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Consider the complex number

z=-(7sqrt2)/(2)-(7sqrt2)/(2)i". "
Which of the following complex numbers best approximates
z^(3) ? Hint:
z has a modulus of 7 and an argument of
225^(@).
Choose 1 answer:
(A)
-14.8-14.8 i
(B)
242.5-242.5 i
(c) -21
(D)
343 i

Consider the complex number $\newline$ $z=-\frac{7 \sqrt{2}}{2}-\frac{7 \sqrt{2}}{2} i \text { }$ $\newline$ Which of the following complex numbers best approximates $z^{3}$ ? Hint: $z$ has a modulus of $7$ and an argument of $225^{\circ}$ . $\newline$ Choose $1$ answer: $\newline$ (A) $-14.8-14.8 i$ $\newline$ (B) $242.5-242.5 i$ $\newline$ (C) $-21$ $\newline$ (D) $343 i$

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Fundamental Theorem of Algebra

Full solution

Q. Consider the complex number

\newline

z=-\frac{7 \sqrt{2}}{2}-\frac{7 \sqrt{2}}{2} i \text { }

\newline

Which of the following complex numbers best approximates

z^{3}

? Hint:

z

has a modulus of

7

and an argument of

225^{\circ}

.

\newline

Choose

1

answer:

\newline

(A)

-14.8-14.8 i

\newline

(B)

242.5-242.5 i

\newline

(C)

-21

\newline

(D)

343 i

Given complex number: We are given the complex number $z = -\frac{7\sqrt{2}}{2} - \frac{7\sqrt{2}}{2}i$ . To find $z^3$ , we can use the polar form of the complex number, which is given by $z = r(\cos(\theta) + i \sin(\theta))$ , where $r$ is the modulus and $\theta$ is the argument of $z$ .
Modulus and argument: The modulus of $z$ is given as $7$ . We can verify this by calculating the modulus: $|z| = \sqrt{(-7\sqrt{2}/2)^2 + (-7\sqrt{2}/2)^2} = \sqrt{2*(7^2/4)} = \sqrt{49/2} = 7$ .
Polar form: The argument of $z$ is given as $225$ degrees. This corresponds to the point in the complex plane where the angle with the positive $x$ -axis (real axis) is $225$ degrees, which is in the third quadrant where both the real and imaginary parts are negative.
De Moivre's theorem: Now we can write $z$ in polar form: $z = 7(\cos(225°) + i \sin(225°))$ . To find $z^3$ , we will use De Moivre's theorem, which states that $(r(\cos(\theta) + i \sin(\theta)))^n = r^n(\cos(n\theta) + i \sin(n\theta))$ .
Applying De Moivre's theorem: Applying De Moivre's theorem to find $z^3$ , we get $z^3 = 7^3(\cos(3\cdot225°) + i \sin(3\cdot225°)) = 343(\cos(675°) + i \sin(675°))$ .
Simplifying angles: Since the sine and cosine functions are periodic with a period of $360°$ , we can simplify the angles by subtracting multiples of $360°$ . So, $\cos(675°) = \cos(675° - 2\times360°) = \cos(-45°)$ and $\sin(675°) = \sin(675° - 2\times360°) = \sin(-45°)$ .
Calculating $z^3$ : The cosine and sine of $-45$ degrees are both $-\sqrt{2}/2$ . Therefore, $z^3 = 343(-\sqrt{2}/2 + i(-\sqrt{2}/2))$ .
Multiplying by $-\sqrt{2}/2$ : Multiplying $343$ by $-\sqrt{2}/2$ , we get $z^3 = 343 \cdot -\sqrt{2}/2 + i(343 \cdot -\sqrt{2}/2) = -343/\sqrt{2} + i(-343/\sqrt{2})$ .
Rationalizing the denominator: To simplify further, we can multiply and divide by $\sqrt{2}$ to rationalize the denominator: $z^3 = \frac{-343}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} + i\left(\frac{-343}{\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}}\right) = \frac{-343\sqrt{2}}{2} + i\left(\frac{-343\sqrt{2}}{2}\right)$ .
Final numerical value: Now we calculate the numerical value: $-343\sqrt{2}/2 \approx -242.5$ . So, $z^3 \approx -242.5 - 242.5i$ .

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