Lesson Upper level problems on complex numbers
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<H2>Upper level problems on complex numbers</H2> <H3>Problem 1</H3>If a complex number z satisfies equation |(z)/(|conj(z)|) -conj(z) | = 1 + | z |, then prove that z is pure imaginary. <B>Solution</B> <pre> The equation in the post is | z | | ---------- - conj(z) | = 1 + |z| (1) | |conj(z)| | Vertical lines denote absolute value, or the modulus of complex numbers. |conj(z)| is the same as |z|. z z So, ---------- is the same as ----, which is the unit vector, co-directed with z. |conj(z)| |z| So, equation (1) tells us that a unit vector, co-directed with z, MINUS conj(z) has the absolute value (the modulus) of 1+|z|. (2) Taking into account the rule of adding/subtracting complex numbers as vectors, we can interpret the statement (2) THIS WAY : we have a triangle with three sides: side "a" is the unit vector, co-directed with z; side "b" is the vector -conj(z); side "c" has the length 1+|z|. Keeping in mind a triangle inequality, we can conclude that this triangle is degenerated and its sides "a", "b" and "c" all are collinear. Namely, +-------------------------------------+ | -conj(z) is co-directed with z | +-------------------------------------+ which may happen IF and ONLY IF z is a purely imaginary complex number. It is the statement to prove, and at this point, the solution is COMPLETE and the problem' statement is proven. </pre> <H3>Problem 2</H3>If z = ({{{1}}},{{{theta}}}), find {{{(z + z^10)/(z - z^10)}}} in polar form ({{{r}}},{{{alpha}}}). <B>Solution</B> <pre> In cis-form, z = {{{cis(1,theta)}}}. It means that the modulus of z is 1 and the argument of z is {{{theta}}}. Then {{{z^10}}} = {{{cis(1,10*theta)}}}. In other words, {{{z^10}}} has the modulus 1 and the argument of {{{10*theta}}}. To find the sum z + {{{z^10}}}, you should use the parallelogram rule of adding complex numbers as vectors. Since both addends, z and {{{z^10}}} have equal modulus, the parallelogram rule in this case becomes the rhombus rule. In rhombus, the diagonal is the bisector of the angle. THEREFORE, in our case the argument of {{{z+z^10}}} is {{{(theta+10*theta)/2)}}} = {{{5.5*theta)}}}. The modulus of {{{z+z^10}}} is the third side of an isosceles triangle with the equal legs of the length 1 and the angle between these legs of {{{10*theta - theta}}} = {{{9*theta}}} A = {{{sqrt(1^2 + 1^2 - 2*cos(9*theta))}}} = {{{sqrt(2-2cos(9*theta))}}}. So, {{{z + z^10}}} = {{{cis(A,5.5*theta)}}}. Similarly, for the denominator {{{z - z^10}}} = {{{cis(B,(theta-10*theta)/2)}}} = {{{cis(B,-4.5*theta))}}}, where B = {{{sqrt(2+2*cos(9*theta))}}}. <<<---=== Note that the angle between z and {{{-z^10}}} is {{{(10*theta+pi)-theta}}} = {{{9*theta+pi}}} and {{{cos(9*theta+pi)}}} = {{{-cos(9*theta)}}}. Thus the ratio {{{(z + z^(10))/(z - z^(10))}}} has the modulus {{{sqrt(2-2cos(9*theta))/sqrt(2+2cos(9*theta))}}} = {{{sqrt(1-cos(9*theta))/sqrt(1+cos(9*theta))}}} and the argument as the difference {{{5.5*theta - (-4.5*theta)}}} = {{{(5.5+4.5)*theta}}} = {{{10*theta}}}. Using formulas {{{sqrt(1-cos(a))/2}}} = {{{sin(a/2)}}}, {{{sqrt(1+cos(a))/2}}} = {{{cos(a/2)}}}, the modulus of the ratio can be simplified {{{sqrt(1-cos(9*theta))/sqrt(1+cos(9*theta))}}} = {{{abs(sin((9*theta)/2)/cos((9*theta)/2))}}} = {{{abs(sin(4.5*theta)/cos(4.5*theta))}}} Therefore, the ratio {{{(z + z^(10))/(z - z^(10))}}} is {{{cis(sqrt(1-cos(9*theta))/sqrt(1+cos(9*theta)),10*theta)}}}, or, which is the same, ({{{abs(sin(4.5*theta)/cos(4.5*theta))}}},{{{10*theta)}}}). <U>ANSWER</U>. In polar form, {{{(z + z^(10))/(z - z^(10))}}} is ({{{abs(sin(4.5*theta)/cos(4.5*theta))}}},{{{10*theta)}}}). </pre> Solved. The key ideas of solution are - use cis-form of complex numbers; - use the parallelogram rule to add and to subtract complex numbers; - use the fact that the parallelogram rule in the case of equal modulus becomes the rhombus rule; - use the fact that in rhombus a diagonal bisects the angle of the rhombus - to find the modulus of the numerator and denominator, use the cosine law. </pre> The rest is just arithmetic (quite tight, though). =========================== This problem is special. Its level is much high than the average high school Math; it is higher than teachers teach in Math schools, higher than average Math circles level, different from regular Math Olympiads. It is the level of Math Olympiads among undergraduate Math students of Math departments of renowned universities/colleges, like the Putnam competition. It does not require "flight of thought", but requires solid and firm knowledge of complex numbers in all relevant aspects and firm knowledge of relevant adjacent Math subjects, together with perfect and firm Math technique. My other lessons on complex numbers in this site are - <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> - <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 problem 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/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.
