Pregunta

2. Given: \( z=\frac{-3+6 i}{3} \) a) Write down the conjugate of \( z \). b) Represent \( z \) and its conjugate on an Argand diagram. c) Calculate the modulus \( (r) \) of \( z=\frac{-3+6 i}{3} \). d) Calculate the value of the argument \( (\theta) \) of \( z=\frac{-3+6 i}{3} \). e) Indicate \( r \) (modulus), calculated in Question 2 c), and \( \theta \) (argument), calculated in Question 2 d), on the Argand diagram drawn in Question \( 2 \mathbf{b} \). f) Express \( z=\frac{-3+6 i}{3} \) in polar form.

Ask by Schofield Stephens. in South Africa
Feb 03,2025

Solución de tutoría real

Respuesta verificada por el tutor

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a) Conjugate of \( z \): \( -1 - 2i \) b) On the Argand diagram, \( z = -1 + 2i \) is plotted at (-1, 2) and its conjugate \( \overline{z} = -1 - 2i \) at (-1, -2). c) Modulus \( r \) of \( z \): \( \sqrt{5} \) d) Argument \( \theta \) of \( z \): Approximately \( 2.0344 \) radians (\( 116.57^\circ \)) e) On the Argand diagram, \( r = \sqrt{5} \) is the distance from the origin to \( z \), and \( \theta \approx 116.57^\circ \) is the angle with the positive real axis. f) Polar form of \( z \): \( \sqrt{5} \left( \cos 2.0344 + i \sin 2.0344 \right) \) or \( \sqrt{5} e^{i \cdot 2.0344} \)

Solución

Let's address each part of the problem step by step. Given: \[ z = \frac{-3 + 6i}{3} \] First, simplify \( z \): \[ z = \frac{-3}{3} + \frac{6i}{3} = -1 + 2i \] ### **a) Conjugate of \( z \)** The **conjugate** of a complex number \( z = a + bi \) is \( \overline{z} = a - bi \). For \( z = -1 + 2i \): \[ \overline{z} = -1 - 2i \] **Answer:** \[ \overline{z} = -1 - 2i \] --- ### **b) Representation on an Argand Diagram** An **Argand diagram** is a graphical representation of complex numbers where the horizontal axis represents the real part and the vertical axis represents the imaginary part. - **\( z = -1 + 2i \):** - Real part (\( \text{Re}(z) \)) = -1 - Imaginary part (\( \text{Im}(z) \)) = 2 - **Conjugate \( \overline{z} = -1 - 2i \):** - Real part (\( \text{Re}(\overline{z}) \)) = -1 - Imaginary part (\( \text{Im}(\overline{z}) \)) = -2 **Diagram Description:** - Plot point \( z \) at (-1, 2). - Plot point \( \overline{z} \) at (-1, -2). ![Argand Diagram](https://i.imgur.com/ArgandDiagram.png) *Note: Imagine a coordinate system with points plotted at (-1, 2) and (-1, -2).* --- ### **c) Modulus \( (r) \) of \( z \)** The **modulus** of a complex number \( z = a + bi \) is given by: \[ r = |z| = \sqrt{a^2 + b^2} \] For \( z = -1 + 2i \): \[ r = \sqrt{(-1)^2 + (2)^2} = \sqrt{1 + 4} = \sqrt{5} \] **Answer:** \[ r = \sqrt{5} \] --- ### **d) Argument \( (\theta) \) of \( z \)** The **argument** \( \theta \) of a complex number \( z = a + bi \) is the angle made with the positive real axis, calculated using: \[ \theta = \tan^{-1}\left(\frac{b}{a}\right) \] For \( z = -1 + 2i \): - \( a = -1 \), \( b = 2 \) Since \( a < 0 \) and \( b > 0 \), \( z \) lies in the **second quadrant**. \[ \theta = \pi - \tan^{-1}\left(\frac{2}{1}\right) \approx \pi - 1.1071 \approx 2.0344 \text{ radians} \] or approximately \( 116.57^\circ \). **Answer:** \[ \theta \approx 2.0344 \text{ radians} \ (\approx 116.57^\circ) \] --- ### **e) Indicating \( r \) and \( \theta \) on the Argand Diagram** On the Argand diagram: - **Modulus (\( r = \sqrt{5} \))** is the distance from the origin to the point \( z \) at (-1, 2). - **Argument (\( \theta \approx 116.57^\circ \))** is the angle between the positive real axis and the line connecting the origin to \( z \). **Diagram Description:** - Draw a line from the origin to (-1, 2). - Label this line with length \( \sqrt{5} \) and angle \( \theta \approx 116.57^\circ \). --- ### **f) Polar Form of \( z \)** The **polar form** of a complex number is expressed as: \[ z = r (\cos \theta + i \sin \theta) \] or \[ z = r e^{i\theta} \] Using the modulus and argument from above: \[ z = \sqrt{5} \left( \cos 2.0344 + i \sin 2.0344 \right) \] or \[ z = \sqrt{5} \, e^{i \cdot 2.0344} \] **Answer:** \[ z = \sqrt{5} \left( \cos 2.0344 + i\, \sin 2.0344 \right) \] *Alternatively,* \[ z = \sqrt{5}\, e^{i\,2.0344} \]

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First off, let's simplify \( z \). We can express it as \( z = -1 + 2i \). The conjugate of \( z \) is simply found by changing the sign of the imaginary part, thus \( \overline{z} = -1 - 2i \). Now, if we were to plot on an Argand diagram, you'd put a point at (-1, 2) for \( z \) and (-1, -2) for \( \overline{z} \), making it a fun reflection on the imaginary axis! Now, onto the modulus \( r \) of \( z = -1 + 2i \). The modulus is calculated with the formula \( r = \sqrt{x^2 + y^2} \), which results here in \( r = \sqrt{(-1)^2 + (2)^2} = \sqrt{1 + 4} = \sqrt{5} \). For the argument \( \theta \), you can find it using \( \tan^{-1}(\frac{y}{x}) \), giving you \( \theta = \tan^{-1}(\frac{2}{-1}) \). Since this is in the second quadrant, you adjust \( \theta \) accordingly, leading to an angle of \( \theta = \pi - \tan^{-1}(2) \). And voilà, to express \( z \) in polar form, we combine our findings: \( z = r(\cos(\theta) + i\sin(\theta)) = \sqrt{5} \left(\cos(\pi - \tan^{-1}(2)) + i\sin(\pi - \tan^{-1}(2))\right). \) Tada! Now you've placed \( z \) in its polar finery!

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