Класс 11 RD Sharma Solutions – Глава 13 Комплексные числа – Упражнение 13.2 | Набор 1
Вопрос 1. Запишите следующие комплексные числа в стандартной форме a + ib:
(i) (1 + i) (1 + 2i)
Решение:
We have, z = (1 + i) (1 + 2i)
= 1 (1 + 2i) + i (1 + 2i)
= 1 + 2i + i + 2i2
= 1 + 3i + 2(−1)
= 1 + 3i − 2
= −1 + 3i
Therefore, the standard form is −1 + 3i where a = −1 and b = 3.
(ii) 
Решение:
We have, z =
=
=
=
=
=
Therefore, the standard form is
where a = −4/5 and b = −7/5.
(iii) 
Решение:
We have, z =
=
=
=
=
=
Therefore, the standard form is
where a = 3/25 and b = −4/25.
(4) 
Решение:
We have, z =
=
=
=
= −i
Therefore, the standard form is −i where a = 0 and b = −1.
(в) 
Решение:
We have, z =
=
=
=
=
=
=
Therefore, the standard form is
where a = 37/13 and b = 16/13.
(мы) 
Решение:
We have, z =
=
=
=
=
=
=
= –√3 + i
Therefore, the standard form is –√3 + i where a = –√3 and b = 1.
(vii) 
Решение:
We have, z =
=
=
=
Therefore, the standard form is
where a = 23/41 and b = 2/41.
(VIII) 
Решение:
We have, z =
=
=
=
=
=
= –3 – i
Therefore, the standard form is –3 – i where a = –3 and b = – 1.
(ix) (1 + 2i) -3
Решение:
We have z = (1 + 2i)-3
=
=
=
=
=
=
Therefore, the standard form is
where a = –3/13 and b = 2/13.
(Икс) 
Решение:
We have, z =
=
=
=
=
=
=
Therefore, the standard form is
where a = –1/4 and b = –3/4.
(xi) 
Решение:
We have, z =
=
=
=
=
=
=
=
=
Therefore, the standard form is
where a = 478/884 and b = 928/884.
(xii) 
Решение:
We have, z =
=
=
=
= 1+ 2√2i
Therefore, the standard form is 1+ 2√2i where a = 1 and b = 2√2.
Вопрос 2. Найдите действительные значения x и y, если
(i) (x + iy) (2 – 3i) = 4 + i
Решение:
We have,
=> (x + iy) (2 – 3i) = 4 + i
=> 2x – 3xi + 2yi – 3yi2 = 4 + i
=> 2x + (–3x+2y)i + 3y = 4 + i
=> (2x+3y) + i(–3x+2y) = 4 + i
On comparing real and imaginary parts on both sides, we get,
2x + 3y = 4 . . . . (1)
And –3x + 2y = 1 . . . . (2)
On multiplying (1) by 3 and (2) by 2 and adding, we get
=> 6x – 6x – 9y + 4y = 12 + 2
=> 13y = 14
=> y = 14/13
On putting y = 14/13 in (1), we get
=> 2x + 3(14/13) = 4
=> 2x = 4 – (42/13)
=> 2x = 10/13
=> x = 5/13
Therefore, the real values of x and y are 5/13 and 14/13 respectively.
(ii) (3x – 2iy) (2 + i) 2 = 10(1 + i)
Решение:
We have,
=> (3x – 2iy) (2 + i)2 = 10(1 + i)
=> (3x – 2yi) (4 + i2 + 4i) = 10 + 10i
=> (3x – 2yi) (3 + 4i) = 10+10i
=> 3x – 2yi =
=> 3x – 2yi =
=> 3x – 2yi =
=> 3x – 2yi =
On comparing real and imaginary parts on both sides, we get,
=> 3x = 70/25 and –2y = –10/25
=> x = 70/75 and y = 1/5
Therefore, the real values of x and y are 70/75 and 1/5 respectively.
(iii) 
Решение:
We have,
=>
=>
=>
=> (4+2i) x − 3i − 3 + (9−7i)y = 10i
=> (4x+9y−3) + i(2x−7y−3) = 10i
On comparing real and imaginary parts on both sides, we get,
4x + 9y − 3 = 0 . . . . (1)
And 2x − 7y − 3 = 10 . . . . (2)
On multiplying (1) by 7 and (2) by 9 and adding, we get,
=> 28x + 18x + 63y – 63y = 117 + 21
=> 46x = 117 + 21
=> 46x = 138
=> x = 3
On putting x = 3 in (1), we get
=> 4x + 9y − 3 = 0
=> 9y = −9
=> y = −1
Therefore, the real values of x and y are 3 and −1 respectively.
(iv) (1 + i) (x + iy) = 2 – 5i
Решение:
We have,
=> (1 + i) (x + iy) = 2 – 5i
=> x + iy =
=> x + iy =
=> x + iy =
=> x + iy =
On comparing real and imaginary parts on both sides, we get,
=> x = −3/2 and y = −7/2
Therefore, the real values of x and y are −3/2 and −7/2 respectively.
Вопрос 3. Найдите сопряжения следующих комплексных чисел:
(и) 4 – 5и
Решение:
We know the conjugate of a complex number (a + ib) is (a – ib).
Therefore, the conjugate of (4 – 5i) is (4 + 5i).
(ii) 
Решение:
We have, z =
=
=
=
We know the conjugate of a complex number (a + ib) is (a – ib).
Therefore, the conjugate of
is
.
(iii) 
Решение:
We have, z =
=
=
=
We know the conjugate of a complex number (a + ib) is (a – ib).
Therefore, the conjugate of
is
.
(4) 
Решение:
We have, z =
=
=
=
=
=
= 2 – 4i
We know the conjugate of a complex number (a + ib) is (a – ib).
Therefore, the conjugate of
is 2 + 4i.
(в) 
Решение:
We have, z =
=
=
=
=
=
=
We know the conjugate of a complex number (a + ib) is (a – ib).
The conjugate of
is
.
(мы) 
Решение:
We have, z =
=
=
=
=
=
We know the conjugate of a complex number (a + ib) is (a – ib).
Therefore, the conjugate of
is
.
Вопрос 4. Найдите мультипликативное обратное значение следующих комплексных чисел:
(и) 1 – я
Решение:
We have z = 1 – i
We know the multiplicative inverse of a complex number z is 1/z. So, we get,
=
=
=
=
Therefore, the multiplicative inverse of (1 – i) is
(ii) (1 + я √3) 2
Решение:
We have, z = (1 + i √3)2
= 1 + 3i2 + 2 i√3
= 1 + 3(−1) + 2 i√3
= 1 – 3 + 2 i√3
= −2 + 2 i√3
We know the multiplicative inverse of a complex number z is 1/z. So, we get,
=
=
=
=
=
Therefore, the multiplicative inverse of (1 + i √3)2 is
.
(iii) 4 – 3и
Решение:
We have z = 4 – 3i
We know the multiplicative inverse of a complex number z is 1/z. So, we get,
=
=
=
=
Therefore, the multiplicative inverse of 4 – 3i is
.
(4) √5 + 3i
Решение:
We have z = √5 + 3i
We know the multiplicative inverse of a complex number z is 1/z. So, we get,
=
=
=
=
Therefore, the multiplicative inverse of √5 + 3i is
.
Вопрос 5. Если z 1 = 2 − i, z 2 = 1 + i, найти 
.
Решение:
Given z1 = 2 − i, z2 = 1 + i, we get,
=
=
=
= 2√2
Therefore, the value of
Вопрос 6. Если z 1 = (2 – i), z 2 = (–2 + i), найти
(и) Ре 
Решение:
Given z1 = (2 – i), z2 = (–2 + i), we get,
=
=
=
=
=
=
=
Therefore, Re
=
.
(ii) Я 
Now,
=
=
=
=
=
Therefore, Im
= 0.
Вопрос 7. Найдите модуль 
.
Решение:
We have, z =
=
=
=
= 2i
So, modulus of z =
= 2.
Therefore, the modulus of
Вопрос 8. Если x + iy = 
, докажите, что x 2 + y 2 = 1.
Решение:
We have,
=> x + iy =
On applying modulus on both sides we get,
=> |x + iy| =
=> |x + iy| =
=>
=>
= 1
=> x2 + y2 = 1
Hence proved.
Вопрос 9. Найдите наименьшее положительное целое значение n, при котором 
это реально.
Решение:
We have, z =
=
=
=
= in
For n = 2, we have in = i2 = −1, which is real
Therefore, the least positive integral value of n for which
Вопрос 10. Найдите действительные значения θ, при которых комплексное число 
чисто реально.
Решение:
We have, z =
=
=
=
=
For a complex number to be purely real, the imaginary part should be equal to zero.
So, we get,
= 0
=> cos θ = 0
=> cos θ = cos π/2
=> 2nπ ± π/2, for n ∈ Z
Therefore, the values of θ for the complex number to be purely real are 2nπ ± π/2, for n ∈ Z.
Вопрос 11. Найдите наименьшее натуральное значение n, для которого 
является действительным числом.
Решение:
We have, z =
=
=
=
=
=
=
= in × (−2i)
= −2in+1
For n = 1, we have z = −2i1+1
= −2i2
= 2, which is real
Therefore, the smallest positive integer value of n for which is a real number
Вопрос 12. Если 
, найти (х, у).
Решение:
We have,
=>
=>
=>
=> i3 – (–i3) = x + iy
=> 2i3 = x + iy
=> x + iy = −2i
On comparing real and imaginary parts on both sides, we get,
=> (x, y) = (0, −2)
Вопрос 13. Если 
, найти х + у.
Решение:
We have,
=>
=>
=>
=>
=>
On comparing real and imaginary parts on both sides, we get,
=> x = −2/5 and y = 4/5
So, x + y = −2/5 + 4/5
= (−2+4)/5
= 2/5
Therefore, the value of (x + y) is 2/5.
