Complex Numbers ProblemsDate: 12/23/97 at 06:02:30 From: abecina Subject: A complex problem Hello Dr math . I am having difficulty with my complex numbers ...please help ? 1. On an Argand diagram the points P and Q represent the numbers z1 and z2 respectively. OPQ is an equilateral triangle. Show that z12+z22 = z1z2 2. Show that | |z1|-|z2| | < |z1+z2|. State the condition for equality to hold. 3. Show that |z1+z2+....zn|<|z1|+|z2|+....+|zn| 4. On an Argand diagram the points A and B represent the numbers z1 and z2 respectively. I is the point {1,0}. D is the point such that triangle OID is similar to triangle OBA. Show that D represents z1/z2. Thanks a lot. Mike Date: 12/29/97 at 16:33:54 From: Doctor Anthony Subject: Re: A Complex problem > 1. On an Argand diagram the points P and Q represent the numbers z1 and z2 respectively. OPQ is an equilateral triangle. Show that z12+z22 = z1z2 Draw a diagram and consider the triangle OPQ. Let OP = OQ = PQ = r and the angle between OP and OQ = pi/3 I shall be using the form z = r.e^(i.theta) for a general complex number. If you are not familiar with this you can use the form z = r(cos(theta) + i.sin(theta)) instead, going through the same steps as shown below. Let z1 = r.e^(i.theta) then z2 = r.e^[i(theta+pi/3)] z1.z2 = r^2.e^[i(2.theta + pi/3)] ..........(1) And then: z1^2 + z2^2 = r^2.e^(i.2.theta) + r^2.e^[i(2.theta) + 2.pi/3)] = r^2.e^(i.2.theta)[1 + e^(i.2.pi/3)] = r^2.e^(i.2.theta)[1 + cos(2.pi/3) + i.sin(2pi/3)] = r^2.e^(i.2.theta)[1 - 1/2 + i.sqrt(3)/2] = r^2.e^(i.2.theta)[1/2 + i.sqrt(3)/2] = r^2.e^(i.2.theta)e^(i.pi/3) = r^2.e^[i(2.theta + pi/3)] ............(2) and comparing (1) and (2) we see that the two expressions are the same. So we have z1^2 + z2^2 = z1.z2 > 2. Show that | |z1|-|z2| | < |z1+z2|. State the condition for equality to hold. This can be shown very easily on the Argand diagram. Suppose |z1| > |z2|, but the argument can be shown just as easily if they are the other way around. Draw in the vector z1. Now from the point z1 draw in the vector z2, (z1 is nose to tail with z2). Then z1+z2 is the vector joining the origin to the end of the second vector z2. The distance from the origin to z1 by the direct route is ||z1|-|z2|| + |z2|, whereas going from the origin to z1 via the point (z1+z2) the total distance is |z1+z2| + |z2| and this will be two sides of a triangle and so greater than the direct route. It follows, comparing these, that ||z1|-|z2|| + |z2| < |z1+z2| + |z2| so ||z1|-|z2|| < |z1+z2| Equality arises if z1 and z2 are in a straight line with the origin and on opposite sides of the origin. > 3. Show that |z1+z2+....zn|<|z1|+|z2|+....+|zn| This is again proved with the 'triangle inequality' - two sides of a triangle are always greater than the third side. In this case we have a polygon which is formed by adding the n vectors nose to tail. The lefthand side is the magnitude of the straight line joining the origin to the end of the last vector, thereby closing the polygon. The righthand side is the sum of all the individual sides of the polygon, (except the closing line) and is clearly greater than the single closing line to the end of the last vector. > 4. On an Argand diagram the points A & B reprsent the numbers z1 and z2 respectively. I is the point {1,0}. D is the point such that triangle OID is similar to triangle OBA. Show that D represents z1/z2. Draw the diagram with OA of length r1 and argument alpha, and draw z2 with modulus r2 and argument (alpha+theta). When dividing complex numbers we divide the moduli and subtract the arguments. So mod(z1/z2) = r1/r2 arg(z1/z2) = alpha - (alpha+theta) = -theta Now OD/1 = r1/r2 and the argument of OD is -theta. It follows that OD represents the complex number z1/z2. -Doctor Anthony, The Math Forum Check out our web site! http://mathforum.org/dr.math/ Date: 12/02/97 at 06:44:20 From: Abecina Subject: More complex numbers Please help me solve the following ? z E ( that "epsilon" symbol ......sort of ) C such that Re z = 2Im z and z2-4i is real. Find z Thank you so much once again! Mike Date: 12/02/97 at 08:03:06 From: Doctor Jerry Subject: Re: More complex numbers Hi Mike, I'll try to help you get started, but the idea of our service is not to "solve the question to the end." We can let z=x+i*y, where x and y are real. Re(z)=x and Im(z)=y. So, Re(z)=2*Im(z) means that x=2y. Also, you say tht z^2-4i is real. All you have to do is to form z^2 -4i and look at the imaginary part. Set it equal to zero. Now you have two equations in two unknowns. -Doctor Jerry, The Math Forum Check out our web site! http://mathforum.org/dr.math/ Date: 12/02/97 at 16:59:57 From: Doctor Anthony Subject: Re: More complex numbers The first expression you mention is simply stating that z is a complex number. Suppose we put z = x + iy then x is the real part and y the imaginary part. We are told that x = 2y and z^2 - 4i is real x^2 - y^2 + 2ixy - 4i is real. This requires 2xy = 4 So we have x = 2y and xy = 2 2y^2 = 2 and so y^2 = 1 y = +1 or -1 and then x = 2 or -2 Two possible values of z are 2 + i and -2 - i -Doctor Anthony, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994- The Math Forum at NCTM. All rights reserved.
http://mathforum.org/dr.math/