11类NCERT解决方案-第5章复数和二次方程式-练习5.2

找出每个复数i的模数和自变量。练习1至2。

问题1。z = – 1 – i√3

解决方案：

We have, 

z = -1 – i√3

We know that, z = r (cosθ + i sinθ) 

Therefore,

r cosθ = -1  —(1)

r sinθ = -√3  —-(2)

On Squaring and adding (1) and (2), we obtain 

r² (cos ²θ + sin ²θ) = (-1)² + (-√3)²  

r² = 1 + 3

r = √4

Since r has to positive, Therefore r = 2

Putting r = 2 on (1) and (2), we get

cosθ = -1 / 2 and sinθ = -√3 / 2

Therefore, θ = – 2π / 3 (Since cosθ and sinθ both are negative, therefore θ lies in third quadrant)

Hence, modulus and argument of z = -1 – i√3 are 2 and – 2π / 3 respectively.

问题2. z =-√3+ i

解决方案：

We have,

z = -√3 + i

We know that, z = r (cosθ + i sinθ)

Therefore,

r cosθ = -√3  —(1)

r sinθ = 1  —-(2)

On Squaring and adding (1) and (2), we obtain

r² (cos ²θ + sin ²θ) = (-√3)² + (1)²  

r²  = 3 + 1   (Since, cos ²θ + sin ²θ = 1)

r² = 3 + 1

r = √4

Since r has to positive, Therefore r = 2

Putting r = 2 on (1) and (2), we get

cosθ = -√3 / 2 and sinθ = 1 / 2

Therefore, θ = 5π / 6 (Since cosθ negative and sinθ positive, therefore θ lies on second quadrant)

Hence, modulus and argument of z = -√3 + i are 2 and 5π / 6 respectively.

将练习3中给出的每个复数以极形式转换为：

问题3. 1 –我

解决方案：

We have z = 1 – i,

Let  r cosθ  = 1   —(1)  and,

      r sinθ  = -1 —(2)

On Squaring and adding  (1) and (2) , we obtain

r² (  cos ²θ + sin ²θ  )  =  (1)² + (-1)²  

r² =  2

r = √2 ( Since r has to be positive )

Putting r = √2 on (1) and (2) , we get

cosθ = 1 / √2  and  sinθ = -1 / √2

Therefore, θ =  – π / 4 ( Since cosθ positive and sinθ  negative, therefore θ is negative as it lies on fourth quadrant)

Hence , z in polar form:  z =  r cosθ  + i  r sinθ = √2 (cos (- π / 4) + i sin (- π / 4)).

问题4. -1 + i

解决方案：

We have z = -1 + i,

Let  r cosθ  = -1   —(1)  and,

      r sinθ  = 1 —(2)

On Squaring and adding (1) and (2), we obtain

r² (cos ²θ + sin ²θ)  = (-1)² + (1)²  

r² = 2

r = √2 (Since r has to be positive)

Putting r = √2 on (1) and (2), we get

cosθ = -1 / √2 and sinθ = 1 / √2

Therefore, θ = 3π / 4 (Since cosθ negative and sinθ positive, therefore θ is positive as it lies on second quadrant)

Hence, z in polar form:  z = r cosθ + i  r sinθ = √2 (cos (3π / 4) + i sin (3π / 4)).

问题5. -1 – i

解决方案：

We have z = -1 – i,

Let  r cosθ  = -1   —(1)  and,

     r sinθ  = -1 —(2)

On Squaring and adding  (1) and (2), we obtain

r² (cos ²θ + sin ²θ)  =  (-1)² + (-1)²  

r² =  2

r = √2 ( Since r has to be positive )

Putting r = √2 on (1) and (2) , we get

cosθ = -1 / √2  and  sinθ = -1 / √2

Therefore,  θ =  -3π / 4 (Since  cosθ  negative and sinθ  negative, therefore θ is negative as it lies on third quadrant)

Hence, z in polar form:  z =  r cosθ  + i  r sinθ  = √2 (cos (-3π / 4) + i sin (-3π / 4)).

问题6. -3

解决方案：

We have z = -3,

Let  r cosθ  = -3   —(1)  and,

      r sinθ  = 0 —(2)

On Squaring and adding (1) and (2), we obtain

r² (cos ²θ + sin ²θ) =  (-3)² + (0)²  

r² = 9

r = 3 (Since r has to be positive)

Putting r = 3 on (1) and (2), we get

cosθ = -3 / 3  and  sinθ = 0 / 3

Therefore,  θ =  π

Hence, z in polar form:  z =  r cosθ  + i  r sinθ = 3(cos (π) + i sin (π)).

问题7.√3+ i

解决方案：

We have z = √3 + i,

Let  r cosθ  = √3  —(1)  and,

      r sinθ  = 1 —(2)

On Squaring and adding  (1) and (2) , we obtain

r² (cos ²θ + sin ²θ) = (√3)² + (1)²  

r² = 4

r = 2 (Since r has to be positive )

Putting r = 2 on (1) and (2), we get

cosθ = √3 / 2 and  sinθ = 1 / 2

Therefore, θ =  π / 6 ( Since  cosθ  positive and sinθ  positive, therefore θ is positive as it lies on first quadrant)

Hence, z in polar form:  z =  r cosθ  + i  r sinθ  = 2 (cos (π / 6) + i sin (π / 6)).

问题8.我

解决方案：

We have z = i,

Let  r cosθ  = 0 —(1)  and,

      r sinθ  = 1 —(2)

On Squaring and adding  (1) and (2) , we obtain

r² (cos ²θ + sin ²θ)  =  (0)² + (1)²  

r² =  1

r = 1 (Since r has to be positive)

Putting r = 1 on (1) and (2), we get

cosθ = 0 / 1 and sinθ = 1 / 1

Therefore, θ = π / 2

Hence, z in polar form: z = r cosθ + i r sinθ = cos (π / 2) + i sin (π / 2)