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

Ask by Martin Dickson. in the United States
Dec 15,2024

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Answer

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

Solution

To express \((-1 - i)^3\) in both rectangular form \(x + yi\) and exponential form \(re^{i\theta}\), we will follow these steps: ### Step 1: Convert to Polar Form First, we need to convert \(-1 - i\) into polar form. 1. **Magnitude (r)**: \[ r = | -1 - i | = \sqrt{(-1)^2 + (-1)^2} = \sqrt{1 + 1} = \sqrt{2} \] 2. **Argument (\(\theta\))**: 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 \(-1 - i\) is in the third quadrant, we add \(\pi\): \[ \theta = \pi + \frac{\pi}{4} = \frac{5\pi}{4} \] Thus, we can express \(-1 - i\) in polar form as: \[ -1 - i = \sqrt{2} e^{i \frac{5\pi}{4}} \] ### Step 2: Raise to the Power of 3 Now we raise this expression 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 argument: \[ \frac{15\pi}{4} = 3\pi + \frac{3\pi}{4} = 3\pi + \frac{3\pi}{4} - 2\pi = \pi + \frac{3\pi}{4} = \frac{7\pi}{4} \] 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: \[ e^{i \frac{7\pi}{4}} = \cos\left(\frac{7\pi}{4}\right) + i \sin\left(\frac{7\pi}{4}\right) \] 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: \[ e^{i \frac{7\pi}{4}} = \frac{\sqrt{2}}{2} - i \frac{\sqrt{2}}{2} \] Now multiply by \(2\sqrt{2}\): \[ 2\sqrt{2} \left(\frac{\sqrt{2}}{2} - i \frac{\sqrt{2}}{2}\right) = 2 - 2i \] ### 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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To convert the expression \( (-1-i)^{3} \) to rectangular form, we first find the modulus and argument of the complex number \( -1 - i \). The modulus \( r \) is given by \( r = \sqrt{(-1)^2 + (-1)^2} = \sqrt{2} \). The argument \( \theta \) can be calculated using \( \tan \theta = \frac{-1}{-1} = 1 \), giving \( \theta = \frac{3\pi}{4} \) since the point lies in the third quadrant. Using the polar to exponential conversion, we have: \[ -1 - i = \sqrt{2} e^{i \frac{5\pi}{4}} \] Now we raise this to the power of 3: \[ (-1 - i)^{3} = \left(\sqrt{2} e^{i \frac{5\pi}{4}}\right)^{3} = (\sqrt{2})^{3} e^{i \frac{15\pi}{4}} = 2\sqrt{2} e^{i \frac{15\pi}{4}} \] Now, \( \frac{15\pi}{4} \) can be simplified since it is equivalent to \( \frac{15\pi}{4} - 2\pi = \frac{7\pi}{4} \). Now converting back to rectangular form: \[ (-1 - i)^{3} = 2\sqrt{2} \left(\cos \frac{7\pi}{4} + i \sin \frac{7\pi}{4}\right) = 2\sqrt{2} \left(\frac{\sqrt{2}}{2} - i \frac{\sqrt{2}}{2}\right) = 2 - 2i \] So, in rectangular form, \( (-1 - i)^{3} = 2 - 2i \) and in exponential form, it can be expressed as \( 2\sqrt{2} e^{i \frac{7\pi}{4}} \).

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