# PROPERTIES OF MODULUS OF COMPLEX NUMBERS

Following are the properties of modulus of a complex number z.

1.  Let z = a + ib, where a and b are real numbers. Then,

|z| = √(a2 + b2)

2.  |z| = |conjugate of z|

3.  |z1 + z2≤ |z1| +|z2| (Triangle Inequality)

4.  |z1 - z2 |z1| - |z2|

5.  |z1z2| = |z1||z2|

6.  |z1/z2| = |z1|/|z2|

7.  |zn| = |z|n, where n is an integer.

8.  Re(z) ≤ |z|

9.  Im(z) ≤ |z|

10. The distance between the two points z1 and z2 in complex plane is |z1 -  z2|.

## Practice Questions

Questions 1-4 : Find the modulus of each of the following complex numbers

Question 1 :

2/(3 + 4i)

|(2/(3 + 4i)| = |2|/|(3 + 4i)|

= 2/√(32 + 42)

= 2/√(9 + 16)

= 2/√25

= 2/5

Question 2 :

(2 - i)/(1 + i) + (1 - 2i)/(1 - i)

= |(2 - i)/(1 + i) + (1 - 2i)/(1 - i)|

= |(2 - i)|/|(1 + i)| + |(1 - 2i)|/|(1 - i)| ----(1)

|(2 - i)| = √(22 + 12) = √5

|1 + i| = √(12 + 12) = √2

|1 - 2i| = √(12 + 22) = √5

|1 - i| = √(12 + 12) = √2

Substitute the values in (1).

= (√5/√2) + (√5/√2)

= 2√5/√2

= √2√5

√10

Question 3 :

(1 - i)10

|zn|  =  |z|n

(1 - i)10 =  {(1 - i)2}5

= (12 + i2 - 2i)5

= (1 - 1 - 2i)5

= (- 2i)5

= -32i5

= |-32i|

√(-32)2

= 32

Question 4 :

2i(3− 4i)(4 − 3i)

|2i(3− 4i)(4 − 3i)| = |2i||3 - 4i||4 - 3i|

√22 √32 + (-4)2√4+ (-3)

√4√25√25

= 2(5)(5)

= 50

Question 5 :

For any two complex numbers z1 and z2 , such that |z1| = |z2|  =  1 and z1 z2 ≠ -1, then show that z1 + z2/(1 + z1 z2) is a real number.

Let z1  =  1 and z =  i

|z1|  =  √1+ 0 =  1

|z2|  =  √0 + 1 =  1

z1 z2  =  1 + i

z z1  =   i

By applying the  values of zz2 and z zin the given statement, we get

z1 + z2/(1 + z1 z2)    =  (1 + i)/(1 + i)  =  1

1 is real. Hence it is proved.

Question 6 :

Which one of the points 10 − 8i , 11 + 6i is closest to 1 + i

Let the given points as A(10 - 8i), B (11 + 6i) and C (1 + i).

To find which point is more closer, we have to find the distance between the points AC and BC.

 AC  =  √(1-10)2 + (1+8)2   =  √92 + 92  =  √(81 + 81)AC  =  √162 BC  =  √(1-11)2 + (1-6)2   =  √102 + (-5)2  =  √(100 + 25)BC  =  √125

√162 > √125

Hence the point B is closer to C.

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