Q: What are the integer powers of the imaginary unit?
A:
$j^0=1$
$j^1=j$
$j^2=-1$
$j^3=-j$
This basic four-part pattern continues for consecutive integer powers above and below this arbitrarily selected sequence
Showing posts with label complex numbers. Show all posts
Showing posts with label complex numbers. Show all posts
Monday, January 5, 2015
Complex Conjugates
Given: $z = a + jb = |z|e^{j\theta}$
Q3: How can you simplify the expression $zz^*$?
A3: Although the multiplication suggests that you use polar form to simplify the product, Cartesian form yields an interesting result.
$zz^*=(a+jb)(a-jb)$
Notice that the cross-terms (i.e. the O and I products of FOIL) negate each other.
$zz^*=a^2+b^2$
$zz^*={|z|}^2$
Q1: What is the complex conjugate of $z$ in Cartesian form?
A1: $z^* = a - jb$
Q2: What is the complex conjugate of $z$ in polar form?
A2: Note that $z^*$ is simply the reflection of $z$ over the real axis of the complex plane. It then follows that:
1.The magnitude of $z^*$ is the same as the magnitude of $z$
2. The angle that $z^*$ makes with respect to the positive real axis is the negative value of the angle that $z$ makes with the positive real axis.
Hence, $z^*= |z|e^{-j\theta}$
Q3: How can you simplify the expression $zz^*$?
A3: Although the multiplication suggests that you use polar form to simplify the product, Cartesian form yields an interesting result.
$zz^*=(a+jb)(a-jb)$
Notice that the cross-terms (i.e. the O and I products of FOIL) negate each other.
$zz^*=a^2+b^2$
$zz^*={|z|}^2$
Complex Number Exponentiation
Given: $z$ a complex number
Q1: What is $z^n$? Assume $n$ is an integer.
A1: Since exponentiation is repeated multiplication, it's convenient to use the complex polar form of $z$:
$z=\left|z\right|e^{j\theta}$
$z^n={\left|z\right|}^ne^{jn\theta}$
This result is known as De Moivre's formula.
Q2: Suppose you don't have $z$ raised to an integer power. Instead, you have $z^{\frac{1}{n}}$. How do you find the $n$ roots of this expression?
Q3: To answer Q2, let's look at the specific case where $n$ is 3. What are the 3 roots of $z^{\frac{1}{3}}$?
A3: Following the logic from A1,
Root 1: $z^{\frac{1}{3}}={\left|z\right|}^{\frac{1}{3}}e^{\frac{j\theta}{3}}$.
But this is only 1 root. We know we need to find 2 more. Note that $2\pi$ can be added to the angle $\theta$ without changing the result. We can write 2 more cases:
Root 2: $z^{\frac{1}{3}}={\left|z\right|}^{\frac{1}{3}}e^{\frac{j(\theta+2\pi)}{3}}$.
Root 3: $z^{\frac{1}{3}}={\left|z\right|}^{\frac{1}{3}}e^{\frac{j(\theta+4\pi)}{3}}$.
Q4: This gives us 3 roots - just what we want. But why not continue adding 2$\pi$ to the angle again? Couldn't this produce a 4th root?
A4: Suppose you added another 2$\pi$. Then you'd get the fourth case:
Root 4: $z^{\frac{1}{3}}={\left|z\right|}^{\frac{1}{3}}e^{\frac{j(\theta+6\pi)}{3}}$.
But note that this reduces to $z^{\frac{1}{3}}={\left|z\right|}^{\frac{1}{3}}e^{\frac{j\theta}{3}+2\pi}={\left|z\right|}^{\frac{1}{3}}e^{\frac{j\theta}{3}}$.
which is the same as Root 1. So there's no need to continue adding 2$\pi$ past the Root 3. Any other roots found by adding 2$\pi$ past this point are redundant.
A2: To answer the general case of Q2, we substitute $n$ for 3 to find that the $n$ roots of $z^{\frac{1}{n}}$ are:
$z^{\frac{1}{n}}={\left|z\right|}^{\frac{1}{n}}e^{\frac{j(\theta+2k\pi)}{n}}$ where $k$ is an integer in the range 0, 1, ... , $n-1$
Q3: To answer Q2, let's look at the specific case where $n$ is 3. What are the 3 roots of $z^{\frac{1}{3}}$?
A3: Following the logic from A1,
Root 1: $z^{\frac{1}{3}}={\left|z\right|}^{\frac{1}{3}}e^{\frac{j\theta}{3}}$.
But this is only 1 root. We know we need to find 2 more. Note that $2\pi$ can be added to the angle $\theta$ without changing the result. We can write 2 more cases:
Root 2: $z^{\frac{1}{3}}={\left|z\right|}^{\frac{1}{3}}e^{\frac{j(\theta+2\pi)}{3}}$.
Root 3: $z^{\frac{1}{3}}={\left|z\right|}^{\frac{1}{3}}e^{\frac{j(\theta+4\pi)}{3}}$.
Q4: This gives us 3 roots - just what we want. But why not continue adding 2$\pi$ to the angle again? Couldn't this produce a 4th root?
A4: Suppose you added another 2$\pi$. Then you'd get the fourth case:
Root 4: $z^{\frac{1}{3}}={\left|z\right|}^{\frac{1}{3}}e^{\frac{j(\theta+6\pi)}{3}}$.
But note that this reduces to $z^{\frac{1}{3}}={\left|z\right|}^{\frac{1}{3}}e^{\frac{j\theta}{3}+2\pi}={\left|z\right|}^{\frac{1}{3}}e^{\frac{j\theta}{3}}$.
which is the same as Root 1. So there's no need to continue adding 2$\pi$ past the Root 3. Any other roots found by adding 2$\pi$ past this point are redundant.
A2: To answer the general case of Q2, we substitute $n$ for 3 to find that the $n$ roots of $z^{\frac{1}{n}}$ are:
$z^{\frac{1}{n}}={\left|z\right|}^{\frac{1}{n}}e^{\frac{j(\theta+2k\pi)}{n}}$ where $k$ is an integer in the range 0, 1, ... , $n-1$
Complex Multiplication/Division
Given:
$z_1=a_1+jb_1=\left|z_1\right|e^j\theta_1$
$z_2=a_2+jb_2=\left|z_2\right|e^j\theta_2$
Q1: What is $z_1z_2$ in Cartesian form?
A1: Using the FOIL method and collecting the real and imaginary parts:
$z_1z_2=(a_1a_2-b_1b_2)+j(a_1b_2+a_2b_1)$
Q2: What is $z_1z_2$ in polar form?
A2: $z_1z_2=\left|z_1\right|\left|z_2\right|e^{j(\theta_1+\theta_2)}$
Q3: What is $\frac{z_1}{z_2}$ in Cartesian form?
A3: Multiplying both the numerator and denominator by $z_2^*$:
$\frac{z_1}{z_2}=\frac{(a_1a_2+b_1b_2)+j(a_2b_1-a_1b_2)}{a_2^2+b2_2}$
Q4: What is $\frac{z_1}{z_2}$ in polar form?
A4: $\frac{z_1}{z_2}=\frac{\left|z_1\right|}{\left|z_2\right|}e^{j(\theta_1-\theta_2)}$
$z_1=a_1+jb_1=\left|z_1\right|e^j\theta_1$
$z_2=a_2+jb_2=\left|z_2\right|e^j\theta_2$
Q1: What is $z_1z_2$ in Cartesian form?
A1: Using the FOIL method and collecting the real and imaginary parts:
$z_1z_2=(a_1a_2-b_1b_2)+j(a_1b_2+a_2b_1)$
Q2: What is $z_1z_2$ in polar form?
A2: $z_1z_2=\left|z_1\right|\left|z_2\right|e^{j(\theta_1+\theta_2)}$
Q3: What is $\frac{z_1}{z_2}$ in Cartesian form?
A3: Multiplying both the numerator and denominator by $z_2^*$:
$\frac{z_1}{z_2}=\frac{(a_1a_2+b_1b_2)+j(a_2b_1-a_1b_2)}{a_2^2+b2_2}$
Q4: What is $\frac{z_1}{z_2}$ in polar form?
A4: $\frac{z_1}{z_2}=\frac{\left|z_1\right|}{\left|z_2\right|}e^{j(\theta_1-\theta_2)}$
Complex Addition/Subtraction
Given:
$z_1=a_1+jb_1$
$z_2=a_2+jb_2$
Q1: How do I add complex numbers in Cartesian form?
A1: Complex numbers are added component-wise:
$z_1+z_2=(a_1+a_2)+j(b_1+b_2)$
Q2: How do I subtract complex numbers in Cartesian form?
A2: Complex numbers are subtracted component-wise:
$z_1-z_2=(a_1-a_2)+j(b_1-b_2)$
$z_1=a_1+jb_1$
$z_2=a_2+jb_2$
Q1: How do I add complex numbers in Cartesian form?
A1: Complex numbers are added component-wise:
$z_1+z_2=(a_1+a_2)+j(b_1+b_2)$
Q2: How do I subtract complex numbers in Cartesian form?
A2: Complex numbers are subtracted component-wise:
$z_1-z_2=(a_1-a_2)+j(b_1-b_2)$
Cartesian and Polar Forms of Complex Numbers
Q1: What is a complex number?
A1: A complex number $z$ is a number with:
1. A real component $a$
2. An imaginary component $b$
where $z=a+jb$ where $j=\sqrt{-1}$
This is the Cartesian form of the complex number.
Q2: How else may a complex number be expressed?
A2: Alternatively, $z$ may be expressed in polar form with:
1. A magnitude $\left|z\right|$
2. An angle $\theta$
where $z=\left|z\right|e^j\theta$
Q3: Why should I bother learning Cartesian and polar form of complex numbers?
A3: Cartesian form is convenient for complex addition/subtraction while polar form is convenient for complex multiplication/division.
Q4: How do I switch from Cartesian to polar form?
A4: Use the relations:
1. $\left|z\right|=\sqrt{a^2+b^2}$
2. $\tan{\theta}=\frac {b}{a}$
Q5: How do I switch from polar to Cartesian form?
A5: Use the relations:
1. $a=\left|z\right|\cos{\theta}$
2. $b=\left|z\right|\sin{\theta}$
A1: A complex number $z$ is a number with:
1. A real component $a$
2. An imaginary component $b$
where $z=a+jb$ where $j=\sqrt{-1}$
This is the Cartesian form of the complex number.
Q2: How else may a complex number be expressed?
A2: Alternatively, $z$ may be expressed in polar form with:
1. A magnitude $\left|z\right|$
2. An angle $\theta$
where $z=\left|z\right|e^j\theta$
Q3: Why should I bother learning Cartesian and polar form of complex numbers?
A3: Cartesian form is convenient for complex addition/subtraction while polar form is convenient for complex multiplication/division.
Q4: How do I switch from Cartesian to polar form?
A4: Use the relations:
1. $\left|z\right|=\sqrt{a^2+b^2}$
2. $\tan{\theta}=\frac {b}{a}$
Q5: How do I switch from polar to Cartesian form?
A5: Use the relations:
1. $a=\left|z\right|\cos{\theta}$
2. $b=\left|z\right|\sin{\theta}$
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