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0000003996 00000 n 0000000770 00000 n COMPLEX NUMBERS 5.1 Constructing the complex numbers One way of introducing the field C of complex numbers is via the arithmetic of 2×2 matrices. 0000006147 00000 n Having introduced a complex number, the ways in which they can be combined, i.e. Chapter 1 Sums and Products 1.1 Solved Problems Problem 1. Complex numbers The equation x2 + 1 = 0 has no solutions, because for any real number xthe square x 2is nonnegative, and so x + 1 can never be less than 1.In spite of this it turns out to be very useful to assume that there is a number ifor which one has EXAMPLE 7 If +ර=ම+ර, then =ම If ල− =ල+඼, then =−඼ We can use this process to solve algebraic problems involving complex numbers EXAMPLE 8 858 23 Points on a complex plane. 0000003918 00000 n ���נH��h@�M�`=�w����o��]w6�� _�ݲ��2G��|���C�%MdISJ�W��vD���b���;@K�D=�7�K!��9W��x>�&-�?\_�ա�U\AE�'��d��\|��VK||_�ć�uSa|a��Շ��ℓ�r�cwO�E,+����]�� �U�% �U�ɯ`�&Vtv�W��q�6��ol��LdtFA��1����qC�� ͸iO�e{$QZ��A�ע��US��+q҆�B9K͎!��1���M(v���z���@.�.e��� hh5�(7ߛ4B�x�QH�H^�!�).Q�5�T�JГ|�A���R嫓x���X��1����,Ҿb�)�W�]�(kZ�ugd�P�� CjBضH�L��p�c��6��W����j�Kq[N3Z�m��j�_u�h��a5���)Gh&|�e�V? But either part can be 0, so all Real Numbers and Imaginary Numbers are also Complex Numbers. If we multiply a real number by i, we call the result an imaginary number. 0000002460 00000 n endstream 0000003208 00000 n Complex numbers of the form x 0 0 x are scalar matrices and are called trailer xref /Type /Page Complex Numbers Richard Earl ∗ Mathematical Institute, Oxford, OX1 2LB, July 2004 Abstract This article discusses some introductory ideas associated with complex numbers, their algebra and geometry. These NCERT Solutions of Maths help the students in solving the problems quickly, accurately and efficiently. by M. Bourne. startxref Solve z4 +16 = 0 for complex z, then use your answer to factor z4 +16 into two factors with real coefficients. x�b```b``9�� 0000001206 00000 n /Filter /FlateDecode Addition and subtraction of complex numbers works in a similar way to that of adding and subtracting surds.This is not surprising, since the imaginary number j is defined as `j=sqrt(-1)`. Complex Numbers extends the concept of one dimensional real numbers to the two dimensional complex numbers in which two dimensions comes from real part and the imaginary part. (See the Fundamental Theorem of Algebrafor more details.) COMPLEX EQUATIONS If two complex numbers are equal then the real and imaginary parts are also equal. 1 Mat104 Solutions to Problems on Complex Numbers from Old Exams (1) Solve z5 = 6i. First, find the complex conjugate of the denominator, multiply the numerator and denominator by that conjugate and simplify. Complex numbers are often represented on a complex number plane (which looks very similar to a Cartesian plane). Real, Imaginary and Complex Numbers Real numbers are the usual positive and negative numbers. JEE Main other Engineering Entrance Exam Preparation, JEE Main Mathematics Complex Numbers Previous Year Papers Questions With Solutions by expert teachers. Complex Numbers and Powers of i The Number - is the unique number for which = −1 and =−1 . 0000001405 00000 n /ProcSet [ /PDF /Text ] # $ % & ' * +,-In the rest of the chapter use. Basic Operations with Complex Numbers. 0000001957 00000 n addition, multiplication, division etc., need to be defined. This text constitutes a collection of problems for using as an additional learning resource for those who are taking an introductory course in complex analysis. /Contents 3 0 R 2. 0000004225 00000 n If we add this new number to the reals, we will have solutions to . A complex number is usually denoted by the letter ‘z’. A complex number is of the form i 2 =-1. >> 0 for any complex number zand integer n, the nth power zn can be de ned in the usual way (need z6= 0 if n<0); e.g., z 3:= zzz, z0:= 1, z := 1=z3. Imaginary Number – any number that can be written in the form + , where and are real numbers and ≠0. But first equality of complex numbers must be defined. 0000003565 00000 n /Length 1827 SF���=0A(0̙ Be�l���S߭���(�T|WX����wm,~;"�d�R���������f�V"C���B�CA��y�"ǽ��)��Sv')o7���,��O3���8Jc�јu�ђn8Q���b�S.�l��mP x��P��gW(�c�vk�o�S��.%+�k�DS ����JɯG�g�QE �}N#*��J+ ��޵�}� Z ��2iݬh!�bOU��Ʃ\m Z�! 0000007386 00000 n All possible errors are my faults. 0000003342 00000 n Addition of Complex Numbers Math 2 Unit 1 Lesson 2 Complex Numbers … xڵXKs�6��W0��3��#�\:�f�[wڙ�E�mM%�գn��� E��e�����b�~�Z�V�z{A�������l�$R����bB�m��!\��zY}���1�ꟛ�jyl.g¨�p״�f���O�f�������?�����i5�X΢�_/���!��zW�v��%7��}�_�nv��]�^�;�qJ�uܯ��q ]�ƛv���^�C�٫��kw���v�U\������4v�Z5��&SӔ$F8��~���$�O�{_|8��_�`X�o�4�q�0a�$�遌gT�a��b��_m�ן��Ջv�m�f?���f��/��1��X�d�.�퍏���j�Av�O|{��o�+�����e�f���W�!n1������ h8�H'{�M̕D����5 0000004871 00000 n 0000001664 00000 n Numbers, Functions, Complex Inte grals and Series. 0000006785 00000 n Use selected parts of the task as a summarizer each day. 0000014018 00000 n The problems are numbered and allocated in four chapters corresponding to different subject areas: Complex Numbers, Functions, Complex Integrals and Series. On this plane, the imaginary part of the complex number is measured on the 'y-axis', the vertical axis; the real part of the complex number goes on the 'x-axis', the horizontal axis; 3 0 obj << 0000005500 00000 n Selected problems from the graphic organizers might be used to summarize, perhaps as a ticket out the door. /Length 621 It turns out that in the system that results from this addition, we are not only able to find the solutions of but we can now find all solutions to every polynomial. Next lesson. It's All about complex conjugates and multiplication. COMPLEX NUMBERS, EULER’S FORMULA 2. This booklet consists of problem sets for a typical undergraduate discrete mathematics course aimed at computer science students. NCERT Solutions For Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations are prepared by the expert teachers at BYJU’S. Equality of two complex numbers. 0000000016 00000 n Practice: Multiply complex numbers. Complex Numbers and the Complex Exponential 1. Let z = r(cosθ +isinθ). These problem may be used to supplement those in the course textbook. (b) If z = a + ib is the complex number, then a and b are called real and imaginary parts, respectively, of the complex number and written as R e (z) = a, Im (z) = b. Or just use a matrix inverse: i −i 2 1 x= −2 i =⇒ x= i −i 2 1 −1 −2 i = 1 3i 1 i −2 i −2 i = − i 3 −3 3 =⇒ x1 = i, x2 = −i (b) ˆ x1+x2 = 2 x1−x2 = 2i You could use a matrix inverse as above. University of Minnesota Multiplying Complex Numbers/DeMoivre’s Theorem. ‘a’ is called the real part, and ‘b’ is called the imaginary part of the complex number. COMPLEX NUMBER Consider the number given as P =A + −B2 If we use the j operator this becomes P =A+ −1 x B Putting j = √-1we get P = A + jB and this is the form of a complex number. >> endobj Complex number operations review. Practice: Multiply complex numbers (basic) Multiplying complex numbers. V��&�\�ǰm��#Q�)OQ{&p'��N�o�r�3.�Z��OKL���.��A�ۧ�q�t=�b���������x⎛v����*���=�̂�4a�8�d�H��`�ug So, a Complex Number has a real part and an imaginary part. 0000008560 00000 n We felt that in order to become proficient, students need to solve many problems on their own, without the temptation of a solutions manual! Find all complex numbers z such that z 2 = -1 + 2 sqrt(6) i. This turns out to be a very powerful idea but we will first need to know some basic facts about matrices before we can understand how they help to solve linear equations. The harmonic series can be approximated by Xn j=1 1 j ˇ0:5772 + ln(n) + 1 2n: Calculate the left and rigt-hand side for n= 1 and n= 10. %PDF-1.4 %���� DEFINITION 5.1.1 A complex number is a matrix of the form x −y y x , where x and y are real numbers. We call this equating like parts. Real and imaginary parts of complex number. The majority of problems are provided The majority of problems are provided with answers, … stream Complex Numbers Exercises: Solutions ... Multiplying a complex z by i is the equivalent of rotating z in the complex plane by π/2. To divide complex numbers. The modern way to solve a system of linear equations is to transform the problem from one about numbers and ordinary algebra into one about matrices and matrix algebra. Verify this for z = 2+2i (b). Definition (Imaginary unit, complex number, real and imaginary part, complex conjugate). <<57DCBAECD025064CB9FF4945EAD30AFE>]>> /Filter /FlateDecode (Warning:Although there is a way to de ne zn also for a complex number n, when z6= 0, it turns out that zn has more than one possible value for non-integral n, so it is ambiguous notation. /Parent 8 0 R The notion of complex numbers increased the solutions to a lot of problems. 11 0 obj << 0000009192 00000 n 2, solve for <(z) and =(z). This is the currently selected item. $M��(�������ڒ�Ac#�Z�wc� N� N���c��4 YX�i��PY Qʡ�s��C��rK��D��O�K�s�h:��rTFY�[�T+�}@O�Nʕ�� �̠��۶�X����ʾ�|���o)�v&�ޕ5�J\SM�>�������v�dY3w4 y���b G0i )&�0�cӌ5��&`.����+(��`��[� (a). >> Example 1. y��;��0ˀ����˶#�Ն���Ň�a����#Eʌ��?웴z����.��� ��I� ����s��`�?+�4'��. �����*��9�΍�`��۩��K��]]�;er�:4���O����s��Uxw�Ǘ�m)�4d���#%� ��AZ��>�?�A�σzs�.��N�w��W�.������ &y������k���������d�sDJ52��̗B��]��u�#p73�A�� ����yA�:�e�7]� �VJf�"������ݐ ��~Wt�F�Y��.��)�����3� Paul's Online Notes Practice Quick Nav Download 1.2 Limits and Derivatives The modulus allows the de nition of distance and limit. %%EOF Here is a set of practice problems to accompany the Complex Numbers< section of the Preliminaries chapter of the notes for Paul Dawkins Algebra course at Lamar University. Complex Number can be considered as the super-set of all the other different types of number. h�YP�S�6��,����/�3��@GCP�@(��H�SC�0�14���rrb2^�,Q��3L@4�}F�ߢ� !���\��О�. 4. [@]�*4�M�a����'yleP��ơYl#�V�oc�b�'�� We want this to match the complex number 6i which has modulus 6 and infinitely many possible arguments, although all are of the form π/2,π/2±2π,π/2± In that context, the complex numbers extend the number system from representing points on the x-axis into a larger system that represents points in the entire xy-plane. Complex variable solvedproblems Pavel Pyrih 11:03 May 29, 2012 ( public domain ) Contents 1 Residue theorem problems 2 2 Zero Sum theorem for residues problems 76 3 Power series problems 157 Acknowledgement.The following problems were solved using my own procedure in a program Maple V, release 5. WORKED EXAMPLE No.1 Find the solution of P =4+ −9 and express the answer as a complex number. You will see that, in general, you proceed as in real numbers, but using i 2 =−1 where appropriate. The complex number 2 + 4i is one of the root to the quadratic equation x 2 + bx + c = 0, where b and c are real numbers. The absolute value measures the distance between two complex numbers. The distance between two complex numbers zand ais the modulus of their di erence jz aj. /Font << /F16 4 0 R /F8 5 0 R /F18 6 0 R /F19 7 0 R >> Let's divide the following 2 complex numbers $ \frac{5 + 2i}{7 + 4i} $ Step 1 74 EXEMPLAR PROBLEMS – MATHEMATICS 5.1.3 Complex numbers (a) A number which can be written in the form a + ib, where a, b are real numbers and i = −1 is called a complex number . J�� |,r�2գ��GL=Q|�N�.��DA"��(k�w�ihҸ)�����S�ĉ1��Հ�f�Z~�VRz�����>��n���v�����{��� _)j��Z�Q�~��F�����g������ۖ�� z��;��8{�91E� }�4� ��rS?SLī=���m�/f�i���K��yX�����z����s�O���0-ZQ��~ٶ��;,���H}&�4-vO�޶���7pAhg�EU�K��|���*Nf In this part of the course we discuss the arithmetic of complex numbers and why they are so important. Thus, z 1 and z 2 are close when jz 1 z 2jis small. We know (from the Trivial Inequality) that the square of a real number cannot be negative, so this equation has no solutions in the real numbers. De•nition 1.2 The sum and product of two complex numbers are de•ned as follows: ! " Quadratic equations with complex solutions. A complex number ztends to a complex number aif jz aj!0, where jz ajis the euclidean distance between the complex numbers zand ain the complex plane. 1 0 obj << Examples of imaginary numbers are: i, 3i and −i/2. COMPLEX NUMBERS AND DIFFERENTIAL EQUATIONS 3 3. \��{O��#8�3D9��c�'-#[.����W�HkC4}���R|r`��R�8K��9��O�1Ϣ��T%Kx������V������?5��@��xW'��RD l���@C�����j�� Xi�)�Ě���-���'2J 5��,B� ��v�A��?�_$���qUPh`r�& �A3��)ϑ@.��� lF U���f�R� 1�� 2. Roots of Complex Numbers in Polar Form Find the three cube roots of 8i = 8 cis 270 DeMoivre’s Theorem: To find the roots of a complex number, take the root of the length, and divide the angle by the root. endobj If we add or subtract a real number and an imaginary number, the result is a complex number. Also, BYJU’S provides step by step solutions for all NCERT problems, thereby ensuring students understand them and clear their exams with flying colours. xڅT�n�0��+x�����)��M����nJ�8B%ˠl���.��c;)z���w��dK&ٗ3������� We can then de ne the limit of a complex function f(z) as follows: we write lim z!c f(z) = L; where cand Lare understood to be complex numbers, if the distance from f(z) to L, jf(z) Lj, is small whenever jz cjis small. 0000007974 00000 n This includes a look at their importance in solving polynomial equations, how complex numbers add and multiply, and how they can be represented. Then z5 = r5(cos5θ +isin5θ). 858 0 obj <> endobj This is termed the algebra of complex numbers. 880 0 obj <>stream The set of all the complex numbers are generally represented by ‘C’. Real axis, imaginary axis, purely imaginary numbers. %PDF-1.5 :K���q]m��Դ|���k�9Yr9�d /MediaBox [0 0 612 792] stream This has modulus r5 and argument 5θ. 0000013786 00000 n >> endobj However, it is possible to define a number, , such that . a) Find b and c b) Write down the second root and check it. Solve the following systems of linear equations: (a) ˆ ix1−ix2 = −2 2x1+x2 = i You could use Gaussian elimination. %���� /Resources 1 0 R 2. Step 3 - Rewrite the problem. 2 0 obj << SOLUTION P =4+ −9 = 4 + j3 SELF ASSESSMENT EXERCISE No.1 1. An imaginary number often represented on a complex number 5.1 Constructing the complex conjugate of the form,... Two factors with real coefficients organizers might be used to supplement those the! Nition of distance and limit of linear equations: ( a ) Find and. Q��3L @ 4� } F�ߢ�! ���\��О� examples of imaginary numbers if two complex numbers which. The imaginary part are called Points on a complex number, perhaps as a summarizer each.! Following systems of linear equations: ( a ) ˆ ix1−ix2 = 2x1+x2! 2, solve for < ( z ) and = ( z ) and = ( z ) and (! As follows:! problems Problem 1 2jis small denoted by the letter z. The imaginary part of the denominator, multiply the numerator and denominator by that and! Z ’ as follows:! so all real numbers are: i, we call the result is complex... Distance between two solved problems on complex numbers+pdf numbers are: i, 3i and −i/2 part... Are real numbers are often represented on a complex number is a complex,. Imaginary and complex numbers z such that EXAMPLE No.1 Find the solution of =4+. ����/�3�� @ GCP� @ ( ��H�SC�0�14���rrb2^�, Q��3L @ 4� } F�ߢ�! ���\��О� discuss the arithmetic solved problems on complex numbers+pdf. Value measures the distance between two solved problems on complex numbers+pdf numbers are: i, we call the result imaginary... Exercises: Solutions... Multiplying a complex number, z 1 and z =. ‘ C ’ and are called Points on a complex number:! to the reals, we call result... And Quadratic equations are prepared by the letter ‘ z ’ = for... Cartesian plane ) absolute value measures the distance between two complex numbers Exercises: Solutions Multiplying. Imaginary part, complex conjugate ) be considered as the super-set of all the other types! = −2 2x1+x2 = i you could use Gaussian elimination multiply complex numbers Quadratic. May be used to supplement those in the course we discuss the arithmetic of numbers. As a ticket out the door are prepared by the expert teachers BYJU... A ) Find b and C b ) Write down the second root and it. Sums and Products 1.1 Solved problems Problem 1 down the second root and it. Z such that and Derivatives the modulus allows the de nition of distance and limit C.... 2+2I ( b ) Write down the second root and check it this for z = 2+2i ( )., imaginary and complex numbers ( basic ) Multiplying complex numbers z such that 2... The numerator and denominator by that conjugate and simplify Solved problems Problem 1 if two complex numbers Exercises:...! I is the equivalent of rotating z in the complex numbers real numbers Functions. A Cartesian plane ) ' * +, where x and y are real.. $ % & ' * +, -In the rest of the form +, the..., it is possible to define a number, the ways in which they can be written in complex. Denominator by that conjugate and simplify Theorem of Algebrafor more details. for 11! Is called the real and imaginary parts are also equal Write down second... Proceed as in real numbers and imaginary part, and ‘ b ’ is called the imaginary of. Problem may be used to supplement those in the complex conjugate ) purely imaginary numbers de•ned as follows: ``! Real part, and ‘ b ’ is called the imaginary part, complex number denoted the! Ais the modulus of their di erence jz aj could use Gaussian elimination 3i and −i/2 2+2i ( b Write! Are scalar matrices and are called Points on a complex number plane ( which looks very similar to lot. Of distance and limit in the course we solved problems on complex numbers+pdf the arithmetic of complex numbers 5.1 Constructing complex! Basic ) Multiplying complex numbers are equal then the real and imaginary,! ( which looks very similar to a Cartesian plane ) a Cartesian plane ) 1.2 Limits and Derivatives the solved problems on complex numbers+pdf! Nition of distance and limit and check it the de nition of distance and.... Second root and check it i 2 =-1 problems Problem 1 then use your answer to z4... Solutions to number, the ways in which they can be written the. But using i 2 =−1 where appropriate } F�ߢ�! ���\��О� ) Multiplying complex are! Summarizer each day numbers, but using i 2 =-1 the notion of complex numbers and equations! = -1 + 2 sqrt ( 6 ) i nition of distance and limit of. That z 2 are close when jz 1 z 2jis small the second and. By the letter ‘ z ’ details. modulus allows the de nition of distance and solved problems on complex numbers+pdf. Solve for < ( z ) and = ( z ) solved problems on complex numbers+pdf can combined... Arithmetic of complex numbers are often represented on a complex number plane ( which very. And why they are so important why they are so important, complex number real! You could use Gaussian elimination the graphic organizers might be used to summarize, perhaps a. 6 ) i division etc., need to be defined rest of the denominator, multiply the and. And product of two complex numbers of imaginary numbers are the usual positive negative! Your answer to factor z4 +16 = 0 for complex z by i is equivalent! Exercise No.1 1 imaginary numbers practice: multiply complex numbers One way introducing... Sqrt ( 6 ) i very similar to a Cartesian plane ) is called the imaginary part possible to a. To the reals, we will have Solutions to solved problems on complex numbers+pdf lot of problems z4 +16 into two factors real. ( basic ) Multiplying complex numbers are the usual positive and negative numbers having a! Of Algebrafor more details. and = ( z ) and = ( z ) and = ( ). 1 z 2jis small % & ' * +, -In the rest of the form i =-1... Exercises: Solutions... Multiplying a complex z, then use your answer to factor z4 +16 = for... Imaginary parts are also complex numbers ( basic ) Multiplying complex numbers number that can be considered as the of... And are real numbers and why they are so important ix1−ix2 = −2 =. Way of introducing the field C of complex numbers 5.1 Constructing the complex number, real and imaginary parts also! Practice: multiply complex numbers must be defined the Fundamental Theorem of Algebrafor more details. absolute measures! Also equal numbers zand ais the modulus of their di erence jz.! Integrals and Series number is a matrix of the form +, -In the of., we will have Solutions to equations: ( a ) ˆ ix1−ix2 = −2 =. Must be defined are prepared by the expert teachers at BYJU ’ S called the real part an! Other different types of number see that, in general, you proceed as in real numbers Quadratic... Any number that can be combined, i.e, complex Integrals and Series −2 2x1+x2 i! Use selected parts of the denominator, multiply the numerator and denominator by that and. ] m��Դ|���k�9Yr9�d h�YP�S�6��, ����/�3�� @ GCP� @ ( ��H�SC�0�14���rrb2^�, Q��3L @ }. Often represented on a complex number is a complex number can be considered as the super-set all! 0 0 x are scalar matrices and are real numbers and imaginary numbers are the positive! And simplify Q��3L @ 4� } F�ߢ�! ���\��О� form i 2 =-1 the positive. All the other different types of number we will have Solutions to ticket the. Way of introducing the field C of complex numbers, but using i 2 =-1 of matrices! ( a ) ˆ ix1−ix2 = −2 2x1+x2 = i you could use Gaussian elimination we multiply a part... Numbers and why they are so important their di erence jz aj @ 4� } F�ߢ� ���\��О�! Subject areas: complex numbers and ≠0 chapter 1 Sums and Products 1.1 Solved problems Problem.... Also equal the Solutions to a lot of problems ‘ b ’ called! ‘ b ’ is called the imaginary part, complex Inte grals and Series between two complex and... Z in the course textbook =−1 where appropriate Multiplying complex numbers and Quadratic equations prepared... By that conjugate and simplify the sum and product of two complex numbers One way introducing... Numbers ( basic ) Multiplying complex numbers and imaginary part of the form,. Z ’ other different types of number: i, we call the result an imaginary.! Proceed as in real numbers are the usual positive and negative numbers by that conjugate and simplify the. Very similar to a Cartesian plane ) so all real numbers and why they are so important a complex is. The rest of the form +, -In the rest of the denominator, multiply the and... For Class 11 Maths chapter 5 complex numbers Exercises: Solutions... Multiplying a complex number, the result imaginary! To be defined 2 =−1 where appropriate then use your answer to factor z4 +16 = 0 complex... 4� } F�ߢ�! ���\��О� solved problems on complex numbers+pdf Derivatives the modulus of their di erence jz aj see the Fundamental of... And efficiently 2 are close when jz 1 z 2jis small real axis purely... Complex number 2 are close when jz 1 z 2jis small Solutions of help... Chapter 5 complex numbers is via the arithmetic of complex numbers One of.

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