Class 11 RD Sharma解决方案–第13章复数–练习13.2 |套装2

问题14.如果$\left(\frac{1-i}{1+i}\right)^{100}=a+ib$ ，找到(a，b)。

解决方案：

We have,

=>$\left(\frac{1-i}{1+i}\right)^{100}=a+ib$

=>$\left[\frac{(1-i)^2}{(1+i)(1-i)}\right]^{100}=a+ib$

=>$\left(\frac{1+i^2-2i}{1-i^2}\right)^{100}=a+ib$

=>$\left(\frac{-2i}{2}\right)^{100}=a+ib$

=> (−i)¹⁰⁰ = a + ib

=> a + ib = 1

On comparing real and imaginary parts on both sides, we get,

=> (a, b) = (1, 0)

问题15。如果a = cosθ+ i sinθ，则求出值$\frac{1+a}{1-a}$ 。

解决方案：

Given a = cos θ + i sin θ, we get,

$\frac{1+a}{1-a}$=$\frac{1+cosθ+isinθ}{1-cosθ-isinθ}$

=$\frac{(1+cosθ+isinθ)(1-cosθ+isinθ)}{(1-cosθ-isinθ)(1-cosθ+isinθ)}$

=$\frac{(1+isinθ)^2-cos^2θ}{(1-cosθ)^2-i^2sin^2θ}$

=$\frac{1+i^2sin^2θ+2isinθ-cos^2θ}{1+cos^2θ-2cosθ-i^2sin^2θ}$

=$\frac{1+2isinθ-sin^2θ-cos^2θ}{1-2cosθ+cos^2θ+sin^2θ}$

=$\frac{1+2isinθ-1}{1-2cosθ+1}$

=$\frac{2isinθ}{2-2cosθ}$

=$\frac{isinθ}{1-cosθ}$

=$\frac{2isin\frac{θ}{2}cos\frac{θ}{2}}{2sin^2\frac{θ}{2}}$

=$icot\frac{θ}{2}$

Therefore, the value of$\frac{1+a}{1-a}$is$icot\frac{θ}{2}$.

问题16：评估以下内容：

(i)当x =(3-5i)/ 2时^{2x 3} + 2x ^{2 − 7x + 72}

解决方案：

We have, x = (3−5i)/2

=> 2x = 3 − 5i

=> 2x − 3 = −5i

=> (2x − 3)² = 25i²

=> 4x² + 9 − 12x = −25

=> 4x² − 12x + 34 = 0

=> 2x² − 6x + 17 = 0

Now, 2x³ + 2x² − 7x + 72 = x (2x² − 6x + 17) + 6x² − 17x + 2x² − 7x + 72

= x (0) + 8x² − 24x + 72

= 4 (2x² − 6x + 17) + 4

= 4 (0) + 4

= 4

Therefore, the value of 2x³ + 2x² − 7x + 72 is 4.

(ii)当x = 3 + 2i时， ^{x 4} − 4x ³ + 4x ^{2 + 8x +44}

解决方案：

We have, x = 3 + 2i

=> x − 3 = 2i

=> (x − 3)² = (2i)²

=> x² + 9 − 6x = 4i²

=> x² − 6x + 9 + 4 = 0

=> x² − 6x + 13 = 0

Now, x⁴ − 4x³ + 4x² + 8x + 44 = x² (x² − 6x + 13) + 6x³ − 13x² − 4x³ + 4x² + 8x + 44

= 2x³ − 9x² + 8x + 44

= 2x (x² − 6x + 13) + 12x² − 26x − 9x² + 8x + 44

= 3x² − 18x + 44

= 3 (x² − 6x + 13) + 5

= 5

Therefore, the value of x⁴ − 4x³ + 4x² + 8x + 44 is 5.

(iii)x ⁴ + 4x ³ + 6x ² + 4x + 9(当x = −1 +i√2时)

解决方案：

We have, x = −1 + i√2

=> x + 1 = i√2

=> (x + 1)² = 2i²

=> x² + 1 + 2x = −2

=> x² + 2x + 3 = 0

Now, x⁴ + 4x³ + 6x² + 4x + 9 = x² (x² + 2x + 3) − 2x³ − 3x² + 4x³ + 6x² + 4x + 9

= 2x³ + 3x² + 4x + 9

= 2x (x² + 2x + 3) − 4x² − 6x + 3x² + 4x + 9

= − x² − 2x + 9

= − (x² + 2x + 3) + 3 + 9

= 3 + 9

= 12

Therefore, the value of x⁴ + 4x³ + 6x² + 4x + 9 is 12.

(iv)x ⁶ + x ⁴ + x ² +1，当x =(1 + i)/√2时

解决方案：

We have, x = (1+i)/√2

=> √2x = 1 + i

=> 2x² = 1 + i² + 2i

=> 2x² = 2i

=> 4x⁴ = 4i²

=> x4 = −1

=> x⁴ + 1 = 0

Now, x⁶ + x⁴ + x² + 1 = (x⁶ + x²) + (x⁴ +1)

= x⁶ + x²

= x² (x⁴ + 1)

= 0

Therefore, the value of x⁶ + x⁴ + x² + 1 is 0.

(v)当x = −2 −√3i时为^{2x 4} + 5x ³ + 7x ^{2 − x + 41}

解决方案：

We have, x = −2 − √3i

x² = (−2 − √3i)² = 4 + 4√3i + 3i² = 1 + 4√3i

x³ = (1 + 4√3i) (−2 − √3i) = −2 − √3i − 8√3i −12i² = 10 − 9√3i

x⁴ = (1 + 4√3i)² = 1 + 8√3i + 48i² = −47 + 8√3i

Now, 2x⁴ + 5x³ + 7x² − x + 41 becomes,

= 2(−47 + 8√3i) + 5(10 − 9√3i) + 7(1 + 4√3i) − (−2 − √3i) + 41

= −94 + 16√3i + 50 − 45√3i + 7 + 28√3i + 2 + √3i + 41

= 6

Therefore, the value of 2x⁴ + 5x³ + 7x² − x + 41 is 6.

问题17.对于正整数n，找到(1-i) ⁿ (1-1 / i) ^n的值。

解决方案：

We have,

(1−i)ⁿ (1−1/i)ⁿ = (1−i)ⁿ$(\frac{i-1}{i})^n$

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

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

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

= 2ⁿ

Therefore, the value of (1−i)ⁿ (1−1/i)ⁿ is 2ⁿ.

问题18.如果(1 + i)z =(1-i) $\bar{z}$ ，则表明z = −i $\bar{z}$ 。

解决方案：

We have,

=> (1+i)z = (1−i)$\bar{z}$

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

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

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

=> z =$\left[\frac{-2i}{2}\right]\bar{z}$

=> z = −i$\bar{z}$

Hence proved.

问题19：解方程组：Re(z ² )= 0，| z | = 2。

解决方案：

Let z = x + iy.

Now z² = (x + iy)²

= x² + i²y² + 2xyi

= x² − y² + 2xyi

We have, Re(z²) = 0

=> x² − y² = 0 . . . . (1)

Also, it is given, |z| = 2.

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

=> x² + y² = 4 . . . . (2)

Solving (1) and (2), we get, x = ±√2 and y = ±√2.

Therefore, x + iy = ±√2 ± √2i .

问题20.如果$\frac{z-1}{z+1}$是纯虚数(z≠−1)，请找到| z |的值。

解决方案：

Let z = x + iy

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

=$\frac{x-1+iy}{x+1+iy}$

=$\frac{(x-1+iy)(x+1-iy)}{(x+1+iy)(x+1-iy)}$

=$\frac{x^2+x-ixy-x-1+iy+ixy+iy+y^2}{(x+1)^2-(iy)^2}$

=$\frac{x^2+y^2-1+2iy}{x^2+y^2+1+2xy}$

=$\frac{x^2+y^2-1}{x^2+y^2+1+2xy}+\frac{2yi}{x^2+y^2+1+2xy}$

As the complex number is purely imaginary, therefore,

=> Re(z) = 0

=>$\frac{x^2+y^2-1}{x^2+y^2+1+2xy}$= 0

=> x² + y² = 1

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

=> |z| = 1

Therefore, the value of |z| is 1.

问题21.如果z ₁是非-1的复数，则| z ₁ | = 1和z ₂ = $\frac{z_1-1}{z_1+1}$ ，则表明z ₂的实部为零。

解决方案：

Given |z| = 1

=> |z|² = 1

=> x² + y² = 1 . . . . (1)

Let z₁ = x + iy and z₂ = a + ib.

According to the question, we have,

=> z₂ =$\frac{z_1-1}{z_1+1}$

=> a + ib =$\frac{x+iy-1}{x+iy+1}$

=> a + ib =$\frac{(x-1+iy)(x+1-iy)}{(x+1+iy)(x+1-iy)}$

=> a + ib =$\frac{x^2-1+y^2-iyx+iy+iyx+iy}{(x+1)^2-(iy)^2}$

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

Using (1) we get,

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

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

On comparing the real and imaginary parts on both sides, we get a = 0.

Therefore, the real parts of z₂ is 0. Hence proved.

问题22：如果| z + 1 | = z + 2(1 + i)，找到z。

解决方案：

Let z = x + iy. According to the question, we have,

=> |x + iy + 1| = x + iy + 2(1 + i)

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

On comparing the real and imaginary parts, we get

=> y + 2 = 0

=> y = −2

And also,

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

=> (x + 2)² = (x+1)² + y²

=> x² + 4 + 4x = x² + 2x + 1+ y²

=> 2x = y² − 3

=> 2x = 4 − 3

=> 2x = 1

=> x = 1/2

Therefore, z = x + iy = 1/2 −2i.

问题23.解方程：| z | = z + 1 + 2i。

解决方案：

Let z = x + iy. According to the question, we have,

=> |z| = z + 1 + 2i

=> |x + iy| = x + iy + 1 + 2i

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

=> x² + y² = (x+1)² + (y+2)²i² + 2 (x+1) (y+2)i

=> x² + y² = x²+1 + 2x − y² − 1 + 2y + 2 (x+1) (y+2)i

=> 2y² − 2x + 4y + 4 = 2i (x+1) (y+2)

=> y² − x + 2y + 2 = i (x+1) (y+2)

On comparing both sides, we get,

=> (x+1) (y+2) = 0

=> x = −1 and y = −2

Also, y² − x + 2y + 2 = 0

Taking x = −1, we get y² − (−1) + 2y + 2 = 0

=> y² + 2y + 3 = 0, which doesn’t have a solution as the roots are imaginary.

Taking y = −2, (4 − x −4 + 2) = 0

=> x = 2

Therefore, z = x + iy = 2 − 2i.

问题24.(1 + i) ²ⁿ =(1-i) ²ⁿ的最小正整数n是多少?

解决方案：

We are given,

=> (1+i)²ⁿ = (1−i)²ⁿ

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

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

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

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

=> i²ⁿ = 1

=> i²ⁿ = i⁴

=> 2n = 4

=> n = 2

Therefore, the smallest positive integer n for which (1+i)²ⁿ = (1−i)²ⁿ is 2.

问题25.如果z ₁ ，z ₂ ，z ₃是复数，使得| z ₁ | = | z ₂ | = | z ₃ | = $|\frac{1}{z_1}+\frac{1}{z_2}+\frac{1}{z_3}|$ = 1，然后找到| z ₁ + z ₂ + z ₃ |的值。

解决方案：

We are given,

|z₁| = |z₂| = |z₃| =$|\frac{1}{z_1}+\frac{1}{z_2}+\frac{1}{z_3}|$= 1

Now, |z₁ + z₂ + z₃| =$|\frac{z_1\bar{z_1}}{\bar{z_1}}+\frac{z_2\bar{z_2}}{\bar{z_2}}+\frac{z_3\bar{z_3}}{\bar{z_3}}|$

=$|\frac{|z_1|^2}{\bar{z_1}}+\frac{|z_2|^2}{\bar{z_2}}+\frac{|z_3|^2}{\bar{z_3}}|$

=$|\frac{1}{z_1}+\frac{1}{z_2}+\frac{1}{z_3}|$

= 1

Therefore, the value of |z₁ + z₂ + z₃| is 1.

问题26.找出z ² + | z |的解数² = 0。

解决方案：

Let z = x + iy. We have,

=> z² + |z|² = 0

=> (x + iy)² + |x + iy|² = 0

=> x² + i²y² + 2xyi + x² + y² = 0

=> x² − y² + 2xyi + x² + y² = 0

=> 2x² + 2xyi = 0

On comparing the real and imaginary parts on both sides, we get

=> 2x² = 0 and 2xy = 0

=> x = 0 and y ∈ R

Therefore, z = 0 + iy, where y ∈ R.