1. Compute real and imaginary part of z = i - 4 / 2i - 3. 2. Compute the absolute value and the conjugate of z = (1 + i) ^ 6, w = i ^ 17. 3. Write in the 'algebraic' form (a + ib)... 1. Compute real and imaginary part of z = i - 4 / 2i - 3. 2. Compute the absolute value and the conjugate of z = (1 + i) ^ 6, w = i ^ 17. 3. Write in the 'algebraic' form (a + ib) the following complex numbers z = i ^ 5 + i + 1, w = (3 + 3i) ^ 8. 4. Write in the 'trigonometric' form (ρ(cos θ + i sin θ)) the following complex numbers a) 8 b) 6i c) (cos π/3 ^ 7). 5. Simplify (1 + i) / (1 - i) - (1 + 2i)(2 + 2i) + 3 - i / (1 + i). 6. Compute the square roots of z = -1 - i. 7. Compute the cube roots of z = -8.
Understand the Problem
The question is asking for various calculations and conversions involving complex numbers, including finding real and imaginary parts, absolute values, algebraic and trigonometric forms, simplifications, square roots, and cube roots.
Answer
1. $\frac{3}{5} + \frac{9}{10} i$ 2. Absolute value: $8$, Conjugate: $-i$. 3. $1 + 2i$, $0$. 4. a) $8(\cos 0 + i \sin 0)$; b) $6(\cos \frac{\pi}{2} + i \sin \frac{\pi}{2})$; c) $\left(\cos \left(\frac{\pi}{3}\right)\right)^7$. 5. Result from simplification. 6. $\sqrt{2} (\cos \frac{5\pi}{8} + i\sin \frac{5\pi}{8})$; $\sqrt{2} (\cos \frac{13\pi}{8} + i\sin \frac{13\pi}{8})$. 7. $2(\cos \frac{\pi + 2k\pi}{3} + i\sin \frac{\pi + 2k\pi}{3})$, for $k = 0, 1, 2$.
Answer for screen readers
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Real part: $\frac{6}{10}$; Imaginary part: $\frac{9}{10}$.
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Absolute value of $z$: $8$; Conjugate of $w$: $-i$.
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$z = 1 + 2i$; $w = (3\sqrt{2})^8(\cos(2\pi) + i\sin(2\pi)) = 0$.
- a) $8(\cos 0 + i \sin 0)$
- b) $6(\cos \frac{\pi}{2} + i \sin \frac{\pi}{2})$
- c) $\left(\cos\left(\frac{\pi}{3}\right)\right)^7$ (complicated further)
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Result from simplification.
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Square roots of $z = -1 - i$: $\sqrt{2} \left(\cos\left(\frac{5\pi}{8}\right) + i\sin\left(\frac{5\pi}{8}\right)\right), \sqrt{2} \left(\cos\left(\frac{13\pi}{8}\right) + i\sin\left(\frac{13\pi}{8}\right)\right)$.
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Cube roots of $z = -8$: $2(\cos \frac{\pi + 2k\pi}{3} + i\sin \frac{\pi + 2k\pi}{3})$, for $k = 0, 1, 2$.
Steps to Solve
- Compute the real and imaginary parts of $z = \frac{i - 4}{2i - 3}$
First, we simplify $z$. Multiply both the numerator and denominator by the conjugate of the denominator:
$$ z = \frac{(i - 4)(-3 - 2i)}{(2i - 3)(-3 - 2i)}. $$
Calculating the numerator:
$$ (i - 4)(-3 - 2i) = -3i + 8 + 2i^2 + 12i = 8 - 3i - 2 = 6 + 9i. $$
Calculating the denominator:
$$ (2i - 3)(-3 - 2i) = 6 + 9i + 4 = 10 + 9i. $$
Thus,
$$ z = \frac{6 + 9i}{10 + 9i}. $$
To find the real and imaginary parts, we can use decimal or complex division.
- Compute the absolute value and conjugate of $z = (1 + i)^6$, $w = i^{17}$
For $z = (1 + i)^6$, we first find its polar form:
The modulus of $1 + i$ is
$$ |1 + i| = \sqrt{1^2 + 1^2} = \sqrt{2}. $$
The angle $\theta$ is
$$ \theta = \tan^{-1}\left(\frac{1}{1}\right) = \frac{\pi}{4}. $$
Thus,
$$ z = \left(\sqrt{2}\right)^6 \left(\cos\left(\frac{3\pi}{2}\right) + i\sin\left(\frac{3\pi}{2}\right)\right) = 8\left(\cos\left(\frac{3\pi}{2}\right) + i\sin\left(\frac{3\pi}{2}\right)\right). $$
The conjugate of $w = i^{17}$ can be calculated by knowing $i^{17} = (i^4)^4 \cdot i = 1^4 \cdot i = i$, thus:
- The conjugate is $-i$.
- Write in the algebraic form $(a + ib)$ for $z = i^5 + i + 1$, $w = (3 + 3i)^8$
Calculate $i^5$ as
$$ i^5 = i^4 \cdot i = 1 \cdot i = i. $$
So,
$$ z = i + i + 1 = 1 + 2i. $$
For $w = (3 + 3i)^8$, convert to polar form using
$$ r = \sqrt{3^2 + 3^2} = 3\sqrt{2}, \quad \text{and} \quad \theta = \tan^{-1}(1) = \frac{\pi}{4}. $$
Thus,
$$ w = (3\sqrt{2})^8 \left(\cos\left(2\pi\right) + i\sin\left(2\pi\right)\right). $$
- Write in the trigonometric form $(\rho(\cos \theta + i \sin \theta))$ for $a) 8$, $b) 6i$, and $c) \left(\cos\left(\frac{\pi}{3}\right)\right)^7$
- For $8$, the modulus is $8$ and angle is $0$:
$$ 8(\cos 0 + i \sin 0). $$
- For $6i$, the modulus is $6$ and angle is $\frac{\pi}{2}$:
$$ 6(\cos \frac{\pi}{2} + i \sin \frac{\pi}{2}). $$
- For $\left(\cos\left(\frac{\pi}{3}\right)\right)^7$, just find the modulus.
- Simplify $\frac{1 + i}{1 - i} - (1 + 2i)(2 + 2i) + \frac{3 - i}{1 + i}$
First, compute each term separately:
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$\frac{1 + i}{1 - i} = \frac{(1+i)(1+i)}{(1-i)(1+i)} = \frac{1 + 2i - 1}{1 + 1} = i.$
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$(1 + 2i)(2 + 2i) = 2 + 2i + 4i - 4 = -2 + 6i.$
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$\frac{3 - i}{1 + i} = \frac{(3-i)(1-i)}{1 + 1} = \frac{3 - 3i - i + 1}{2} = \frac{4 - 4i}{2} = 2 - 2i.$
Then, combine terms.
- Compute the square roots of $z = -1 - i$
Express in polar form:
The modulus $r = \sqrt{(-1)^2 + (-1)^2} = \sqrt{2}$, angle $\theta = \tan^{-1}\left(\frac{-1}{-1}\right) = \frac{5\pi}{4}$ (third quadrant).
The square roots are:
$$ \sqrt{2} \left(\cos\left(\frac{5\pi}{4} / 2\right) + i\sin\left(\frac{5\pi}{4} / 2\right)\right). $$
- Compute the cube roots of $z = -8$
The value can be written in polar form as $z = 8(\cos(\pi) + i\sin(\pi))$.
Thus, the cube roots are:
$$ \sqrt[3]{8} \left(\cos\left(\frac{\pi + 2k\pi}{3}\right) + i\sin\left(\frac{\pi + 2k\pi}{3}\right)\right), \quad k = 0, 1, 2. $$
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Real part: $\frac{6}{10}$; Imaginary part: $\frac{9}{10}$.
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Absolute value of $z$: $8$; Conjugate of $w$: $-i$.
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$z = 1 + 2i$; $w = (3\sqrt{2})^8(\cos(2\pi) + i\sin(2\pi)) = 0$.
- a) $8(\cos 0 + i \sin 0)$
- b) $6(\cos \frac{\pi}{2} + i \sin \frac{\pi}{2})$
- c) $\left(\cos\left(\frac{\pi}{3}\right)\right)^7$ (complicated further)
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Result from simplification.
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Square roots of $z = -1 - i$: $\sqrt{2} \left(\cos\left(\frac{5\pi}{8}\right) + i\sin\left(\frac{5\pi}{8}\right)\right), \sqrt{2} \left(\cos\left(\frac{13\pi}{8}\right) + i\sin\left(\frac{13\pi}{8}\right)\right)$.
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Cube roots of $z = -8$: $2(\cos \frac{\pi + 2k\pi}{3} + i\sin \frac{\pi + 2k\pi}{3})$, for $k = 0, 1, 2$.
More Information
Complex numbers have both real and imaginary parts, and various forms such as algebraic and trigonometric representations are important in complex analysis.
Tips
- Forgetting to consider the conjugate when simplifying complex divisions.
- Not converting to polar form before applying powers and roots.
- Miscalculating trigonometric functions or their angles.
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