Complex Numbers Problems

Date: 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/