if z1= 3+3i and z2=7(cos(5pi/9) + i sin (5pi/9)), then z1/z2= blank

Answers

Answer 1

[tex]z1=\stackrel{a}{3}+\stackrel{b}{3}i~~ \begin{cases} r = \sqrt{a^2+b^2}\\ r = \sqrt{18}\\[-0.5em] \hrulefill\\ \theta =\tan^{-1}\left( \frac{b}{a} \right)\\ \theta =\frac{\pi }{4} \end{cases}~\hfill z1=\sqrt{18}\left[\cos\left( \frac{\pi }{4} \right) i\sin\left( \frac{\pi }{4} \right) \right] \\\\[-0.35em] ~\dotfill[/tex]

[tex]\cfrac{z1}{z2}\implies \cfrac{\sqrt{18}\left[\cos\left( \frac{\pi }{4} \right) i\sin\left( \frac{\pi }{4} \right) \right]} {7\left[\cos\left( \frac{5\pi }{9} \right) i\sin\left( \frac{5\pi }{9} \right) \right]} \\\\[-0.35em] ~\dotfill\\\\ \qquad \textit{division of two complex numbers} \\\\ \cfrac{r_1[\cos(\alpha)+i\sin(\alpha)]}{r_2[\cos(\beta)+i\sin(\beta)]}\implies \cfrac{r_1}{r_2}[\cos(\alpha - \beta)+i\sin(\alpha - \beta)] \\\\[-0.35em] ~\dotfill[/tex]

[tex]\cfrac{z1}{z2}\implies \cfrac{\sqrt{18}}{7}\left[\cos\left( \frac{\pi }{4}-\frac{5\pi }{9} \right)+i\sin\left( \frac{\pi }{4}-\frac{5\pi }{9} \right) \right] \\\\\\ \cfrac{\sqrt{18}}{7}\left[\cos\left( \frac{-11\pi }{36} \right) +i\sin\left( \frac{-11\pi }{36} \right) \right]\implies \cfrac{\sqrt{18}}{7}\left[\cos\left( \frac{83\pi }{36} \right) +i\sin\left( \frac{83\pi }{36} \right) \right] \\\\[-0.35em] ~\dotfill\\\\ ~\hfill \cfrac{z1}{z2}\approx 0.348~~ + ~~0.496i~\hfill[/tex]

Answer 2

The value of z1/z2 is √18/ 7 (cos ( 11π/36 ) - isin ( 11π/36 )).

What is complex number?

"A complex number is the sum of a real number and an imaginary number and it is of the form x + iy and is usually represented by z".

For the given situation,

z1= 3+3i and

z2= 7(cos(5π/9) + i sin (5π/9))

To divide the complex numbers, both should be in same form.

Convert z1 in polar form.

z is of the form x+iy, so, r=[tex]\sqrt{x^{2}+y^{2} }[/tex]

⇒[tex]r=\sqrt{3^{2}+3^{2} }[/tex]

⇒[tex]r=\sqrt{18}[/tex]

θ = [tex]tan^{-1}(\frac{b}{a} )[/tex]

⇒[tex]tan^{-1}(\frac{3}{3} )[/tex]

⇒[tex]tan^{-1}(1)[/tex]

⇒[tex]45[/tex]°

The polar form is of the form, z= r (cosθ + i sinθ),

⇒ z1 = [tex]\sqrt{18}[/tex] (cosπ/4 + isinπ/4)

The formula for dividing complex number is

z1/z2 = r1(cos θ1 + isin θ1) / r2(cos θ2 + isin θ2)

⇒ z1/z2 = r(cosθ + isinθ)

where, r = r1/r2 and θ = (θ1 - θ2)

z1/z2 = [tex]\sqrt{18}[/tex] (cos π/4 + isin π/4) / 7 (cos 5π/9 + isin 5π/9)

⇒ r = [tex]\sqrt{18}[/tex] / 7 and

θ = (π/4 - 5π/9 )

⇒ θ = (-11π/36)

cos(-θ) = cos θ

⇒cos( -11π/36 ) = cos ( 11π/36 )

sin(-θ) = -sin θ

⇒ sin ( -11π/36 ) = -sin ( 11π/36 )

Thus, z1/z2 = [tex]\sqrt{18}[/tex] / 7 (cos ( 11π/36 ) - isin ( 11π/36 ))

Hence we can conclude that the value of z1/z2 is

√18/ 7 (cos ( 11π/36 ) - isin ( 11π/36 )).

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A similar problem is given at https://brainly.com/question/12517818

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Answers

Answer:

option B is correct answer. ∆SVZ~=∆YVX, SSS

Step-by-step explanation:

reasons,

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therefore, ∆SVZ~=∆YVX.

hope this helps you.

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Step-by-step explanation:

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Step-by-step explanation:

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