Class 11 RD Sharma解决方案–第13章复数–练习13.2 |套装1

问题1.在标准形式a + ib中表示以下复数：

(i)(1 + i)(1 + 2i)

解决方案：

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) $\frac{3+2i}{−2+i}$

解决方案：

We have, z =$\frac{3+2i}{−2+i}$

=$\frac{(3+2i)(-2-i)}{(-2+i)(-2-i)}$

=$\frac{3(-2-i) + 2i (-2-i)}{(-2)^2-(i)^2}$

=$\frac{-6-3i-4i-2i^2}{4-i^2}$

=$\frac{-6-7i+2}{4+1}$

=$\frac{-4 -7i}{5}$

Therefore, the standard form is$\frac{-4 -7i}{5}$where a = −4/5 and b = −7/5.

(iii) $\frac{1}{(2 + i)^2}$

解决方案：

We have, z =$\frac{1}{(2 + i)^2}$

=$\frac{1}{4+i^2+4i}$

=$\frac{1}{3+4i}$

=$\frac{3-4i}{(3+4i)(3-4i)}$

=$\frac{3-4i}{9+16}$

=$\frac{3-4i}{25}$

Therefore, the standard form is$\frac{3-4i}{25}$where a = 3/25 and b = −4/25.

(iv) $\frac{1-i}{1+i}$

解决方案：

We have, z =$\frac{1-i}{1+i}$

=$\frac{(1-i)(1-i)}{(1+i)(1-i)}$

=$\frac{1+i^2-2i}{1-i^2}$

=$\frac{-2i}{2}$

= −i

Therefore, the standard form is −i where a = 0 and b = −1.

(v) $\frac{(2+i)^3}{2+3i}$

解决方案：

We have, z =$\frac{(2+i)^3}{2+3i}$

=$\frac{8+i^3+12i+6i^2}{2+3i}$

=$\frac{8-i+12i-6}{2+3i}$

=$\frac{2+11i}{2+3i}$

=$\frac{(2+11i)(2-3i)}{(2+3i)(2-3i)}$

=$\frac{4-6i+22i-33i^2}{4+9}$

=$\frac{37+16i}{13}$

Therefore, the standard form is$\frac{37+16i}{13}$where a = 37/13 and b = 16/13.

(六) $\frac{(1+i)(1+\sqrt{3}i)}{1-i}$

解决方案：

We have, z =$\frac{(1+i)(1+\sqrt{3}i)}{1-i}$

=$\frac{1+\sqrt{3}i+i+\sqrt{3}i^2}{1-i}$

=$\frac{(1-\sqrt{3})+(1+\sqrt{3})i}{(1-i)}$

=$\frac{[(1-\sqrt{3})+(1+\sqrt{3})i](1+i)}{(1-i)(1+i)}$

=$\frac{[1-\sqrt{3}+(1-\sqrt{3})i+(1+\sqrt{3})i+(1+\sqrt{3})i^2]}{(1-(-1))}$

=$\frac{[(1-\sqrt{3})+(1-\sqrt{3}+1+\sqrt{3})i+(1+\sqrt{3})(-1)]}{2}$

=$\frac{-2\sqrt{3}+2i}{2}$

= –√3 + i

Therefore, the standard form is –√3 + i where a = –√3 and b = 1.

(vii) $\frac{2+3i}{4+5i}$

解决方案：

We have, z =$\frac{2+3i}{4+5i}$

=$\frac{(2+3i)(4-5i)}{(4+5i)(4-5i)}$

=$\frac{8-10i+12i-15i^2}{16+25}$

=$\frac{23+2i}{41}$

Therefore, the standard form is$\frac{23+2i}{41}$where a = 23/41 and b = 2/41.

(viii) $\frac{(1-i)^3}{1-i^3}$

解决方案：

We have, z =$\frac{(1-i)^3}{1-i^3}$

=$\frac{1-3i+3i^2-i^3}{1-(-1)i}$

=$\frac{-2-4i}{1+i}$

=$\frac{(-2-4i)(1-i)}{(1+i)(1-i)}$

=$\frac{-2+2i-4i+4i^2}{1-(-1)}$

=$\frac{-6-2i}{2}$

= –3 – i

Therefore, the standard form is –3 – i where a = –3 and b = – 1.

(ix)(1 + 2i) ^-3

解决方案：

We have z = (1 + 2i)^-3

=$\frac{1}{1+6i+4i^2+8i^3}$

=$\frac{1}{1+6i-4-8i}$

=$\frac{-1}{3+2i}$

=$\frac{-1(3-2i)}{(3+2i)(3-2i)}$

=$\frac{-3+2i}{9+4}$

=$\frac{-3+2i}{13}$

Therefore, the standard form is$\frac{-3+2i}{13}$where a = –3/13 and b = 2/13.

(X) $\frac{3-4i}{(4-2i)(1+i)}$

解决方案：

We have, z =$\frac{3-4i}{(4-2i)(1+i)}$

=$\frac{3-4i}{4+4i-2i-2i^2}$

=$\frac{3-4i}{6+2i}$

=$\frac{(3-4i)(6-2i)}{(6+2i)(6-2i)}$

=$\frac{18-6i-24i+8i^2}{36-4i^2}$

=$\frac{10-30i}{40}$

=$\frac{1-3i}{4}$

Therefore, the standard form is$\frac{1-3i}{4}$where a = –1/4 and b = –3/4.

(xi) $\left(\frac{1}{1-4i}-\frac{2}{1+i}\right)\frac{3-4i}{5+i}$

解决方案：

We have, z =$\left(\frac{1}{1-4i}-\frac{2}{1+i}\right)\frac{3-4i}{5+i}$

=$\left(\frac{1+i-2+8i}{(1-4i)(1+i)}\right)\frac{3-4i}{5+i}$

=$\left(\frac{-1+9i}{1+i-4i-4i^2}\right)\frac{3-4i}{5+i}$

=$\frac{(-1+9i)(3-4i)}{(5-3i)(5+i)}$

=$\frac{-3+4i+27i-9i^2}{25+5i-15i-3i^2}$

=$\frac{6+31i}{28-10i}$

=$\frac{(6+31i)(28+10i)}{(28-10i)(28+10i)}$

=$\frac{168+60i+868i+310i^2}{784+100}$

=$\frac{478+928i}{884}$

Therefore, the standard form is$\frac{478+928i}{884}$where a = 478/884 and b = 928/884.

(xii) $\frac{5+\sqrt{2}i}{1-\sqrt{2}i}$

解决方案：

We have, z =$\frac{5+\sqrt{2}i}{1-\sqrt{2}i}$

=$\frac{(5+\sqrt{2}i)(1+\sqrt{2}i)}{(1-\sqrt{2}i)(1+\sqrt{2}i)}$

=$\frac{5+5\sqrt{2}i+\sqrt{2}i+2i^2}{1-2i^2}$

=$\frac{3+6\sqrt{2}i}{3}$

= 1+ 2√2i

Therefore, the standard form is 1+ 2√2i where a = 1 and b = 2√2.

问题2。找到x和y的实值，如果

(i)(x + iy)(2 – 3i)= 4 + i

解决方案：

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)

解决方案：

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 =$\frac{10+10i}{3+4i}$

=> 3x – 2yi =$\frac{(10+10i)(3-4i)}{(3+4i)(3-4i)}$

=> 3x – 2yi =$\frac{30-40i+30i-40i^2}{9+16}$

=> 3x – 2yi =$\frac{70-10i}{25}$

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) $\frac{(1+i)x-2i}{3+i}+\frac{(2-3i)y+i}{3-i}=i$

解决方案：

We have,

=>$\frac{(1+i)x-2i}{3+i}+\frac{(2-3i)y+i}{3-i}=i$

=>$\frac{((1+i)(3-i)x)-2i(3-i)+((2-3i)(3+i)y)+i(3+i)}{(3+i)(3-i)}=i$

=>$\frac{(3-i+3i-i^2)x-6i+2i^2+(6+2i-9i-3i^2)y+3i+i^2}{9+1}=i$

=> (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

解决方案：

We have,

=> (1 + i) (x + iy) = 2 – 5i

=> x + iy =$\frac{2-5i}{1+i}$

=> x + iy =$\frac{(2-5i)(1-i)}{(1+i)(1-i)}$

=> x + iy =$\frac{2-i-5i+5i^2}{1+1}$

=> x + iy =$\frac{-3-7i}{2}$

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.

问题3.查找以下复数的共轭：

(i)4 – 5i

解决方案：

We know the conjugate of a complex number (a + ib) is (a – ib).

Therefore, the conjugate of (4 – 5i) is (4 + 5i).

(ii) $\frac{1}{3+5i}$

解决方案：

We have, z =$\frac{1}{3+5i}$

=$\frac{3-5i}{(3+5i)(3-5i)}$

=$\frac{3-5i}{9+25}$

=$\frac{3-5i}{34}$

We know the conjugate of a complex number (a + ib) is (a – ib).

Therefore, the conjugate of$\frac{1}{3+5i}$is$\frac{3+5i}{34}$.

(iii) $\frac{1}{1+i}$

解决方案：

We have, z =$\frac{1}{1+i}$

=$\frac{1-i}{(1+i)(1-i)}$

=$\frac{1-i}{1+1}$

=$\frac{1-i}{2}$

We know the conjugate of a complex number (a + ib) is (a – ib).

Therefore, the conjugate of$\frac{1}{1+i}$is$\frac{1+i}{2}$.

(iv) $\frac{(3-i)^2}{2+i}$

解决方案：

We have, z =$\frac{(3-i)^2}{2+i}$

=$\frac{9+i^2-6i}{2+i}$

=$\frac{8-6i}{2+i}$

=$\frac{(8-6i)(2-i)}{(2+i)(2-i)}$

=$\frac{16-8i-12i+6i^2}{4-i^2}$

=$\frac{10-20i}{5}$

= 2 – 4i

We know the conjugate of a complex number (a + ib) is (a – ib).

Therefore, the conjugate of$\frac{(3-i)^2}{2+i}$is 2 + 4i.

(v) $\frac{(1+i)(2+i)}{3+i}$

解决方案：

We have, z =$\frac{(1+i)(2+i)}{3+i}$

=$\frac{2+i+2i+i^2}{3+i}$

=$\frac{1+3i}{3+i}$

=$\frac{(1+3i)(3-i)}{(3+i)(3-i)}$

=$\frac{3-i+9i-3i^2}{9+1}$

=$\frac{6+8i}{10}$

=$\frac{3+4i}{5}$

We know the conjugate of a complex number (a + ib) is (a – ib).

The conjugate of$\frac{(1+i)(2+i)}{3+i}$is$\frac{3-4i}{5}$.

(六) $\frac{(3-2i)(2+3i)}{(1+2i)(2-i)}$

解决方案：

We have, z =$\frac{(3-2i)(2+3i)}{(1+2i)(2-i)}$

=$\frac{6+9i-4i-6i^2}{2-i+4i-2i^2}$

=$\frac{12+5i}{4+3i}$

=$\frac{(12+5i)(4-3i)}{(4+3i)(4-3i)}$

=$\frac{48-36i+20i-15i^2}{16+9}$

=$\frac{63-16i}{25}$

We know the conjugate of a complex number (a + ib) is (a – ib).

Therefore, the conjugate of$\frac{(3-2i)(2+3i)}{(1+2i)(2-i)}$is$\frac{63+16i}{25}$.

问题4.找到以下复数的乘法逆：

(i)1 – i

解决方案：

We have z = 1 – i

We know the multiplicative inverse of a complex number z is 1/z. So, we get,

=$\frac{1}{1-i}$

=$\frac{1+i}{(1-i)(1+i)}$

=$\frac{1+i}{1-i^2}$

=$\frac{1+i}{2}$

Therefore, the multiplicative inverse of (1 – i) is$\frac{1+i}{2}$.

(ii)(1 +我√3) ²

解决方案：

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,

=$\frac{1}{−2 + 2 i\sqrt{3}}$

=$\frac{−2-2i\sqrt{3}}{(−2+2i\sqrt{3})(−2-2i\sqrt{3})}$

=$\frac{−2-2i\sqrt{3}}{4-12i^2}$

=$\frac{−2-2i\sqrt{3}}{16}$

=$\frac{−1-i\sqrt{3}}{8}$

Therefore, the multiplicative inverse of (1 + i √3)² is$\frac{−1-i\sqrt{3}}{8}$.

(iii)4 – 3i

解决方案：

We have z = 4 – 3i

We know the multiplicative inverse of a complex number z is 1/z. So, we get,

=$\frac{1}{4-3i}$

=$\frac{4+3i}{(4-3i)(4+3i)}$

=$\frac{4+3i}{16+9}$

=$\frac{4+3i}{25}$

Therefore, the multiplicative inverse of 4 – 3i is$\frac{4+3i}{25}$.

(iv)√5+ 3i

解决方案：

We have z = √5 + 3i

We know the multiplicative inverse of a complex number z is 1/z. So, we get,

=$\frac{1}{\sqrt{5}+3i}$

=$\frac{\sqrt{5}-3i}{(\sqrt{5}+3i)(\sqrt{5}-3i)}$

=$\frac{\sqrt{5}-3i}{5+9}$

=$\frac{\sqrt{5}-3i}{14}$

Therefore, the multiplicative inverse of √5 + 3i is$\frac{\sqrt{5}-3i}{14}$.

问题5.如果z ₁ = 2 − i，z ₂ = 1 + i，则找到$|\frac{z_1+z_2+1}{z_1-z_2+i}|$ 。

解决方案：

Given z₁ = 2 − i, z₂ = 1 + i, we get,

$|\frac{z_1+z_2+1}{z_1-z_2+i}|$=$|\frac{2-i+1+i+1}{2-i-1-i+i}|$

=$\frac{|4|}{|1-i|}$

=$\frac{\sqrt{4^2+0^2}}{\sqrt{1^2+(-1)^2}}$

=$\frac{4}{\sqrt{2}}$

= 2√2

Therefore, the value of$|\frac{z_1+z_2+1}{z_1-z_2+i}|$is 2√2.

问题6。如果z ₁ =(2 – i)，z ₂ =(–2 + i)，则找到

(i)重新$(\frac{z_1z_2}{\bar{z_1}})$

解决方案：

Given z₁ = (2 – i), z₂ = (–2 + i), we get,

$\frac{z_1z_2}{\bar{z_1}}$=$\frac{z^2_1z_2}{\bar{z_1}z_1}$

=$\frac{z^2_1z_2}{|z_1|^2}$

=$\frac{(2-i)^2(-2+i)}{2^2+(-1)^2}$

=$\frac{(4+i^2-4i)(-2+i)}{4+1}$

=$\frac{(3-4i)(-2+i)}{5}$

=$\frac{-6+3i+8i+4}{5}$

=$\frac{-2+11i}{5}$

Therefore, Re$(\frac{z_1z_2}{\bar{z_1}})$=$\frac{-2}{5}$.

(ii)我$(\frac{1}{z_1\bar{z_1}})$

Now,$\frac{1}{z_1\bar{z_1}}$=$\frac{1}{|z_1|^2}$

=$\frac{1}{|2-i|^2}$

=$\frac{1}{2^2+(-1)^2}$

=$\frac{1}{4+1}$

=$\frac{1}{5}+0i$

Therefore, Im$(\frac{1}{z_1\bar{z_1}})$= 0.

问题7.求出模数$\frac{1+i}{1-i}-\frac{1-i}{1+i}$ 。

解决方案：

We have, z =$\frac{1+i}{1-i}-\frac{1-i}{1+i}$

=$\frac{(1+i)^2-(1-i)^2}{1-i^2}$

=$\frac{1+i^2+2i-1-i^2+2i}{1-(-1)}$

=$\frac{4i}{2}$

= 2i

So, modulus of z =$\sqrt{0^2+2^2}$= 2.

Therefore, the modulus of$\frac{1+i}{1-i}-\frac{1-i}{1+i}$is 2.

问题8.如果x + iy = $\frac{a+ib}{a−ib}$ ，证明x ² + y ² = 1。

解决方案：

We have,

=> x + iy =$\frac{a+ib}{a−ib}$

On applying modulus on both sides we get,

=> |x + iy| =$|\frac{a+ib}{a−ib}|$

=> |x + iy| =$\frac{|a+ib|}{|a−ib|}$

=>$\sqrt{x^2+y^2}=\frac{\sqrt{a^2+b^2}}{\sqrt{a^2+b^2}}$

=>$\sqrt{x^2+y^2}$= 1

=> x² + y² = 1

Hence proved.

问题9.求n的最小正整数值$\left[\frac{1+i}{1-i}\right]^n$是真实的。

解决方案：

We have, z =$\left[\frac{1+i}{1-i}\right]^n$

=$\left[\frac{(1+i)^2}{(1-i)(1+i)}\right]^n$

=$\left[\frac{1+i^2+2i}{1-i^2}\right]^n$

=$\left[\frac{2i}{2}\right]^n$

= iⁿ

For n = 2, we have iⁿ = i² = −1, which is real

Therefore, the least positive integral value of n for which$\left[\frac{1+i}{1-i}\right]^n$is real is 2.

问题10.求出复数θ的实值$\frac{1 + i cos θ}{1 - 2i cos θ}$是纯粹的真实。

解决方案：

We have, z =$\frac{1 + i cos θ}{1 - 2i cos θ}$

=$\frac{(1+icosθ)(1+2icosθ)}{(1-2icosθ)(1+2icosθ)}$

=$\frac{1+2icosθ+icosθ+2i^2cos^2θ}{1-4i^2cos^2θ}$

=$\frac{1-2cos^2θ+3icosθ}{1+4cos^2θ}$

=$\frac{1-2cos^2θ}{1+4cos^2θ}+\frac{3cosθi}{1+4cos^2θ}$

For a complex number to be purely real, the imaginary part should be equal to zero.

So, we get,$\frac{3cosθ}{1+4cos^2θ}$= 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.

问题11.求n的最小正整数值$\frac{(1+i)^n}{(1-i)^{n-2}}$是一个实数。

解决方案：

We have, z =$\frac{(1+i)^n}{(1-i)^{n-2}}$

=$\frac{(1+i)^n(1-i)^2}{(1-i)^{n-2}(1-i)^2}$

=$\frac{(1+i)^n}{(1-i)^n}×(1-i)^2$

=$\left(\frac{1+i}{1-i}\right)^n×(1+i^2-2i)$

=$\left[\frac{(1+i)^2}{(1-i)(1+i)}\right]^n×(-2i)$

=$\left[\frac{1+i^2+2i}{1-i^2}\right]^n×(-2i)$

=$\left(\frac{2i}{2}\right)^n×(-2i)$

= 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$\frac{(1+i)^n}{(1-i)^{n-2}}$is 1.

问题12。 $\left(\frac{1+i}{1-i}\right)^3-\left(\frac{1-i}{1+i}\right)^3=x+iy$ ，找到(x，y)。

解决方案：

We have,

=>$\left[\frac{(1+i)^2}{(1-i)(1+i)}\right]^3-\left[\frac{(1-i)^2}{(1+i)(1-i)}\right]^3=x+iy$

=>$\left[\frac{1+i^2+2i}{1-i^2}\right]^3-\left[\frac{1+i^2-2i}{1-i^2}\right]^3=x+iy$

=>$\left[\frac{2i}{2}\right]^3-\left[\frac{-2i}{2}\right]^3=x+iy$

=> 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)

问题13 $\frac{(1+i)^2}{2-i} = x + iy$ ，找到x + y。

解决方案：

We have,

=>$\frac{(1+i)^2}{2-i} = x + iy$

=>$\frac{(1+i^2+2i)(2+i)}{(2-i)(2+i)} = x + iy$

=>$\frac{2i(2+i)}{4+1} = x + iy$

=>$\frac{2i^2+4i}{5} = x + iy$

=>$\frac{-2}{5}+\frac{4i}{5} = x + iy$

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.