Mathematics : Complex Numbers: Modulus of a Complex Number: Solved Example Problems with Answers, Solution

If *z*_{1} = 3 + 4*i*, *z*_{2} = 5 -12*i*, and *z*_{3} = 6 + 8*i*, find
|*z*_{1}| , |*z*_{2}|, |*z*_{3}|, |z_{1} + z_{2}| ,| z_{2} - z_{3}|, and |z_{1} + z_{3}|.

Using the given values for z_{1}, z_{2} and z_{3} we get |*z*_{1}*| = |3 + 4i| =* âˆš[3^{2} +4^{2}] =5

|z_{2}| = |5 -12*i*| = âˆš[5^{2} + (-12)^{2}] = 13

z_{3} = |6 + 8*i*| = âˆš[6^{2} + 8^{2}] = 10

|z_{1} + z_{2}| = |(3 + 4*i* ) + (5 -12*i*)| = |8 - 8*i*| =
âˆš128 = 8âˆš2

|z_{2} - z_{3}| = |(5 -12*i* ) - (6 + 8*i* )| = |-1- 20*i*| =
âˆš401

|z_{1} + z_{3}| = |(3 + 4*i*) + (6 + 8*i*)| = |9 +12*i*| =
âˆš225 = 15

Note that the triangle inequality is satisfied in all the cases.

|z_{1} + z_{3}| = |z_{1}| + |z_{3}|= 15 (why?)

**Example 2.10**

Find the following

**Solution**

**Example 2.11**

Which one of the points *i, *-2 + *i , *and 3 is
farthest from the origin?

**Solution**

The distance between origin to *z *= *i*, -2 + *i*,
and 3 are

| *z *| = | *i *| = 1

| *z *| = | -2 + *i *|= âˆš[(-2)^{2}+(1)^{2}] = âˆš5

= | z | = | 3 | = 3

Since 1 < âˆš5 < 3 , the farthest point from the origin is 3

**Example 2.12**

If *z*_{1}, *z*_{2}, and *z*_{3} are complex numbers such that |*z*_{1}| = |*z*_{2}| =|*z*_{3}| = |*z*_{1}* *+* z*_{2 }+* z*_{3}| = 1, find the value of .

**Solution**

Since, |z_{1}| = |z_{1}| = |z_{1}| = 1,

**Example 2.13**

If |z| = 2 show that 3 â‰¤ |z + 3 + 4i| â‰¤ 7

**Solution**

|z + 3 + 4i| â‰¤ |z| + |3 + 4*i*| = 2 + 5 = 7

|z + 3 + 4*i*| â‰¤ 7 (1)

|z + 3 + 4*i*| â‰¥ | |z| - | 3 + 4*i*| | = |2-5| = 3

|z + 3 + 4*i*| â‰¥ 3
(2)

From (1) and (2), we get 3 â‰¤ |z + 3 + 4*i*| â‰¤ 7

**Note**

To find the lower bound and upper bound use | |z_{1}| - |z_{2}| | â‰¤ |z_{1} + z_{2} | â‰¤ |z_{1}| + |z_{2}|.

**Example 2.14**

Show that the points 1, are the vertices of an equilateral triangle.

**Solution**

It is enough to prove that the sides of the triangle are equal.

Let z = 1,

The length of the sides of the triangles are

Since the sides are equal, the given points form an equilateral
triangle.

**Example 2.15**

Let z_{1} , z_{2} , and z_{3} be complex numbers such that |z_{1}| = |z_{2}| =|z_{3}| = r > 0 and z_{1} + z_{2} + z_{3} â‰ 0 .

Prove that .

**Solution**

**Example 2.16**

Show that the equation z^{2} = has four solutions.

**Solution**

We have,

It has 3 non-zero solutions. Hence including zero solution, there
are four solutions.

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12th Mathematics : UNIT 2 : Complex Numbers : Modulus of a Complex Number: Solved Example Problems | with Answers, Solution

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