4} \\|z|=\sqrt{8}\\|z|=2\sqrt{2}$

For the argument:

$\theta = tan^{-1}\frac{y}{x}\\ \theta = tan^{-1}\frac{-2}{\sqrt{-2} }\\\theta=tan^{-1}1\\\theta =45^0\\$

The complex number -2-2i in the polar form will be $2\sqrt{2}(cos45^0+isin45^0)$

Part II: Taking the product of wz

(√3 - 1)(-2-2i)

Expand

wz = -2√3-2√3 i + 2 + 2i

wz = (-2√3+2) +(-2√3+2)i

Get the modulus

$=\sqrt{(-2\sqrt{3}+2)^2+(-2\sqrt{3}+2)^2} \\=\sqrt{12-8\sqrt{3}+4 + 12-8\sqrt{3}+4 } \\=\sqrt{24-16\sqrt{3}+8}\\=\sqrt{32-16\sqrt{3} }\\=\sqrt{16(2-\sqrt{3} )}\\=4\sqrt{2-\sqrt{3} }$

Get the argument:

$\theta = tan^{-1}\frac{-2\sqrt{3}+2}{-2\sqrt{3}+2}\\ \theta = tan^{-1}1\\\theta =45^0$

Expression in polar form is $4\sqrt{2-\sqrt{3} } (cos45^0+isin45^0)$

Part III: To calculate z², we will simply square the polar form of z and apply De Moivre's theorem as shown:

Since z = $2\sqrt{2}(cos45^0+isin45^0)$

$z^2 = [2\sqrt{2}(cos\frac{\pi}{4}+isin\frac{\pi}{4} )]^2\\z^2 = (2\sqrt{2})^2(cos\frac{1}{2} * \frac{\pi}{4 }+isin\frac{1}{2} * \frac{\pi}{4} )]\\z^2 = 8(cos\frac{\pi}{8 }+isin \frac{\pi}{8} )\\$

This shows that the expression of z² value in polar form is $z^2 = 8(cos\frac{\pi}{8 }+isin \frac{\pi}{8} )\\$

Part 4: We will also use De Moivre's theorem to get w⁴

Since w= 2(cos210^0+isin210^0)

$w^4 = [2(cos210+isin210)]^4\\w^4 = 2^4(cos\frac{210}{4} +isin\frac{210}{4} )\\w^4=16(cos52.5^0+isin52.5^0)$

Hence the expression of w^4 in polar form is expressed as 16(cos52.5^0+isin52.5^0)

Learn more about complex numbers here: link

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The lengths of two line segments are shown. Use the ruler provided to measure the length of a third line segment to the nearest 12 inches. Which statement is true about these e three-line segments?

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