Associated Topics || Dr. Math Home || Search Dr. Math

### 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

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/
```
Associated Topics:
High School Imaginary/Complex Numbers

Search the Dr. Math Library:

 Find items containing (put spaces between keywords):   Click only once for faster results: [ Choose "whole words" when searching for a word like age.] all keywords, in any order at least one, that exact phrase parts of words whole words

Submit your own question to Dr. Math
Math Forum Home || Math Library || Quick Reference || Math Forum Search