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We call a the real part of the complex number c and refer to b as the imaginary part of c The numbers a and b are ordinary real numbers Now let c = a + ib and k = m + in be two complex numbers Here m and n are also two arbitrary real numbers (not integers, we use m and n because I am running out of symbols to use) We can form the sum and difference of two complex numbers by adding (subtracting) their real and imaginary parts independently That is: c + k = a + ib + m + in = ( a + m ) + i ( b + n ) c k = a + ib ( m + in ) = (a m ) + i ( b n ) a To multiply two complex numbers, we just multiply out the real and imaginary parts term by term and use i 2 = 1, then group real and imaginary parts at the end: ck = ( a + ib )( m + in ) = am + ian + ibm + i 2 bn = am + ian + ibm bn = ( am bn ) + i ( an + bm ) To divide two complex numbers and write the result in the form c = a + ib, we re going to need a new concept, called the complex conjugate We form the complex conjugate of any complex number by letting i i The complex conjugate is indicated by putting a bar on top of the number or variable Again, let c = a + ib Then the complex conjugate is c = a ib (13)

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CHAPTER 15:

It s easy to see that if c is purely real, that is, c = a, then the complex conjugate is c = a = a On the other hand, if c is purely imaginary, then c = ib This means that c = ib = ib = c Taking the complex conjugate twice gives back the original number: c = a ib = a + ib = c

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Menus are the navigational controls for your movies Menus can be linked to video clips, to additional menus, and to chapter points within video clips By creating links to chapter points you make it possible for viewers to jump to specific points within a video clip not just to the beginning of the clip

Taking the limit as y 0, once again we obtain partial derivatives This time, however, they are with respect to y: u( x 0 , y0 + y) u( x 0 , y0 ) [ v ( x 0 , y0 + y) v ( x 0 , y0 )] w + lim = lim y 0 z y 0 y 0 i y y lim = i u v + y y

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CHAPTER 13:

This must agree with our previous result It can only do so if the real and imaginary parts of both limits are equal Imposing this condition on the real part of each limit we obtain the rst of the Cauchy-Riemann equations: u v = x y Equating imaginary parts, we nd the second of the Cauchy-Riemann equations: v u = x y EXAMPLE 36 Is the function f ( z ) = z 2 analytic SOLUTION Writing the function in terms of real and imaginary parts, we have f ( z ) = z 2 = ( x + iy)( x + iy) = x 2 y 2 + i 2 xy Hence, in this case u ( x , y) = x 2 y 2 v ( x , y) = 2 xy

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Now let s compute the relevant partial derivatives We nd u = 2x x v = 2y x u = 2 y y v = 2x y

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The menus you create in Sonic DVDit! are far more flexible than the menus you can create in any of the template-based editors we ve looked at earlier In Sonic DVDit! you have complete control over the layouts of your menus, and you can add up to 36 buttons on a single menu if you like (but I don t suggest that you go quite so far overboard, since too many buttons on a single menu can make navigation more rather than less difficult) To create a menu, follow these steps:

The Cauchy-Riemann equations are satis ed, so we conclude the function is analytic EXAMPLE 37 2 Is f ( z ) = z analytic SOLUTION This time the situation is a little bit different We will see that the function is only differentiable at the origin Again, we write the function in terms of real and imaginary parts: f ( z ) = z = zz = ( x + iy)( x iy) = x 2 + y 2

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