Lesson Complex numbers and arithmetic operations on them
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<H2>Complex numbers and arithmetic operations on them</H2> Not every quadratic equation with real coefficients has the real root, as you know. Probably, the most famous of this kind of equations is the one of the form {{{x^2+1 = 0}}}. It is clear why it has no solutions in real numbers. If the real number {{{x}}} is the solution, then {{{x^2}}} is not negative, hence, {{{x^2+1}}} is positive and can not be equal to zero, we have a contradiction. In order to resolve this problem, mathematicians invented so called "complex numbers". Complex numbers have a form {{{a + b*red(i)}}}, where {{{a}}} and {{{b}}} are the real numbers and {{{red(i)}}} is a new kind of number called <B>imaginary unit</B>. The component {{{a}}} of the complex number {{{a+b*red(i)}}} is called the <B>real part</B> of the complex number {{{a+b*red(i)}}}. The component {{{b}}} of the complex number {{{a+b*red(i)}}} is called the <B>imaginary part</B> of the complex number {{{a+b*red(i)}}}. Two complex numbers, {{{a + b*red(i)}}} and {{{c + d*red(i)}}}, are equal if and only if they have equal real parts and equal imaginary parts: {{{a=c}}} and {{{b=d}}}. Real numbers are actually a subset of the set of complex numbers. You can identify the real number {{{a}}} with the complex number {{{a+0*red(i)}}}. Mathematicians say that real numbers are embedded in the set of complex numbers. We just have defined complex numbers as a set. Now we are going to define arithmetical operations on the set of complex numbers: addition, subtraction, multiplication and division. You will see later that these operations are very similar to well known arithmetical operations over real numbers. <B><U>Addition of complex numbers</B></U> <BLOCKQUOTE><B>Definition</B> The sum of complex numbers {{{a+b*red(i)}}} and {{{c+d*red(i)}}} is a complex number {{{(a+c)+(b+d)*red(i)}}}.</BLOCKQUOTE> <B>Examples</B> {{{(1+5*red(i))+(2+3*red(i)) = 3+8*red(i)}}}. {{{(1+0*red(i))+(2+0*red(i)) = 3+0*red(i)}}}. {{{(0+5*red(i))+(0+3*red(i)) = 0+8*red(i)}}}. {{{(2+5*red(i))+(0+3*red(i)) = 2+8*red(i)}}}. Quite simple, isn't? Note that if you have the complex number in the form {{{2+0*red(i)}}}, you can simply write it as {{{2}}}. You can always distinguish between sorts of numbers if you have or have not the zero imaginary part. Similarly, if you have the complex number of the form {{{0+3*red(i)}}}, you can simply write it as {{{3*red(i)}}}. OK, let us go forward with definitions. <B><U>Subtraction of complex numbers</B></U> <BLOCKQUOTE><B>Definition</B> The difference of complex numbers {{{a+b*red(i)}}} and {{{c+d*red(i)}}} is a complex number {{{(a-c)+(b-d)*red(i)}}}.</BLOCKQUOTE> <B>Examples</B> {{{(1+5*red(i))-(2+3*red(i)) = -1+2*red(i)}}}. {{{(1+0*red(i))-(2+0*red(i)) = -1+0*red(i)}}}. {{{(0+5*red(i))-(0+3*red(i)) = 0+2*red(i)}}}. {{{(2+5*red(i))-(0+3*red(i)) = 2+2*red(i)}}}. <B>Important notes about the properties of addition and subtraction operations:</B> 1) Who plays the role of zero in the set of complex numbers? The answer is clear: the complex number {{{0+0*red(i)}}}. Indeed, if we add {{{0+0*red(i)}}} to any complex number, or if we subtract {{{0+0*red(i)}}} from any complex number, we don't change this complex number. This is the role of zero. 2) For a given complex number {{{a+b*red(i)}}} let us consider the number {{{-a-b*red(i)}}}. The sum of {{{a+b*red(i)}}} and {{{-a-b*red(i)}}} is always equal to zero, so {{{-a-b*red(i)}}} is an <B>opposite</B> to {{{a+b*red(i)}}}, exactly in the same sense as the real number {{{-a}}} is opposite to the real number {{{a}}} in the set (in the additive group) of real numbers. 3) To subtract the complex number {{{c+d*red(i)}}} from the complex number {{{a+b*red(i)}}} is the same as to add an opposite number {{{-c-d*red(i)}}} to the complex number {{{a+b*red(i)}}}. One can describe this property in terms: "subtraction of complex numbers is an operation opposite to addition". 4) The addition operation is commutative: {{{(a+bi) + (c+di) = (c+di) + (a+bi)}}}. This property directly follows from the definition and from commutative property of the addition operation for real numbers. 5) The addition operation is associative: {{{((a+bi) + (c+di)) + (e+fi) = (a+bi) + ((c+di) + (e+fi))}}}. This property directly follows from the definition and from associative property of the addition operation for real numbers. 6) In the advanced course of algebra you may learn about additive groups. The validity of properties 1), 2), 3), 4) and 5) means that the set of complex numbers with introduced operations of addition and subtraction is an additive group. <B><U>Multiplication of complex numbers</B></U> <BLOCKQUOTE><B>Definition</B> The product of complex numbers {{{a+b*red(i)}}} and {{{c+d*red(i)}}} is a complex number {{{(ac-bd)+(ad+bc)*red(i)}}}.</BLOCKQUOTE> <B>Examples</B> {{{(1+5*red(i))*(2+3*red(i)) = (1*2-5*3)+(1*3+5*2)*red(i) = -13+13*red(i)}}}. {{{(1+0*red(i))*(2+0*red(i)) = (1*2-0*0)+(1*0+0*2)*red(i) = 2}}}. {{{(0+5*red(i))*(0+3*red(i)) = (0*0-5*3)+(0*3+5*0)*red(i) = -15}}}. {{{(2+5*red(i))*(0+3*red(i)) = (2*0-5*3)+(2*3+5*0)*red(i) = -15+6*red(i)}}}. Note that in accordance with the last definition, {{{red(i)*red(i)=-1}}}, or {{{red(i)^2=-1}}}. This is the key property of the <B>imaginary unit</B>. This property changes the whole picture. First, it shows that the equation {{{x^2+1 = 0}}}, we started this lesson with, has the solution in complex numbers. Namely, the complex number {{{red(i)}}} is the solution, because {{{red(i)^2=-1}}}. Moreover, it has two solutions, because {{{-red(i)}}} is the solution too. Furthermore, because of this property, the following expression is valid: {{{sqrt(-1)}}} = +/-{{{red(i)}}}. This means that taking square root of a negative real number is possible in the set of complex numbers (which is not possible in the set of real numbers). We will discuss it later in more details in further lessons. Next consequence is in that every quadratic equation with complex coefficients has a root in complex numbers. To be more precise, it has exactly two roots in complex numbers. We will discuss it later at the end of this lesson. <B><U>Operation taking the conjugate</U></B> <BLOCKQUOTE><B>Definition</B> <B><U>The conjugate</B></U> to a given complex number {{{a+b*red(i)}}} is a complex number {{{a-b*red(i)}}}.</BLOCKQUOTE> <B>Examples</B> Calculate conjugate to the complex number {{{2+3*red(i)}}}. <B>Answer:</B> {{{2-3*red(i)}}}. Calculate conjugate to the complex number {{{-2+3*red(i)}}}. <B>Answer:</B> {{{-2-3*red(i)}}}. Calculate conjugate to the complex number {{{2+0*red(i)}}}. <B>Answer:</B> {{{2}}}. Calculate conjugate to the complex number {{{3*red(i)}}}. <B>Answer:</B> {{{-3*red(i)}}}. <B><U>The product of a complex number and its conjugate</U></B> Let's calculate the product of a complex number {{{a+b*red(i)}}} and its conjugate {{{a-b*red(i)}}}. {{{(a+b*red(i))*(a-b*red(i)) = a*a+b*(-b) + red(i)*(a*(-b)+b*a) = a^2 + b^2}}}. You see that the product of a complex number {{{a+b*red(i)}}} and its conjugate {{{a-b*red(i)}}} is equal to a real number {{{a^2 + b^2}}}. Put attention that this product, the real number {{{a^2 + b^2}}}, is always greater than zero, except only one case when both {{{a}}} and {{{b}}} are equal to zero. This is exactly the case then the complex number {{{a+b*red(i)}}} equal to zero. So, if complex number {{{a+b*red(i)}}} is not equal to zero, the product of {{{a+b*red(i)}}} to its conjugate {{{a-b*red(i)}}} is a positive real number {{{a^2 + b^2}}}. <B>Important note about the unit element</B>: The complex number {{{1+0*red(i)}}} also plays a special role. If we multiply any complex number {{{a+b*red(i)}}} by {{{1.0+0*red(i)}}}, we get the same complex number {{{a+b*red(i)}}}, with no change. You can easily check it by direct calculation. This means that complex number {{{1+0*red(i)}}}, or simply {{{1}}}, plays the role of the unit element in the multiplicative group of complex numbers, and in the ring of complex numbers (for those who knows what does it mean). <B><U>An inverse element</U></B> <BLOCKQUOTE><B>Definition</B> For the given complex number {{{a+b*red(i)}}} an <B><U>inverse element</U></B> is such a complex number {{{c+d*red(i)}}} that {{{(a+b*red(i))*(c+d*red(i)) = 1}}}.</BLOCKQUOTE> For the any given complex number {{{a+b*red(i)}}}, not equal to zero, an inverse complex number does exist and is equal to {{{(a-bi)/(a^2+b^2)}}}, in other words, is equal to its conjugate divided to {{{a^2+b^2}}}. To check this statement, we have to multiply a+bi by {{{(a-b*red(i))/(a^2+b^2)}}}. Let us do it. Multiplying {{{a+b*red(i)}}} by {{{a-b*red(i)}}} in the numerator, we get {{{a^2+b^2}}} (we just did it above when calculated the product of the complex number and its conjugate). Canceling numerator and denominator to {{{a^2+b^2}}}, we get 1. So, the statement is valid. <B>Examples</B> Calculate the inverse to complex number {{{2+3*red(i)}}}. <B>Answer:</B> {{{(2-3*red(i))/(2^2+3^2) = (2-3*red(i))/13}}}. Calculate the inverse to complex number {{{-2+3*red(i)}}}. <B>Answer:</B> {{{(-2-3*red(i))/(2^2+3^2) = (-2-3*red(i))/13}}}. Now let us make the last definition. <B><U>Division of complex numbers</B></U> <BLOCKQUOTE><B>Definition</B> To divide complex number {{{a+b*red(i)}}} <B>(dividend)</B> by another complex number {{{c+d*red(i)}}} <B>(divisor)</B> means to find the third complex number {{{x+y*red(i)}}} <B>(quotient)</B> such that <B>when multiplied by divisor, it is equal to the dividend</B>.</BLOCKQUOTE> The quotient usually denotes as a fraction {{{(a+b*red(i))/(c+d*red(i))}}}. <B><U>The general formula for division of complex numbers</U></B> The general formula for division of complex numbers is {{{(a+b*red(i))/(c+d*red(i))}}} = {{{(ac+bd)/(c^2+d^2) + (bc-ad)/(c^2+d^2)*red(i)}}} (providing the divisor is non zero). We can simply check this formula by multiplying its right hand side, {{{(ac+bd)/(c^2+d^2) + (bc-ad)/(c^2+d^2)*red(i)}}}, by the divisor {{{c+d*red(i)}}}. According to the definition, the product should be equal to the dividend, {{{a+b*red(i)}}}. Performing multiplication, we get - for the real part of the numerator: {{{(ac+bd)*c - (bc-ad)*d = acc+bcd-bcd+add = a(c^2+d^2)}}}, and - for the imaginary part of the numerator: {{{(ac+bd)*d + (bc-ad)*c = acd+bdd+bcc-acd = b(c^2+d^2)}}}. Canceling numerator and denominator by the factor {{{(c^2+d^2)}}}, we obtain the product {{{a+b*red(i)}}}, exactly as expected. There is another, more practical way to calculate the quotient. This way is to multiply the numerator and denominator of the quotient by the number {{{c-d*red(i)}}}, the conjugate to the quotient denominator. It gives the same result as the general formula above. Examples below illustrate this way. <B>Example</B> Divide {{{8+red(i)}}} by {{{2-3*red(i)}}}. Let us consider the fraction {{{(8+red(i))/(2-3*red(i))}}} and multiply numerator and denominator by number {{{2+3*red(i)}}}, which is the conjugate to {{{2-3*red(i)}}}. You get {{{((8+red(i))*(2+3*red(i)))/((2-3*red(i))*(2+3*red(i)))}}}={{{(13+26*red(i))/13}}} = {{{1+2*red(i)}}}. <B>One more example</B> Divide {{{1+3*red(i)}}} by {{{2+red(i)}}}. Let us consider the fraction {{{(1+3*red(i))/(2+red(i))}}} and multiply numerator and denominator by number <B>2-i</B>, which is the conjugate to <B>2+i</B>. You get {{{((1+3*red(i))*(2-red(i)))/((2+red(i))*(2-red(i))))}}}={{{(5+5*red(i))/5}}} = {{{1+red(i)}}}. Note that to divide the complex number {{{a+b*red(i)}}} by the complex number {{{c+d*red(i)}}} is the same as to multiply the complex number {{{a+b*red(i)}}} by inverse to the complex number {{{c+d*red(i)}}}. One can describe this property in terms: "division of complex numbers is an operation opposite to multiplication". <B><U>Summary</U></B> The sum of complex numbers {{{a+b*red(i)}}} and {{{c+d*red(i)}}} is the complex number {{{(a+c)+(b+d)*red(i)}}}. The difference of complex numbers {{{a+b*red(i)}}} and {{{c+d*red(i)}}} is the complex number {{{(a-c)+(b-d)*red(i)}}}. The product of complex numbers {{{a+b*red(i)}}} and {{{c+d*red(i)}}} is the complex number {{{(ac-bd)+(ad+bc)*red(i)}}}. The conjugate to the complex number {{{a+b*red(i)}}} is the complex number {{{a-b*red(i)}}}. The product of the complex number {{{a+b*red(i)}}} and its conjugate {{{a-b*red(i)}}} is equal to the real number {{{a^2 + b^2}}}. An inverse to the complex number {{{a+b*red(i)}}}, different from zero, does exist and is the complex number {{{(a-bi)/(a^2+b^2)}}}. The quotient of complex numbers {{{a+b*red(i)}}} and {{{c+d*red(i)}}}, {{{(a+b*red(i))/(c+d*red(i))}}}, is the complex number {{{(ac+bd)/(c^2+d^2) + (bc-ad)/(c^2+d^2)*red(i)}}} (providing the divisor is non zero). Complex numbers are used in many scientific fields including engineering, electromagnetism, quantum physics, applied mathematics and so on. For your convenience, below is the list of my relevant lessons on complex numbers in this site in the logical order. They all are under the current topic <B>Complex numbers</B> in the section <B>Algebra II</B>. - <B>Complex numbers and arithmetic operations on them</B> (this lesson) - <A HREF=http://www.algebra.com/algebra/homework/complex/Complex-plane.lesson> Complex plane</A> - <A HREF=http://www.algebra.com/algebra/homework/complex/Addition-and-subtraction-of-complex-numbers-in-complex-plane.lesson> Addition and subtraction of complex numbers in complex plane</A> - <A HREF=http://www.algebra.com/algebra/homework/complex/Multiplication-and-division-of-complex-numbers-in-complex-plane-.lesson> Multiplication and division of complex numbers in complex plane</A> - <A HREF=http://www.algebra.com/algebra/homework/complex/Raising-a-complex-number-to-an-integer-power.lesson> Raising a complex number to an integer power</A> - <A HREF=http://www.algebra.com/algebra/homework/complex/How-to-take-a-root-of-a-complex-number.lesson> How to take a root of a complex number</A> - <A HREF=http://www.algebra.com/algebra/homework/complex/Solution-of-the-quadratic-equation-with-real-coefficients-on-complex-domain.lesson> Solution of the quadratic equation with real coefficients on complex domain</A> - <A HREF=http://www.algebra.com/algebra/homework/complex/How-to-take-a-square-root-of-a-complex-number.lesson> How to take a square root of a complex number</A> - <A HREF=http://www.algebra.com/algebra/homework/complex/Solution-of-the-quadratic-equation-with-complex-coefficients-on-complex-domain.lesson> Solution of the quadratic equation with complex coefficients on complex domain</A> - <A HREF=https://www.algebra.com/algebra/homework/complex/Solved-problems-on-taking-roots-of-complex-numbers.lesson>Solved problems on taking roots of complex numbers</A> - <A HREF=https://www.algebra.com/algebra/homework/complex/Solved-problems-on-arithmetic-operations-on-complex-numbers.lesson>Solved problems on arithmetic operations on complex numbers</A> - <A HREF=https://www.algebra.com/algebra/homework/complex/Solved-problem-on-taking-square-roots-of-complex-numbers.lesson>Solved problem on taking square root of complex number</A> - <A HREF=https://www.algebra.com/algebra/homework/complex/Solving-polynomial--equations-in-complex-domain.lesson>Solving polynomial equations in complex domain</A> - <A HREF=https://www.algebra.com/algebra/homework/complex/Miscellaneous-problems-on-complex-numbers.lesson>Miscellaneous problems on complex numbers</A> - <A HREF=https://www.algebra.com/algebra/homework/complex/Advanced-problem-in-complex-numbers.lesson>Advanced problems on complex numbers</A> - <A HREF=https://www.algebra.com/algebra/homework/complex/Solved-problems-on-de%27Moivre-formula.lesson>Solved problems on de'Moivre formula</A> - <A HREF=https://www.algebra.com/algebra/homework/complex/Proving-identities-using-complex-numbers.lesson>Proving identities using complex numbers</A> - <A HREF=https://www.algebra.com/tutors/18-Calculating-1sin%281%B0%29%2B2sin%282%B0%29%2B3sin%283%B0%29%2B-%2B180sin%28180%B0%29.lesson>Calculating the sum 1*sin(1°) + 2*sin(2°) + 3*sin(3°) + . . . + 180*sin(180°)</A> - <A HREF=https://www.algebra.com/algebra/homework/complex/An-equation-in-complex-numbers-which-HAS-NO-a-solution.lesson>A curious example of an equation in complex numbers which HAS NO a solution</A> - <A HREF=https://www.algebra.com/algebra/homework/complex/Solving-one-non-standard-equation-in-complex-numbers.lesson>Solving non-standard equations in complex numbers</A> - <A HREF=https://www.algebra.com/algebra/homework/complex/Upper-level-problem-on-complex-numbers.lesson>Upper level problem on complex numbers</A> - <A HREF=https://www.algebra.com/algebra/homework/complex/Determine-locus-of-points-using-complex-numbers.lesson>Determine locus of points using complex numbers</A> - <A HREF=https://www.algebra.com/algebra/homework/complex/Joke-problems-on-complex-numbers.lesson>Joke problems on complex numbers</A> - <A HREF=https://www.algebra.com/algebra/homework/complex/Review-of-lessons-on-complex-numbers.lesson>OVERVIEW of lessons on complex numbers</A> Use this file/link <A HREF=https://www.algebra.com/algebra/homework/complex/ALGEBRA-II-YOUR-ONLINE-TEXTBOOK.lesson>ALGEBRA-II - YOUR ONLINE TEXTBOOK</A> to navigate over all topics and lessons of the online textbook ALGEBRA-II.
