Question $\newline$ Watch Video $\newline$ Show Examples $\newline$ Given $z_{1}=-1-2 i$ and $z_{2}=0+3 i$ , calculate the product $z_{1} z_{2}$ and express all three complex numbers in polar form, rounding all angles to the nearest degree in the interval $0^{\circ} \leq \theta<360^{\circ}$ . Express all moduli in simplest radical form. $\newline$ Answer Attempt $1$ out of $2$

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Watch Video

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Show Examples

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Given

z_{1}=-1-2 i

and

z_{2}=0+3 i

, calculate the product

z_{1} z_{2}

and express all three complex numbers in polar form, rounding all angles to the nearest degree in the interval

0^{\circ} \leq \theta<360^{\circ}

. Express all moduli in simplest radical form.

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Answer Attempt

1

out of

2

Calculate Product of Complex Numbers: First, let's calculate the product of the complex numbers $z_{1}$ and $z_{2}$ .
$z_{1} = -1 - 2i$
$z_{2} = 0 + 3i$
The product $z_{1}z_{2}$ is calculated as follows:
$z_{1}z_{2} = (-1 - 2i)(0 + 3i)$
$z_{1}z_{2} = -1(3i) - 2i(3i)$
$z_{1}z_{2} = -3i - 6i^2$
Since $i^2 = -1$ , we can simplify further:
$z_{1}z_{2} = -3i - 6(-1)$
$z_{2}$ $0$
So, $z_{2}$ $1$ .
Express in Polar Form: Now, let's express $z_{1}$ , $z_{2}$ , and $z_{1}z_{2}$ in polar form. The polar form of a complex number $z = a + bi$ is $r(\cos(\theta) + i \sin(\theta))$ , where $r$ is the modulus and $\theta$ is the argument of $z$ .

First, we find the modulus and argument of $z_{1} = -1 - 2i$ .
The modulus $r$ is given by $z_{2}$ $0$ .
For $z_{1}$ , $z_{2}$ $2$ .
The argument $\theta$ is found using the arctan function, $z_{2}$ $4$ .
For $z_{1}$ , $z_{2}$ $6$ .
Since the complex number is in the third quadrant, we add $z_{2}$ $7$ degrees to the angle.
$z_{2}$ $8$ .
Using a calculator, $z_{2}$ $9$ .
Rounded to the nearest degree, $z_{1}z_{2}$ $0$ .
So, $z_{1}$ in polar form is $z_{1}z_{2}$ $2$ .
Find Modulus and Argument: Next, we find the modulus and argument of $z_{2} = 0 + 3i$ . The modulus $r$ is given by $r = \sqrt{a^2 + b^2}$ . For $z_{2}$ , $r = \sqrt{0^2 + 3^2} = \sqrt{9} = 3$ . The argument $\theta$ is found using the arctan function, $\theta = \arctan(\frac{b}{a})$ . For $z_{2}$ , since $a = 0$ , the complex number lies on the positive imaginary axis, so $\theta = 90°$ . So, $z_{2}$ in polar form is $r$ $1$ .
Find Modulus and Argument: Next, we find the modulus and argument of $z_{2} = 0 + 3i$ . The modulus $r$ is given by $r = \sqrt{a^2 + b^2}$ . For $z_{2}$ , $r = \sqrt{0^2 + 3^2} = \sqrt{9} = 3$ . The argument $\theta$ is found using the arctan function, $\theta = \arctan(\frac{b}{a})$ . For $z_{2}$ , since $a = 0$ , the complex number lies on the positive imaginary axis, so $\theta = 90°$ . So, $z_{2}$ in polar form is $r$ $1$ .Finally, we find the modulus and argument of $r$ $2$ . The modulus $r$ is given by $r = \sqrt{a^2 + b^2}$ . For $r$ $5$ , $r$ $6$ . The argument $\theta$ is found using the arctan function, $\theta = \arctan(\frac{b}{a})$ . For $r$ $5$ , $r = \sqrt{a^2 + b^2}$ $0$ . Since the complex number is in the fourth quadrant, we add $r = \sqrt{a^2 + b^2}$ $1$ degrees to the angle. $r = \sqrt{a^2 + b^2}$ $2$ . Using a calculator, $r = \sqrt{a^2 + b^2}$ $3$ . Rounded to the nearest degree, $r = \sqrt{a^2 + b^2}$ $4$ . So, $r$ $5$ in polar form is $r = \sqrt{a^2 + b^2}$ $6$ .