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Magazine Chapter 3: Problem 9
Cartesian to Polar Conversion Convert the following expressions into polar form. Plot their location in the complex plane \(^{36}\). a) \((1+\sqrt{-3})^{2}\) b) \(3+j^{4}\) c) \(\frac{2-j \frac{6}{\sqrt{3}}}{2+j \frac{6}{\sqrt{3}}}\) d) \(\left(4-j^{3}\right)\left(1+j \frac{1}{2}\right)\) e) \(3 e^{j \pi}+4 e^{j \frac{\pi}{2}}\) f) \((\sqrt{3}+j) 2 \times \sqrt{2} e^{-\left(j \frac{\pi}{4}\right)}\) g) \(\frac{3}{1+j 3 \pi}\)
Short Answer
Expert verified
Expressions in polar form are: \(4 \text{cis}(-\frac{\pi}{3})\), \(4 \text{cis} 0\), \(\frac{2}{3} \text{cis} \pi\), \(\frac{\sqrt{85}}{2} \text{cis}\tan^{-1}(\frac{6}{7})\), \(5 \text{cis}(2.21)\), \(4 \text{cis}(-\frac{\pi}{12})\), \text{and computed for g}.
Step by step solution
01
- Simplify Expression a
First, simplify the expression \((1 + \sqrt{-3})^2\). We know that \(\sqrt{-3} = j\sqrt{3}\), therefore we simplify the expression as \((1 + j\sqrt{3})^2 = (1+j\sqrt{3})(1+j\sqrt{3})\). Using the distributive property, \(1 + 2j\sqrt{3} + (j\sqrt{3})^2 = 1 + 2j\sqrt{3} - 3 = -2 + 2j\sqrt{3}\).
02
- Convert Expression a to Polar Form
With the simplified form \(-2 + 2j\sqrt{3}\), find its modulus and argument. The modulus \(r\) is \(\sqrt{(-2)^2 + (2\sqrt{3})^2} = \sqrt{4 + 12} = 4\). The argument \(\theta\) is \(\tan^{-1}\left(\frac{2\sqrt{3}}{-2}\right)\), which translates to \(-60^\circ\) or \(-\frac{\pi}{3}\). The polar form is \(4 \text{cis} (-\frac{\pi}{3})\).
03
- Simplify Expression b
For \(3 + j^4\), note that \(j^4 = 1\). This means \(3 + j^4 = 3 + 1 = 4\).
04
- Convert Expression b to Polar Form
The number \(4\) is a real number, lying on the positive real axis of the complex plane. Hence, its polar form is \(4 \text{cis} 0\).
05
- Simplify Expression c
The expression is \(\frac{2-j \frac{6}{\sqrt{3}}}{2+j \frac{6}{\sqrt{3}}}\). First, calculate the denominator's modulus and conjugate: \(r = \sqrt{2^2 + (\frac{6}{\sqrt{3}})^2}= 4\). Multiply numerator and denominator by the conjugate of the denominator: \( (2-j \frac{6}{\sqrt{3}}) (2-j \frac{6}{\sqrt{3}}) = -\frac{2}{3}\), so the expression reduces to \(-\frac{2}{3}\).
06
- Convert Expression c to Polar Form
The number \(-\frac{2}{3}\) resides on the negative real axis. Its modulus is \(\frac{2}{3}\) and its argument is \(\pi\). Therefore, the polar form is \(\frac{2}{3} \text{cis} \pi\).
07
- Simplify Expression d
We have \((4-j^3)(1+j\frac{1}{2})\). Since \(j^3 = -j\), substitute to obtain \((4+j)(1+\frac{j}{2})\). Use distribution: \(4(1+\frac{j}{2})+j(1+\frac{j}{2}) = 4 + 2j + j - \frac{1}{2} = \frac{7}{2} + 3j\).
08
- Convert Expression d to Polar Form
For \(\frac{7}{2} + 3j\), the modulus \(r\) is \(\sqrt{\left(\frac{7}{2}\right)^2 + 3^2} = \sqrt{\frac{49}{4} + 9} = \sqrt{\frac{85}{4}} = \frac{\sqrt{85}}{2}\). The argument \(\theta\) is \(\tan^{-1}\left(\frac{3}{\frac{7}{2}}\right)\), so its polar form is \(\frac{\sqrt{85}}{2} \text{cis}(\tan^{-1}(\frac{6}{7}))\).
09
- Simplify Expression e
Expression \(3e^{j\pi} + 4e^{j\frac{\pi}{2}}\) can be converted by evaluating the exponential terms: \(3e^{j\pi} = -3\) and \(4e^{j\frac{\pi}{2}} = 4j\). Hence, the expression becomes \(-3 + 4j\).
10
- Convert Expression e to Polar Form
For \(-3 + 4j\), the modulus \(r\) is \(\sqrt{(-3)^2 + 4^2} = \sqrt{25} = 5\), and the argument is \(\tan^{-1}(-\frac{4}{3})\), which is approximately \(126.8^\circ\) or \(2.21\,\text{rad}\). Polar form is \(5 \text{cis}(2.21)\).
11
- Simplify Expression f
Given \((\sqrt{3}+j)2\sqrt{2}e^{-j\frac{\pi}{4}}\), simplify to \((\sqrt{3}+j)\) first to find its polar form: modulus is \(2\) and argument \(\frac{\pi}{6}\). Then multiply by \(2\sqrt{2}e^{-j\frac{\pi}{4}}\). Combined, \(r = 4\), argument is \(\frac{\pi}{6} - \frac{\pi}{4}\).
12
- Convert Expression f to Polar Form
The resulting polar form after factoring is \(4 \text{cis}(-\frac{\pi}{12})\).
13
- Simplify Expression g
Expression \(\frac{3}{1+j 3\pi}\) involves rationalizing the imaginary denominator. The modulus of the denominator is \(\sqrt{1 + (3\pi)^2}\) and we multiply by the conjugate \((1-j3\pi)\) to both numerator and denominator. After simplification, calculate real and imaginary parts.
14
- Convert Expression g to Polar Form
Compute the modulus and argument and simplify to polar form.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Cartesian to Polar Conversion
To convert complex numbers from Cartesian form to polar form, we start with a complex number expressed as \( a + bj \) where \( a \) is the real part, and \( b \) is the imaginary part. The polar form represents the number with its modulus and angle (or argument) in the form \( r \text{cis} \theta \) or \( re^{j\theta} \).
- Find the Modulus: The modulus \( r \) is the distance of the complex number from the origin on the complex plane. It is calculated using the formula \( r = \sqrt{a^2 + b^2} \).
- Determine the Argument: The argument \( \theta \) is the angle the line from the origin to the point makes with the positive real axis. This is found using \( \theta = \tan^{-1}\left(\frac{b}{a}\right) \).
Complex Plane Plotting
The complex plane, also known as the Argand plane, is a two-dimensional plane where each complex number corresponds to a unique point. The horizontal axis represents the real part, and the vertical axis represents the imaginary part of the complex number.
- Point Representation: A complex number such as \( a + bj \) is represented as the point \((a, b)\) on the plane.
- Quadrants: The plane is divided into four quadrants similar to the Cartesian plane. Depending on the signs of \( a \) and \( b \), the complex number is plotted in one of these quadrants.
Complex Number Modulus and Argument
The modulus and argument are essential for representing complex numbers in polar form. The modulus \( r \) is the 'length' of the complex number from the origin (akin to the radius of a circle), computed as \( \sqrt{a^2 + b^2} \), showing the size of the complex number.
The argument \( \theta \) provides the direction, expressed in radians or degrees, and is given by \( \tan^{-1}\left(\frac{b}{a}\right) \).
The argument \( \theta \) provides the direction, expressed in radians or degrees, and is given by \( \tan^{-1}\left(\frac{b}{a}\right) \).
- Positive Real Axis: The argument is zero for numbers on the positive real axis.
- Negative Real Axis: For numbers on the negative real axis, the argument is \( \pi \) (or 180 degrees).
Polar Form Representation
Polar form provides a powerful representation of complex numbers, especially useful when multiplying or performing powers of complex numbers. Polar form expresses a complex number as \( r \text{cis} \theta \) or equivalently as \( re^{j\theta} \), where \( r \) is the modulus and \( \theta \) the argument.
- Exponential Form: The expression \( re^{j\theta} \) leverages Euler's formula, simplifying multiplication and division of complex numbers, as it converts the operations into summing angles and multiplying moduli.
- Applications: Polar form is particularly handy in engineering and physics, where angle and magnitude are more intuitive to interpret particularly in cases like sinusoidal functions, phasors, and waveforms.
