**Complex Numbers with Inequality Problems :**

Here we are going to see some example problems based on complex numbers with inequality.

**Question 1 :**

If | z |= 3, show that 7 ≤ | z + 6 − 8i | ≤ 13.

**Solution :**

Using the property of complex numbers

|z_{1} + z_{2}| ≤ |z_{1}| + |z_{2}|

|z_{1} + z_{2}| ≥ |z_{1}| - |z_{2}|

| z + 6−8i | ≤ | [|z| + |6 − 8i|] |

≤ | 3 + √6^{2} + (-8)^{2}|

≤ |3 + √(36 + 64)|

≤ |3 + √100|

≤ |3 + 10|

≤ 13

| z + 6−8i | ≥ |z| - |6 − 8i|

≥ |3 - √(36 + 64)|

≥ |3 - √100|

≥ |3 - 10|

≥ 7

7 ≤ | z + 6−8i | ≤ 13

**Question 2 :**

If |z| = 1, show that 2 ≤ |z^{2} - 3 | ≤ 4

**Solution :**

|z^{2} - 3 | ≤ ||z^{2}| + |- 3 | |

|z^{2} - 3 | ≤ ||z|^{2} + 3|

|z^{2} - 3 | ≤ |1^{2} + 3|

|z^{2} - 3 | ≤ 4

|z^{2} - 3 | ≥ ||z^{2}| - | 3||

|z^{2} - 3 | ≥ |1 - 3|

|z^{2} - 3 | ≥ 2

2 ≤ |z^{2} - 3 | ≤ 4

Hence proved.

**Question 3 :**

If | z - (2/z) | = 2, show that the greatest and least value of | z | are √3 + 1 and √3 − 1 respectively

**Solution :**

**|z - (2/z) | = 2**

**|(z ^{2} - 2)/z | = 2**

**Question 4 :**

If z_{1}, z_{2} and z_{3} are three complex numbers such that |z_{1}| = 1, |z_{2}| = 2, |z_{3}| = 3 and |z_{1} + z_{2} + z_{3}| = 1, show that |9 z_{1} z_{2} + 4 z_{1} z_{3} + z_{2} z_{3}| = 6

**Solution :**

Now applying the given values, we get

= 1(2)(3) (1)

= 6

Hence proved.

**Question 5 :**

If the area of the triangle formed by the vertices z, iz , and z + iz is 50 square units, find the value of z .

**Solution :**

Length of three sides

z & iz = √(z-0)² + (0-z)² = |z|√2 = a

z & z + iz = √(z-z)² + (0-z)² = |z| = b

z & z + iz = √(0-z)² + (z-z)² = |z| = c

The sides are |z|√2 , |z| & |z|

s = (a + b + c)/2

Area of triangle = 50 sq units

|z|²/2 = 50

|z|² = 100

|z| = 10

Hence the value of |z| = 10.

Let us look into the next problems on "Complex Numbers with Inequality Problems".

**Question 6 :**

Show that the equation z^{3} + 2z bar = 0 has five solutions.

**Solution :**

z^{3} + 2z bar = 0

z = x + iy

(x + iy)^{3} + 2(x - iy) = 0

x^{3} + 3x^{2}(iy) + 3 x(iy)^{2} + (iy)^{3} + 2x - i2y = 0

x^{3} + i3x^{2}y - 3 xy^{2} - iy + 2x - i2y = 0

(x^{3} - 3 xy^{2} + 2x) + i(3x^{2}y - y - 2y) = 0

(x^{3} - 3 xy^{2} + 2x) + i(3x^{2}y - 3y) = 0 + i0

By equating the real and imaginary parts, we get

x^{3} - 3 xy^{2} + 2x = 0 ----(1) and 3x^{2}y - 3y = 0 ----(2)

3y(x^{2 }- 1) = 0

y = 0, x = 1, -1

By applying the two different values of x in (1), we get 2 different values of y.

Hence it has 5 solutions.

After having gone through the stuff given above, we hope that the students would have understood, "Complex Numbers with Inequality Problems".

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