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|.
Questions 1-4 : Find the modulus of each of the following complex numbers
Question 1 :
2/(3 + 4i)
Answer :
|(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)
Answer :
= |(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
Answer :
|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)
Answer :
|2i(3− 4i)(4 − 3i)| = |2i||3 - 4i||4 - 3i|
= √22 √32 + (-4)2√42 + (-3)2
= √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.
Answer :
Let z1 = 1 and z2 = i
|z1| = √12 + 02 = 1
|z2| = √0 + 12 = 1
z1 + z2 = 1 + i
z1 z1 = i
By applying the values of z1 + z2 and z1 z2 in 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
Answer :
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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