Question

For any complex number \(z=x+i y,\) the real and imaginary parts are defined as the real numbers \(\operatorname{Re}(z)=x\) and \(\operatorname{Im}(z)=y .\) The modulus or absolute value is \(|z|=\sqrt{x^{2}+y^{2}}\) and the phase or angle is the value of \(\theta\) when \(z\) is expressed as \(z=r e^{i \theta} .\) The complex conjugate is \(z^{*}=x-i y\) (This last is the notation used by most physicists; most mathematicians use \(\bar{z}\).) For each of the following complex numbers, find the real and imaginary parts, the modulus and phase, and the complex conjugate, and sketch \(z\) and \(z^{*}\) in the complex plane: (a) \(z=1+i \quad\) (b) \(z=1-i \sqrt{3}\) (d) \(z=5 e^{i \omega t}\) (c) \(z=\sqrt{2} e^{-i \pi / 4}\) In part (d), \(\omega\) is a constant and \(t\) is the time.

Short answer

Real and imaginary parts computed, moduli derived, phases found, and complex conjugates determined for each case.

Step-by-step solution

Real and Imaginary Parts for (a)

Given the complex number \( z = 1 + i \), the real part is \( \operatorname{Re}(z) = 1 \) and the imaginary part is \( \operatorname{Im}(z) = 1 \).

Modulus and Phase for (a)

The modulus is calculated as follows: \[|z| = \sqrt{(1)^2 + (1)^2} = \sqrt{2}\] For the phase \( \theta \), since the real part is positive and imaginary part is positive, \( \theta = \frac{\pi}{4} \).

Complex Conjugate for (a)

The complex conjugate of \( z = 1 + i \) is \( z^* = 1 - i \).

Real and Imaginary Parts for (b)

Given \( z = 1 - i \sqrt{3} \), the real part is \( \operatorname{Re}(z) = 1 \) and the imaginary part is \( \operatorname{Im}(z) = -\sqrt{3} \).

Modulus and Phase for (b)

The modulus is: \[|z| = \sqrt{(1)^2 + (-\sqrt{3})^2} = \sqrt{4} = 2\] The phase \( \theta \) can be calculated as \( \theta = \tan^{-1}\left(\frac{-\sqrt{3}}{1}\right) = -\frac{\pi}{3} \).

Complex Conjugate for (b)

The complex conjugate of \( z = 1 - i \sqrt{3} \) is \( z^* = 1 + i \sqrt{3} \).

Real and Imaginary Parts for (c)

Given \( z = \sqrt{2} e^{-i \pi / 4} \), the exponential form reveals the phase directly. The real part can be found using Euler's formula: \( z = \sqrt{2}(\cos(-\pi/4) + i \sin(-\pi/4)) = 1 - i \).Thus, \( \operatorname{Re}(z) = 1 \) and \( \operatorname{Im}(z) = -1 \).

Modulus and Phase for (c)

The modulus \(|z|\) is given by the coefficient of the exponential, \(\sqrt{2}\), and the phase \(\theta\) is \(-\frac{\pi}{4}\).

Complex Conjugate for (c)

The complex conjugate of \( z = \sqrt{2} e^{-i \pi / 4} \) is \( z^* = \sqrt{2} e^{i \pi / 4}\).

Real and Imaginary Parts for (d)

Given \( z = 5 e^{i \omega t} \), it is in polar form. Using Euler's formula, we write it as \( z = 5 (\cos(\omega t) + i \sin(\omega t)) \). Thus, the real part is \( 5 \cos(\omega t) \) and the imaginary part is \( 5 \sin(\omega t) \).

Modulus and Phase for (d)

The modulus is \(|z| = 5\), and the phase is \(\theta = \omega t \).

Complex Conjugate for (d)

The complex conjugate is \( z^* = 5 e^{-i \omega t} \).

Key concepts

Modulus of a Complex Number

The modulus of a complex number provides a measure of its size or magnitude. For a complex number given by \( z = x + iy \), the modulus is calculated using the formula: \[ |z| = \sqrt{x^2 + y^2} \] This formula is derived from applying the Pythagorean theorem to the complex plane. Imagine plotting \( z \) on the plane with the real part \( x \) on the horizontal axis and the imaginary part \( y \) on the vertical axis.

• The modulus \(|z|\) represents the distance from the origin \((0,0)\) to the point \((x, y)\).
• This distance directly corresponds to the hypotenuse of a right triangle where \(x\) and \(y\) are the perpendicular sides.

For example, consider \( z = 1 + i \). Here, \( |z| = \sqrt{1^2 + 1^2} = \sqrt{2} \). The modulus helps in understanding how far the number is from the origin, irrespective of its direction.

Complex Conjugate

The complex conjugate of a complex number is obtained simply by changing the sign of the imaginary part. If you start with a complex number \( z = x + iy \), its complex conjugate is represented as \( z^* = x - iy \).

• In many contexts, the conjugate is denoted as \( \bar{z} \), especially in mathematics, but physicists often use \( z^* \).
• The real part of the number remains unchanged, but the imaginary part takes the opposite sign.

The complex conjugate is crucial in various mathematical calculations because when a complex number is multiplied by its conjugate, it results in a real number: \[ z \cdot z^* = (x + iy)(x - iy) = x^2 + y^2 \] This product is particularly useful for simplifying division of complex numbers and for deriving their modulus.

Polar Form of Complex Numbers

Polar form is a brilliant way to represent complex numbers using a radius and an angle rather than using Cartesian coordinates. Any complex number \( z = x + iy \) can be expressed in polar form as \( z = r e^{i\theta} \). Here's the breakdown:

• \( r \), the modulus, represents the distance from the origin to the point \((x, y)\).
• \( \theta \), the phase or argument, is the angle made with the positive real axis.

The transformation from Cartesian to polar involves: \[ r = \sqrt{x^2 + y^2} \] \[ \theta = \tan^{-1}(y/x) \] Using Euler’s formula, \( e^{i\theta} = \cos(\theta) + i\sin(\theta) \), you can easily convert to and from this form. This form is particularly useful for multiplication and division of complex numbers, as their moduli multiply and their arguments add. For instance, \( z = \sqrt{2} e^{-i \pi/4} \) is already in polar form. Here, \( \sqrt{2} \) is the modulus and \(-\pi/4\) is the angle.

Complex Plane Sketching

Complex plane sketching provides a visual representation of complex numbers.

• On this plane, the x-axis represents the real part, and the y-axis represents the imaginary part.
• Each complex number corresponds to a point \((x, y)\) on this plane.

This could also be seen as a 2D plane where complex numbers live. When sketching, you plot a number like \( z = 1 + i \) by moving one unit along the real axis and one unit up the imaginary axis. The complex conjugate \( z^* = 1 - i \) would be one unit down from the real point on the plane.

• Such sketching helps quickly visualize operations, like addition and subtraction, by considering the vector movement in the plane.
• By drawing, you can also understand how the modulus denotes distance and the argument shows direction.

Sketching provides a useful tool both for practical calculations and for theoretical understanding.