Question

More math question please help!!
Answer all correctly

Answer 1

Answer:

Step-by-step explanation:

Complex numbers polar multiplication form:

$\boxed{z_{1}z_{2}=r_{1}*r_{2}(Cos \ (\theta_{1}+ \theta_{2})+iSin \ (\theta_{1}+ \theta_{2})}$

z₁ = 2 (Cos 60 + i Sin 60)    & z₂ = 8(Cos 150 + i Sin 150)

z₁z₂= 2*8 (Cos [60+150] + i Sin [60+150])

     = 16 (Cos 210 + i Sin 210)     Option C

2) Point A  :   3 - 5i   Option A

The x-axis represent real part and the y-axis represent imaginary part.

4) z = -5√3 -i 5

x = -5√3  & y = -5

$r=\sqrt{x^{2}+y^{2}}\\\\ = \sqrt{(-5\sqrt{3})^{2}+(-5)^{2}}\\\\ = \sqrt{25*3 + 25}= \sqrt{75+25}\\\\ = \sqrt{100}\\\\r = 10$

$\theta = tan^{-1} \ \dfrac{y}{x}\\\\\\=tan^{-1} \ \dfrac{-5}{-5\sqrt{3}}\\\\\\= tan^{-1} \ \dfrac{1}{\sqrt{3}}$

$\sf \theta = 30^ \circ$

Theta lies in third quadrant

Ф = 180 + 30

Ф = 210°

z = 10(Cos 210° + i sin 210°)

Answer 2

Answer:

See below for answers and explanations

Step-by-step explanation:

Problem 1

To multiply two complex numbers in polar form, we use the rule $z_1z_2=r_1r_2[cos(\theta_1+\theta_2)+isin(\theta_1+\theta_2)]$, so:

$z_1z_2=r_1r_2[cos(\theta_1+\theta_2)+isin(\theta_1+\theta_2)]\\z_1z_2=(2)(8)[cos(60^\circ+150^\circ)+isin(60^\circ+150^\circ)]\\z_1z_2=16(cos210^\circ+isin210^\circ)$

This means that C is the correct answer

Problem 2

We treat a complex plane almost exactly like a Cartesian plane where the x-axis is the real axis and the y-axis is the imaginary axis. Hence, point A is located at (3,-5) if this were a Cartesian plane, but since it is a complex plane, it would be read as 3-5i, making A the correct answer

Problem 3

The first part of the problem is really just using the distance formula, treating the real and imaginary parts of the complex numbers as coordinate points:

$\sqrt{(6-2)^2+(7-(-5))^2}\\\sqrt{(4)^2+(7+5)^2}\\\sqrt{16+(12)^2}\\\sqrt{16+144}\\\sqrt{160}\\4\sqrt{10}$

The second part of the problem is simple enough, again, treating the real and imaginary parts of the complex numbers as coordinate points:

$\bigr(\frac{2+6}{2},\frac{-5+7}{2}\bigr)=\bigr(\frac{8}{2},\frac{2}{2}\bigr)=(4,1)=4+i$

Thus, A is the correct answer

Problem 4

Rectangular Form: $z=a+bi$

Polar/Trigonometric Form: $z=r(cos\theta+isin\theta)$

Conversion Rules: $r=\sqrt{a^2+b^2}\\\theta=tan^{-1}(\frac{b}{a})$

Calculations:

$r=\sqrt{(-5\sqrt{3})^2+(-5)^2}\\r=\sqrt{75+25}\\r=\sqrt{100}\\r=10$

$\theta=tan^{-1}(\frac{-5}{-5\sqrt{3}})\\\theta=tan^{-1}(\frac{1}{\sqrt{3}})\\\theta=30^\circ$

Because the complex number is located in Quadrant III, then the reference angle is $30^\circ$ counterclockwise from the negative x-axis, which is equal to $180^\circ+30^\circ=210^\circ$

Thus, the complex number is trigonometric form is $z=10(cos210^\circ+isin210^\circ)$, making C the correct answer

Problem 5

This is just a simple evaluation:

$6\biggr(cos\frac{5\pi}{6}+isin\frac{5\pi}{6}\biggr)\\ \\6\biggr(-\frac{\sqrt{3}}{2}+\frac{1}{2}i\biggr)\\ \\-3\sqrt{3}+3i$

Treating the real and imaginary parts of the complex number as coordinate points, we can see that the best point is Q.