Lesson How to take a square root of a complex number
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<H2>How to take a square root of a complex number</H2> Taking a root of a complex number was just considered in the lesson <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> in this module. In that lesson the original complex numbers were presented in the trigonometric form {{{z = r(cos(alpha) + red(i)*sin(alpha))}}}, where {{{r}}} was the modulus and {{{alpha}}} was the argument. Now I will explain how to take a square root of a complex number written in the form {{{z = a + red(i)*b}}}, (1) where {{{a}}} and {{{b}}} are real numbers. So, let us suppose that the original complex number {{{z}}} is presented in the form (1), and we will look for the square root in the similar form {{{w = r + red(i)*s}}}, (2) where {{{r}}} and {{{s}}} are real numbers. Our goal is to calculate components {{{r}}} and {{{s}}} via given numbers {{{a}}} and {{{b}}}. First of all, if {{{w = r + red(i)*s}}} is the square root of {{{z = a + red(i)*b}}}, then {{{w^2 = z}}}. You can calculate {{{w^2}}} using the multiplication formula for complex numbers from the lesson <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>: {{{w^2 = (r + red(i)s)*(r + red(i)*s) = r^2 - s^2 + red(i)*(sr+rs) = r^2 - s^2 + 2*rs*red(i)}}}. From the other side, {{{w^2 = z = a + red(i)*b}}}. Comparing the real part and the imaginary part of these two expressions for {{{w^2}}} you get two following equations: {{{system (r^2 - s^2 = a, 2*rs = b)}}}. (3) We should solve this system of equations to find unknown real numbers {{{r}}} and {{{s}}} via given real numbers {{{a}}} and {{{b}}}. To solve the system, express {{{s}}} from the second equation: {{{s = b/(2*r)}}} (4) and substitute it to the first equation of the system. You will get {{{r^2 - b^2/(4*r^2) = a}}}. Simplify the last equation step by step: {{{4*r^4 - b^2 = 4*a*r^2}}} (after multiplication of both sides by {{{4*r^2}}}); {{{4*r^4 - 4*a*r^2 - b^2 = 0}}} (after moving the term {{{4*a*(r^2)}}} from the right side to the left); {{{4*y^2 - 4*a*y - b^2 = 0}}} (after introducing new variable {{{y = r^2}}}). You got the quadratic equation for {{{y}}}. Apply the <B>quadratic formula</B> to solve this quadratic equation (see the lesson <A HREF = http://www.algebra.com/algebra/homework/quadratic/lessons/Introduction-Into-Quadratics.lesson> Introduction into Quadratic Equations</A>): {{{y = (4a +- sqrt(16a^2 + 16*b^2))/8 = 4(a +-sqrt(a^2 + b^2))/8 = (a +-sqrt(a^2 + b^2))/2}}}. Since {{{y}}} is the square of the real number ({{{y=r^2}}}), it can not be negative; hence {{{y = (a + sqrt(a^2 + b^2))/2}}}. This means that there are two solutions for {{{r}}}: {{{r[1] = sqrt((a + sqrt(a^2 + b^2))/2)}}} and {{{r[2] = -sqrt((a + sqrt(a^2 + b^2))/2)}}}, that differ by the sign only. Now substitute these expressions for {{{r}}} into the first equation of the system (3) to get {{{s}}}: {{{s^2 = r^2 - a = (a + sqrt(a^2 + b^2))/2 - a = (sqrt(a^2 + b^2) - a)/2}}}. This means that there are two solutions for {{{s}}}: {{{s[1] = sqrt((sqrt(a^2 + b^2) - a)/2)}}} and {{{s[2] = -sqrt((sqrt(a^2 + b^2) - a)/2)}}}. They differ by the sign only. The found values for {{{r}}} and {{{s}}} should be combined such a way to provide the correct sign of the product {{{r*s = b/2}}} (see the formula (3)). <H3>Summary</H3> The square root of the complex number {{{z = a + red(i)*b}}} has two values. The first value is the complex number {{{w[1] = r[1] + red(i)*s[1]}}}, where {{{r[1] = sqrt((a + sqrt(a^2 + b^2))/2)}}}, {{{s[1]}}} = +/-{{{sqrt((sqrt(a^2 + b^2) - a)/2)}}}. The second value is the complex number {{{w[2] = r[2] + red(i)*s[2]}}}, where {{{r[2] = -sqrt((a + sqrt(a^2 + b^2))/2)}}}, {{{s[2]}}} = -/+{{{sqrt((sqrt(a^2 + b^2) - a)/2)}}}. The found values for {{{r}}} and {{{s}}} should be combined such a way to provide the correct sign of the product {{{r*s = b/2}}}. Note that the second complex square root is the complex number opposite to the first one: {{{w[2] = -w[1]}}}. <H3>Check</H3> Let us check formulas for the complex square roots. To make the check, calculate {{{w[1]^2}}}: {{{w[1]^2 = r[1]^2 -s[1]^2 + red(i)*(2*r[1]*s[1]) = ((a + sqrt(a^2 + b^2))/2) - ((sqrt(a^2 + b^2) - a)/2) + red(i)*(2*sqrt((a + sqrt(a^2 + b^2))/2)*sqrt((sqrt(a^2 + b^2) - a)/2))}}} = {{{a + red(i)*(2*sqrt(a^2 - (a^2-b^2))/2) = a + red(i)*b}}}. It confirms that {{{w[1]^2 = z}}}. Similar check confirms that {{{w[2]^2 = z}}}. <H3>A comparison with the trigonometric form of the square root</H3> It was shown in the lesson <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> that the modulus of the <B>n-th</B> root of the complex number is the <B>n-th</B> root of the modulus of the original complex number, and the argument is <B>1/n-th</B> of the argument of that number. Let us compare it with what the obtained formulas produce for the square root. For the modulus of the square root the formulas produce {{{r^2 + s^2 = (a + sqrt(a^2 + b^2))/2 + (sqrt(a^2 + b^2) - a)/2 = sqrt(a^2 + b^2)}}}, exactly as the trigonometric form does. Regarding the argument, let us denote it as {{{alpha}}} for the original complex number and as {{{beta}}} for the square root. Then the formulas produce for {{{tan(beta)}}} {{{tan(beta) = s/r = sqrt((sqrt(a^2 + b^2) - a)/2) / sqrt((sqrt(a^2 + b^2) + a)/2)}}} = {{{sqrt( (sqrt(a^2 + b^2) - a) / (sqrt(a^2 + b^2) + a) )}}} = {{{sqrt( (1 - a/sqrt(a^2 + b^2)) / (1 + a/sqrt(a^2 + b^2)) )}}}. Note that {{{a/sqrt(a^2 + b^2) = cos(alpha)}}}, so the formula above can be written as {{{tan(beta) = sqrt( (1-cos(alpha))/ (1+cos(alpha)) )}}}. It is well known fact from <B>Trigonometry</B> that {{{sqrt( (1-cos(alpha))/ (1+cos(alpha)) ) = tan ((alpha)/2)}}}. Thus, we get {{{tan(beta) = tan ((alpha)/2)}}}, which implies {{{beta = alpha/2 + k*pi}}}, k = 0, 1. This is exactly the same as the <B>Trigonometric form</B> produces for the square root. <H3>Examples</H3> <B>Example 1. Calculate</B> {{{sqrt(1)}}}. We have {{{a = 1}}}, {{{b=0}}}. Substitute values {{{a = 1}}}, {{{b=0}}} to the formulas for {{{w[1]}}}. You get {{{r[1] = sqrt((1 + sqrt(1^2 + 0^2))/2) = 1}}}, {{{s[1] = sqrt((sqrt(1^2 + 0^2) - 1)/2) = 0}}}. So, the first complex square root is {{{w[1] = r[1] + red(i)*s[1] = 1 + red(i)*0 = 1}}}. The second complex square root is opposite to the first one: {{{w[2] = -1 + red(i)*0 = -1}}}. Square roots of <B>1</B> are <B>1</B> and <B>-1</B>, as expected. <B>Example 2. Calculate</B> {{{sqrt(-1)}}}. We have {{{a = -1}}}, {{{b=0}}}. Substitute values {{{a = -1}}}, {{{b=0}}} to the formulas for {{{w[1]}}}. You get {{{r[1] = sqrt(((-1) + sqrt((-1)^2 + 0^2))/2) = 0}}}, {{{s[1] = sqrt((sqrt((-1)^2 + 0^2) - (-1))/2) = 1}}}. So, {{{w[1] = r[1] + red(i)*s[1] = 0 + red(i)*1 = red(i)}}}. The second complex square root is opposite to the first one: {{{w[2] = 0 + red(i)*(-1) = -red(i)}}}. Complex square roots of <B>-1</B> are {{{red(i)}}} and {{{-red(i)}}}, as expected. <B>Example 3. Calculate</B> {{{sqrt(-3)}}}. We have {{{a = -3}}}, {{{b=0}}}. Substitute values {{{a = -3}}}, {{{b=0}}} to the formulas for {{{w[1]}}}. You get {{{r[1] = sqrt(((-3) + sqrt((-3)^2 + 0^2))/2) = 0}}}, {{{s[1] = sqrt((sqrt((-3)^2 + 0^2) - (-3))/2) = sqrt(6/2) = sqrt(3)}}}. So, {{{w[1] = r[1] + red(i)*s[1] = 0 + red(i)*sqrt(3) = red(i)*sqrt(3)}}}. The second complex square root is opposite to the first one: {{{w[2] = 0 - red(i)*sqrt(3) = -red(i)*sqrt(3)}}}. Complex square roots of <B>-3</B> are {{{red(i)*sqrt(3)}}} and {{{-red(i)*sqrt(3)}}}. You can check that complex square roots of <B>-3</B> are {{{red(i)*sqrt(3)}}} and {{{-red(i)*sqrt(3)}}}. Simply square {{{red(i)*sqrt(3)}}}. You get {{{(red(i)*sqrt(3))^2 = red(i)^2*sqrt(3)^2 = (-1)*3 = -3}}}. Same for {{{(-red(i)*sqrt(3))^2 = red(i)^2*sqrt(3)^2 = (-1)*3 = -3}}}. <B>Example 4. Calculate</B> {{{sqrt(-d)}}}, where {{{d}}} is positive real number. We have {{{a = -d}}}, {{{b=0}}}. Substitute values {{{a = -d}}}, {{{b=0}}} to the formulas for {{{w[1]}}}. You get {{{r[1] = sqrt(((-d) + sqrt((-d)^2 + 0^2))/2) = 0}}}, {{{s[1] = sqrt((sqrt((-d)^2 + 0^2) - (-d))/2) = sqrt(2d/2) = sqrt(d)}}}. So, {{{w[1] = r[1] + red(i)*s[1] = 0 + red(i)*sqrt(d) = red(i)*sqrt(d)}}}. The second complex square root is opposite to the first one: {{{w[2] = 0 - red(i)*sqrt(d) = -red(i)*sqrt(d)}}}. Complex square roots of <B>-d</B>, where {{{d}}} is positive real number, are {{{red(i)*sqrt(d)}}} and {{{-red(i)*sqrt(d)}}}. You can check that complex square roots of <B>-d</B> are {{{red(i)*sqrt(d)}}} and {{{-red(i)*sqrt(d)}}}. Simply square {{{red(i)*sqrt(d)}}}. You get {{{(red(i)*sqrt(d))^2 = red(i)^2*sqrt(d)^2 = (-1)*d = -d}}}. Same for {{{(-red(i)*sqrt(d))^2 = red(i)^2*sqrt(d)^2 = (-1)*d = -d}}}. <B>Example 5. Calculate</B> {{{sqrt(i)}}}. We have {{{a = 0}}}, {{{b=1}}}. Substitute values {{{a = 0}}}, {{{b=1}}} to the formulas for {{{w[1]}}}. You get {{{r[1] = sqrt((0 + sqrt(0^2 + 1^2))/2) = 1/sqrt(2) = (sqrt(2))/2}}}, {{{s[1] = sqrt((sqrt(0^2 + 1^2) - 0)/2) = 1/sqrt(2) = (sqrt(2))/2}}}. So, {{{w[1] = r[1] + red(i)*s[1] = (sqrt(2))/2 + red(i)*((sqrt(2))/2)}}}. The second complex square root is opposite to the first one: {{{w[2] = -(sqrt(2))/2 - red(i)*((sqrt(2))/2)}}}. Complex square roots of {{{i}}} are {{{(sqrt(2))/2 + red(i)*((sqrt(2))/2)}}} and {{{-(sqrt(2))/2 - red(i)*((sqrt(2))/2)}}}. <B>Example 6. Calculate</B> {{{sqrt(1 + red(i))}}}. We have {{{a = 1}}}, {{{b=1}}}. Substitute values {{{a = 1}}}, {{{b=1}}} to the formulas for {{{w[1]}}}. You get {{{r[1] = sqrt((1 + sqrt(1^2 + 1^2))/2)}}} = {{{sqrt((1+sqrt(2))/2))}}}, {{{s[1] = sqrt((sqrt(1^2 + 1^2) - 1)/2)}}} = {{{sqrt((sqrt(2)-1)/2))}}}. So, {{{w[1] = r[1] + red(i)*s[1] = sqrt((1+sqrt(2))/2) + red(i)*sqrt((sqrt(2)-1)/2)}}}. The second complex square root is opposite to the first one: {{{w[2] = -sqrt((1+sqrt(2))/2) - red(i)*sqrt((sqrt(2)-1)/2)}}}. Complex square roots of {{{1 + red(i)}}} are {{{sqrt((1+sqrt(2))/2) + red(i)*sqrt((sqrt(2)-1)/2)}}} and {{{-sqrt((1+sqrt(2))/2) - red(i)*sqrt((sqrt(2)-1)/2)}}}. <B>Example 7. Calculate</B> {{{sqrt(0)}}}. We have {{{a = 0}}}, {{{b=0}}}. Substitute values {{{a = 0}}}, {{{b=0}}} to the formulas for {{{w[1]}}}. You get {{{r[1] = sqrt((0 + sqrt(0^2 + 0^2))/2)}}} = {{{sqrt(0) = 0}}}, {{{s[1] = sqrt((sqrt(0^2 + 0^2) - 0)/2)}}} = {{{sqrt(0) = 0}}}. So, {{{w[1] = r[1] + red(i)*s[1] = 0 + red(i)*0 = 0}}}. The second complex square root is opposite to the first one: {{{w[2] = -0 = 0}}}. Both complex square roots of <B>0</B> are equal to <B>0</B>. This is the only case when two values of the complex square roots merge to one complex number. 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>. - <A HREF=http://www.algebra.com/algebra/homework/complex/Complex-numbers-and-arithmetical-operations.lesson> Complex numbers and arithmetic operations on them</A> - <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> - <B>How to take a square root of a complex number</B> (this lesson) - <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.
