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**Lecture 2: Lines and circles on the complex plane; complex multiplication and division**

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Key Points Two complex numbers can be multiplied by expressing each number in the form z = x + jy, then using distributivity and the rule j2 = -1 (i.e., j is treated as the square root of -1). The product of two complex numbers with polar coordinates (r1, θ1) and (r2, θ2) is the complex number with polar coordinates (r1r2, θ1 + θ2). The product of a complex number z and its conjugate z* equals the square modulus |z|2. If z has polar coordinates (r, θ), its inverse z-1 has polar coordinates (r-1; -θ).

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Equations and Curves The modulus |z| represents the length of the vector z. Similarly, the expression |z1-z2| gives the distance between the points z1 and z2. Using this concept, we see that the following two equations in the variable z each have a familiar geometrical interpretation: |z – z0|= a, where a is a positive constant, is the set of points z on the complex plane which are at a fixed distance, a, from the point z0. This is a circle of radius a centered at z0. |z - z1| = |z - z2| is the set of points z on the complex plane which are equidistant from the points z1 and z2. This is the same as the perpendicular bisector of the line segment joining z1 and z2. Your task: Sketch the two curves given by the equations |z-3 + 4j| = 5 and |z – 3|= |z + 4j|

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**Multiplying Two Complex Numbers – Cartesian Form**

The most common form for a complex number z incorporates the real and imaginary parts as follows: z = x + jy This form, together with the convention that j x j = j2 = -1, allows us to multiply two complex numbers together. For example, (5 - 2j)(3 - 4j) = j - 6j + 8j2 = j

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**Multiplying Two Complex Numbers – Polar Form**

In polar form, multiplication of complex numbers is simple. If z1 = r1(cos θ1 + j sin θ1) and z2 = r2(cos θ2 + j sin θ2), then z1z2 = r1r2(cos θ1 + j sin θ1)(cos θ2 + j sin θ2) = r1r2[(cos θ1 cos θ2 - sin θ1 sin θ2) + j(cos θ1 sin θ2 + sin θ1 cos θ2)] But cos θ1 cos θ2 - sin θ1 sin θ2 = cos(θ1 + θ2) cos θ1 sin θ2 + sin θ1 cos θ2 = sin(θ1 + θ2) and thus z = z1z2 has modulus equal to the product of the two moduli, and angle equal to the sum of the two angles.

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**Example Let z1 = 5 - 2j and z2 = 3 – 4j The polar forms are**

|z1| = 29; z1 = and |z2| = 5; z2 = (The angles were obtained using the MATLAB ANGLE function, which returns values between - and .) Therefore |z1z2|= 529; z1z2 = – = and, in Cartesian form z1z2 = 529 · (cos( ) + j sin( )) = 7 - j26

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**z15z22 =( 1.1322 105) (cos(-3.7571) + j sin(-3.7571))**

Extension Powers of z1 and z2, and products thereof, can be also computed easily in polar form. Thus |z15z22| = 295/2 · 52 = 105; z15z22 = 5 · ( ) + 2 · ( ) = and z15z22 =( 105) (cos( ) + j sin( )) j65370 where the real and imaginary parts were rounded to the nearest integers. The calculation was deliberately carried out using low precision (four or five digits) in the intermediate results, in order to illustrate the accumulation of roundoff errors. The correct result is j65374.

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**Division of Two Complex Numbers - Cartesian**

Division of two complex numbers makes implicit use of the inverse, i.e., z1/z2 = z1(1/z2) = z1z2-1 This can be carried out in a single step by multiplying both the numerator z1 and the denominator z2 by the complex conjugate of z2 (which differs from z2 in the sign of the imaginary part). Thus If z1 = 5-2j and z2 = 3-j, then Here we had an instance of the identity, zz* = |z|2 i.e., the product of a complex number and its conjugate equals the (real valued) square magnitude of that number.

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**Division of Two Complex Numbers - Polar**

In polar form, dividing two complex numbers corresponds to dividing their magnitudes and subtracting one angle from the other. This is because z-1 = z*/|z|2 and therefore |z-1| = |z|/|z|2 = |z|-1; z-1 = z* = - z Your task: Verify the above results. Plot z, z* and z-1 for z = 3 - 4j.

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SAT II - Math Level 2 Test #03 Solution. 1.If the graph of 3x + y + 5 = 0 is perpendicular to the graph of ax – 2y + 9 = 0, then a equals (A) 3/2 (B)

SAT II - Math Level 2 Test #03 Solution. 1.If the graph of 3x + y + 5 = 0 is perpendicular to the graph of ax – 2y + 9 = 0, then a equals (A) 3/2 (B)

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