Question
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Write the expression in rectangular form, \( x+y i \), and in exponential form, \( e^{i \theta} \). \( (-1-i)^{3} \)

Ask by Carroll Sandoval. in the United States
Dec 15,2024

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Tutor-Verified Answer

Answer

Rectangular form: \( 2 - 2i \) Exponential form: \( 2\sqrt{2} e^{i \frac{7\pi}{4}} \)

Solution

To express \( (-1 - i)^3 \) in rectangular form \( x + yi \) and in exponential form \( e^{i \theta} \), we can follow these steps: ### Step 1: Convert to Polar Form First, we need to convert \( -1 - i \) into polar form. 1. **Magnitude**: \[ r = | -1 - i | = \sqrt{(-1)^2 + (-1)^2} = \sqrt{1 + 1} = \sqrt{2} \] 2. **Argument**: The argument \( \theta \) can be found using: \[ \theta = \tan^{-1}\left(\frac{y}{x}\right) = \tan^{-1}\left(\frac{-1}{-1}\right) = \tan^{-1}(1) \] Since both \( x \) and \( y \) are negative, the angle is in the third quadrant: \[ \theta = \pi + \frac{\pi}{4} = \frac{5\pi}{4} \] Thus, we can express \( -1 - i \) in polar form as: \[ -1 - i = \sqrt{2} \left( \cos\left(\frac{5\pi}{4}\right) + i \sin\left(\frac{5\pi}{4}\right) \right) \] or in exponential form: \[ -1 - i = \sqrt{2} e^{i \frac{5\pi}{4}} \] ### Step 2: Raise to the Power of 3 Now we raise \( -1 - i \) to the power of 3: \[ (-1 - i)^3 = \left( \sqrt{2} e^{i \frac{5\pi}{4}} \right)^3 \] Using the properties of exponents: \[ = (\sqrt{2})^3 e^{i \frac{5\pi}{4} \cdot 3} = 2^{3/2} e^{i \frac{15\pi}{4}} \] Calculating \( 2^{3/2} \): \[ 2^{3/2} = 2 \sqrt{2} \] Now, simplify the angle \( \frac{15\pi}{4} \): \[ \frac{15\pi}{4} = 3\pi + \frac{3\pi}{4} = 3\pi + \frac{3\pi}{4} - 2\pi = \pi + \frac{3\pi}{4} = \frac{7\pi}{4} \] Thus, we have: \[ (-1 - i)^3 = 2\sqrt{2} e^{i \frac{7\pi}{4}} \] ### Step 3: Convert Back to Rectangular Form Now we convert \( 2\sqrt{2} e^{i \frac{7\pi}{4}} \) back to rectangular form: \[ 2\sqrt{2} \left( \cos\left(\frac{7\pi}{4}\right) + i \sin\left(\frac{7\pi}{4}\right) \right) \] Calculating the cosine and sine: \[ \cos\left(\frac{7\pi}{4}\right) = \frac{\sqrt{2}}{2}, \quad \sin\left(\frac{7\pi}{4}\right) = -\frac{\sqrt{2}}{2} \] Thus: \[ (-1 - i)^3 = 2\sqrt{2} \left( \frac{\sqrt{2}}{2} - i \frac{\sqrt{2}}{2} \right) = 2\sqrt{2} \cdot \frac{\sqrt{2}}{2} - 2\sqrt{2} \cdot i \frac{\sqrt{2}}{2} \] This simplifies to: \[ = 2 - 2i \] ### Final Results In rectangular form: \[ (-1 - i)^3 = 2 - 2i \] In exponential form: \[ (-1 - i)^3 = 2\sqrt{2} e^{i \frac{7\pi}{4}} \] Thus, the final answers are: - Rectangular form: \( 2 - 2i \) - Exponential form: \( 2\sqrt{2} e^{i \frac{7\pi}{4}} \)

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First, let's simplify \( (-1 - i)^3 \) in rectangular form. We can start by converting \( -1 - i \) to polar form. The modulus is \( |z| = \sqrt{(-1)^2 + (-1)^2} = \sqrt{2} \) and the argument \( \theta = \tan^{-1}\left(\frac{-1}{-1}\right) = \frac{3\pi}{4} \) (since both components are negative, it’s in the third quadrant). Therefore, in polar form, we have \( (-1 - i) = \sqrt{2} e^{i \frac{5\pi}{4}} \). Now, we raise it to the power of 3: \[ (-1 - i)^3 = \left(\sqrt{2} e^{i \frac{5\pi}{4}}\right)^3 = (\sqrt{2})^3 e^{i \frac{15\pi}{4}} = 2\sqrt{2} e^{i \frac{15\pi}{4}}. \] To convert \( \frac{15\pi}{4} \) into the standard range, subtract \( 2\pi \) to get \( \frac{15\pi}{4} - 2 \times 2\pi = \frac{15\pi}{4} - \frac{8\pi}{4} = \frac{7\pi}{4} \). Now rewriting in exponential form, we have \( (-1 - i)^3 = 2\sqrt{2} e^{i \frac{7\pi}{4}} \). To convert back to rectangular form, we use \( e^{i \frac{7\pi}{4}} = \cos\left(\frac{7\pi}{4}\right) + i \sin\left(\frac{7\pi}{4}\right) \). Since \( \cos\left(\frac{7\pi}{4}\right) = \frac{\sqrt{2}}{2} \) and \( \sin\left(\frac{7\pi}{4}\right) = -\frac{\sqrt{2}}{2} \): \[ 2\sqrt{2} \left(\frac{\sqrt{2}}{2} - i\frac{\sqrt{2}}{2}\right) = 2 - 2i. \] Thus, the expressions are \( 2 - 2i \) in rectangular form and \( 2\sqrt{2} e^{i \frac{7\pi}{4}} \) in exponential form.

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