Question

Express the given complex number in the form \(R(\cos \theta+\) \(i \sin \theta)=R e^{i \theta}\) $$ 1+i $$

Short answer

Question: Express the given complex number \(1+i\) in polar form. Answer: The polar form of the complex number \(1+i\) is \(\sqrt{2}e^{i\frac{\pi}{4}}\).

Step-by-step solution

Find Modulus (R) of the complex number

The modulus of a complex number of the form \(a+bi\) is given by \(R = \sqrt{a^2+b^2}\). In our case, \(a = 1\) and \(b = 1\). So, $$ R = \sqrt{1^2 + 1^2} = \sqrt{1 + 1} = \sqrt{2}. $$

Find argument (angle \(\theta\)) of the complex number

To find the argument of a complex number, we use the formula \(\theta = \text{atan2}(b,a)\), where \(\text{atan2}(y,x)\) is the two-argument arctangent function that returns the angle in radians between the positive x-axis and the point \((x,y)\). For our complex number, \(a = 1\) and \(b = 1\), so $$ \theta = \text{atan2}(1,1) = \frac{\pi}{4}. $$

Write the complex number in polar form

Now that we have found the modulus (R) and the argument (angle \(\theta\)), we can write the complex number in the polar form: $$ 1+i = \sqrt{2} \left( \cos \frac{\pi}{4} + i \sin \frac{\pi}{4} \right) = \sqrt{2}e^{i \frac{\pi}{4}}. $$

Key concepts

Modulus of a Complex Number

Understanding the modulus of a complex number is essential when it comes to grasping the basics of complex analysis. It can be thought of as the distance of the complex number from the origin in the complex plane. To calculate the modulus of a complex number given by the expression \(a + bi\), where \(a\) and \(b\) are real numbers and \(i\) is the imaginary unit, we use the formula \[ R = \sqrt{a^2 + b^2} \]. This formula is actually derived from the Pythagorean theorem, as the complex number in question can be represented as a point \( (a, b) \) in the plane.

For instance, with the complex number \(1 + i\), the modulus is computed by squaring both the real part (1) and imaginary part (1), adding these together, and then taking the square root. Thus, \[ R = \sqrt{1^2 + 1^2} = \sqrt{2}\]. The result, \(\sqrt{2}\), represents the length of the vector pointing from the origin to the point \(1, 1\) on the complex plane.

Argument of a Complex Number

The argument of a complex number is the measure of the angle it makes with the positive real-axis, counter-clockwise, often denoted as \(\theta\) or arg. For a complex number \(a + bi\), the argument can be found by using trigonometry, and it is typically determined using the arctangent function. The preferred function in computing this is \(\text{atan2}(b, a)\), which is the two-argument version of arctangent that takes into account the signs of both \(a\) and \(b\) to provide the correct quadrant for the angle.

Applying this to our example \(1 + i\), we have \(a = b = 1\). Thus, \[ \theta = \text{atan2}(1, 1) = \frac{\pi}{4} \]. This tells us that the line from the origin to the point \(1, 1\) in the complex plane forms a \(45^\circ\) angle with the positive real-axis. Whenever you are calculating the argument, it's important to figure out which quadrant the complex number is in to determine the correct direction of the angle.

Polar and Exponential Form of Complex Numbers

Converting a complex number to its polar form not only simplifies multiplication and division but also provides deeper insight into the properties of the number. The polar form is expressed as \(R(\cos \theta + i \sin \theta)\), where \(R\) is the modulus and \(\theta\) is the argument of the complex number. This form can also be expressed using Euler's formula as \(Re^{i\theta}\), where \(e\) is the base of the natural logarithm. Euler's formula connects the seemingly unrelated exponential functions to trigonometric functions.

For the given example \(1 + i\), after calculating the modulus and the argument, we can represent it in its polar form as \[ \sqrt{2}(\cos \frac{\pi}{4} + i \sin \frac{\pi}{4}) \] or, equivalently, in the exponential form as \[ \sqrt{2}e^{i \frac{\pi}{4}} \]. The exponential form is particularly useful when dealing with complex number powers and roots, offering a straightforward approach to these calculations by manipulating the modulus and argument.