# COMPLEX NUMBERS WITH INEQUALITY PROBLEMS

Complex Numbers with Inequality Problems :

In this section, we will learn, how to solve problems on complex numbers with inequality.

## Complex Numbers with Inequality Problems - Practice Questions

Question 1 :

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

Solution :

Using the property of complex numbers

|z1 + z2 ≤ |z1| + |z2|

|z1 + z2|  ≥ |z1| - |z2|

| z + 6−8i | ≤ | [|z| + |6 − 8i|] |

≤ | 3 + √62 + (-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 ≤ |z2 -  3 | ≤ 4

Solution :

|z2 -  3 | ≤ ||z2| + |- 3 | |

|z2 -  3 | ≤ ||z|2 + 3|

|z2 -  3 | ≤ |12 + 3|

|z2 -  3 | ≤ 4

|z2 -  3 |   ||z2| - | 3||

|z2 -  3 |   |1 - 3|

|z2 -  3 |  2

2 ≤ |z2 -  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

|(z2 - 2)/z | = 2

Question 4 :

If z1, z2 and z3 are three complex numbers such that |z1| = 1, |z2|  =  2, |z3|  =  3 and |z1 + z2 + z3|  =  1, show that |9 z1 z2 + 4 z1 z3 + z2 z3|  =  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.

Question 6 :

Show that the equation z3 + 2z bar = 0 has five solutions.

Solution :

z3 + 2z bar = 0

z = x + iy

(x + iy)3 + 2(x - iy)  =  0

x3 + 3x2(iy) + 3 x(iy)2 + (iy)3 + 2x - i2y  =  0

x3 + i3x2y - 3 xy2 - iy + 2x - i2y  =  0

(x3  - 3 xy2 + 2x)  + i(3x2y - y - 2y)  =  0

(x3  - 3 xy2 + 2x)  + i(3x2y - 3y)  =  0 + i0

By equating the real and imaginary parts, we get

x3  - 3 xy2 + 2x  =  0 ----(1)

3x2y - 3y  =  0 ----(2)

3y(x2 - 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, how to solve complex numbers with inequality problems

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