11类NCERT解决方案–第5章复数和二次方程式–第5章的其他练习|套装2

问题11：如果a + ib = $\frac{(x+i)^2}{2x^2+1}$ ，证明a ² + b ² = $\frac{(x^2+i)^2}{(2x^2+1)^2}$

解决方案：

Given:

a + ib = $\frac{(x+i)^2}{2x^2+1}\\ =\frac{x^2+i^2+2xi}{2x^2+1}\\ =\frac{x^2-1+i2x}{2x^2+1}\\ =\frac{x^2-1}{2x^2+1}+i\left(\frac{2x}{2x^2+1}\right)$

On comparing the real and imaginary parts, we have

a = $\frac{(x-1)}{2x^2+1}$ and b = $\frac{2x}{2x^2+1}$

Therefore,

a² + b² = $\left(\frac{x^2-1}{2x^2+1}\right)^2+\left(\frac{2x}{2x^2+1}\right)^2\\ =\frac{x^4+1-2x^2+4x^2}{(2x+1)^2}\\ =\frac{x^4+1+2x^2}{(2x^2+1)^2}\\ =\frac{(x^2+1)^2}{(2x^2+1)^2}$

Hence, proved,

a² + b² = $\frac{(x^2+1)^2}{(2x^2+1)^2}$

问题12。设z ₁ = 2 – i ，z ₂ = -2 + i 。找

(一世) $Re\left(\frac{z_1z_2}{\overline{z_1}}\right)$

(ii) $Im\left(\frac{1}{z_1\overline{z_2}}\right)$

解决方案：

(i) Given:

z₁ = 2 – i, z₂ = -2 + i

(i) z₁z₂ = (2 – i)(-2 + i) = -4 + 2i + 2i – i² = -4 + 4i – (-1) = -3 + 4i

$\overline{z_1}$ = 2 + i

Therefore,

$\frac{z_1z_2}{\overline{z_1}}=\frac{-3+4i}{2+i}$

On multiplying numerator and denominator by (2 – i), we get

$\frac{z_1z_2}{\overline{z_1}}=\frac{(-3+4i)(2-i)}{(2+i)(2-i)}=\frac{-6+3i+8i-4i^2}{2^2+1^2}=\frac{-6+11i-4(-1)}{2^2+1^2}\\ =\frac{-2+11i}{5}=\frac{-2}{5}+\frac{11}{5}i$

On comparing the real parts, we have

$Re\left(\frac{z_1z_2}{\overline{z_1}}\right)=\frac{-2}{5}$

(ii) $\frac{1}{z_1\overline{z_2}}=\frac{1}{(2-1)(2+i)}=\frac{1}{(2)^2+(1)^2}=\frac{1}{5}\\$

On comparing the imaginary part, we get

$Im\left(\frac{1}{z_1\overline{z_2}}\right)$ = 0

问题13.查找复数的模和自变量$\frac{1+2i}{1-3i}$

解决方案：

Let, z = $\frac{1+2i}{1-3i}$, then

z = $\frac{1+2i}{1-3i}\times\frac{1+3i}{1+3i}=\frac{1+3i+2i+6i^2}{1^2+3^6}=\frac{1+5i+6(-1)}{1+9}\\ =\frac{-5+5i}{10}=\frac{-5}{10}+\frac{5i}{10}=\frac{-1}{2}+\frac{1}{2}i$

Let z = r cosθ + ir sinθ

So,

r cosθ = $\frac{-1}{2}$ and r sinθ = $\frac{1}{2}$

On squaring and adding, we get

r2(cos2θ + sin2θ) = $\left(\frac{-1}{2}\right)^2+\left(\frac{1}{2}\right)^2$

r² = $\frac{1}{4}+\frac{1}{4}=\frac{1}{2}\ \ \ \ \ \ [Conventionally,\ r>0]$

r = $\frac{1}{\sqrt2}$

Now,

$\frac{1}{\sqrt2}$cosθ = $\frac{-1}{2}$ and $\frac{1}{\sqrt2}$sinθ = $\frac{1}{2}$

= cosθ = $\frac{-1}{\sqrt2}$ and sinθ = $\frac{1}{\sqrt2}$

Therefore,

θ = $\pi-\frac{\pi}{4}=\frac{3\pi}{4}$     [As θ lies in the II quadrant]

问题14.如果( x – iy )(3 + 5 i )是– 6 – 24 i的共轭数，请求出实数x和y 。

解决方案：

Let us assume z = (x – iy) (3 + 5i)

z = 3x + xi – 3yi – 5yi² = 3x + 5xi – 3yi + 5y = (3x + 5y) + i(5x – 3y)

Therefore,

$\overline{z}$ =(3x + 5y) – i(5x – 3y)

Also given, $\overline{z}$ = -6 – 24i 

And,

(3x + 5y) – i(5x – 3y) = -6 -24i

After equating real and imaginary parts, we get

3x + 5y = -6 …… (i)

5x – 3y = 24 …… (ii)

After doing (i) x 3 + (ii) x 5, we have

(9x + 15y) + (25x – 15y) = -18 + 120

34x = 102

x = 102/34 = 3

Putting the value of x in equation (i), we get

3(3) + 5y = -6

5y = -6 – 9 = -15

y = -3

Therefore, the values of x and y are 3 and –3 respectively.

问题15：求出模数$\frac{1+i}{1-i}-\frac{1-i}{1+i}$

解决方案：

$\frac{1+i}{1-i}-\frac{1-i}{1+i}=\frac{(1+i)^2-(1-i)^2}{(1-i)(1+i)}\\ =\frac{1+i^2+2i-1-i^2+2i}{1^2+1^2}\\ =\frac{4i}{2}=2i\\ \therefore\left|\frac{1+i}{1-i}-\frac{1-i}{1+i}\right|=|2i|=\sqrt{2^2}=2$

问题16.如果( x + iy ) ³ = u + iv ，则表明$\frac{u}{y}+\frac{v}{y}$ = 4(x ² – y ² )

解决方案：

(x + iy)³ = u + iv

x³ + (iy)³  + 3 × x × iy(x + iy) = u + iv

x³ + i³y³ + 3x²yi + 3xy² = u + iv

x³ – iy³ + 3x²yi – 3xy² = u + iv

(x³ – 3xy²) + i(3x²y – y³) = u + iv

On equating real and imaginary parts, we get

u = x³ – 3xy², v = 3x²y – y³

$\frac{u}{x}+\frac{v}{y}=\frac{x^3-3xy^2}{x}+\frac{3x^2y-y^3}{y}\\ =\frac{x(x^2-3y^2)}{x}+\frac{y(3x^2y-y)}{y}$

= x² – 3y² + 3x² – y²

= 4x² – 4y²

= 4(x² – y²)

$\therefore\frac{u}{x}+\frac{v}{y}=4(x^2-y^2)$

Hence proved

问题17：如果α和β是带|β|的不同复数= 1，然后找到$\left|\frac{\beta-\alpha}{1-\overline{\alpha}\beta}\right|$

解决方案：

Assume α = a + ib and β = x + iy

Given: |β| = 1

So, $\sqrt{x^2+y^2}=1$

= x2 + y2 = 1            ….(1)

$\left|\frac{\beta-\alpha}{1-\overline{\alpha}\beta}\right|=\left|\frac{(x+iy)(a+ib)}{1-(a-ib)(x+iy)}\right|\\ =\left|\frac{(x-a)+i(y-b)}{1-(ax+aiy-ibx+by)}\right|\\ =\left|\frac{(x-a)+i(y-b)}{(1-ax-by)+i(bx-ay)}\right|\\ =\left|\frac{(x-a)+i(y-b)}{(1-ax-by)+i(bx-ay)}\right|\ \ \ \ \ \left[\left|\frac{z_1}{z_2}\right|=\left|\frac{z_1}{z_2}\right|\right]\\ =\frac{\sqrt{(x-a)^2+(y-b)^2}}{\sqrt{(1-ax-by)^2+(bx-ay)^2}}\\ =\frac{\sqrt{x^2+a^2-2ax+y^2+b^2-2by}}{\sqrt{1+a^2x^2+b^2y^2-2ax+2abxy-2by+b^2x^2+a^2y^2-2abxy}}\\ =\frac{\sqrt{(x^2+y^2)+a^2+b^2-2ax-2by}}{\sqrt{1+a^2(x^2+y^2)+b^2(y^2+x^2)-2ax-2by}}\\ =\frac{\sqrt{1+a^2+b^2-2ax-2by}}{\sqrt{1+a^2+b^2-2ax-2by}}\ \ \ \ \ \ \ \ [Using\ (1)]$ 

= 1

$\therefore\left|\frac{\beta-\alpha}{1-\overline{\alpha}\beta}\right|=1$

问题18。找到方程| 1 – i |的非零积分解的数量。 ^x = 2 ^x

解决方案：

|1 – i|^x = 2^t

$(\sqrt{1^2+(-1)^2})^x=2^x\\ (\sqrt2)^x=2^x\\ 2^{\frac{x}{2}}=2^x\\ \frac{x}{2}=x$

x = 2x

2x – x = 0

Thus, ‘0’ is the only integral solution of the given equation.

Therefore, the number of non-zero integral solutions of the given equation is 0.

问题19.如果(a + ib)(c + id)(e +如果)(g + ih)= A + iB，则证明(a ² + b ² )(c ² + d ² )(e ² + f ² )(g ² + h ² )= A ² + B ² 。

解决方案：

Given:

(a + ib)(c + id)(e + if)(g + ih) = A + iB

Therefore,

|(a + ib)(c + id)(e + if)(g + ih)| = |A + iB|

= |(a + ib)| × |(c + id)| × |(e + if)| × |(g + ih)| = |A + iB|

$\sqrt{a^2+b^2}\times\sqrt{c^2+d^2}\times\sqrt{e^2+f^2}\times\sqrt{g^2+h^2}=\sqrt{A^2+B^2}$

On squaring both sides, we get

(a² + b²) (c² + d²) (e² + f²) (g² + h²) = A² + B²

Hence, proved.

问题20.如果，则找到m的最小正整数值。  $\left(\frac{1+i}{1-i}\right)^m=1$

解决方案：

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

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

i^m = 1

Hence, m = 4k, where k is some integer.

Hence, the least positive integer is 1.

Thus, the least positive integral value of m is 4 (= 4 × 1).