11类NCERT解决方案-第5章复数和二次方程式-练习5.1 |套装2

第5章复数和二次方程式–练习5.1 |套装1

对于Q.11至Q.13，找到给定数字的乘法逆

问题11. 4-3i

解决方案：

Let’s denote given number as a,

the complement of a = $\overline{a}$ = $\overline{4-3i}$

$\overline{a}$ = (4+3i)

Modulus of a = (|a|) = √((4)²+(3)²)

|a|= √(16+9)=√(25)

|a| = 5

$\frac{1}{a}$ = $\frac{\overline{a}}{|a|^2}$

$\frac{1}{a}$ = $\frac{(4+3i)}{25}$

问题12.√5+ 3i

解决方案：

Let’s denote given number as a,

the complement of a = $\overline{a} = \overline{√5+3i}$

$\overline{a}$ = √5-3i

Modulus of a (|a|) = √((√5)²+(-3)²)

|a|= √(5+9)=√(14)

|a|=√(14)

$\frac{1}{a} = \frac{\overline{a}}{|a|^2}$

$\frac{1}{a} = \frac{(5-3i)}{14}$

问题13 -i

解决方案：

Let’s denote given number as a,

the complement of a = $\overline{a} = \overline{-i}$

$\overline{a}$ = i

Modulus of a (|a|) = √((0)²+(-1)²)

|a|= √(1)

|a|=1

$\frac{1}{a} = \frac{\overline{a}}{|a|^2}$

$\frac{1}{a} = \frac{(i)}{1}$ = i

以a + ib的形式表达以下表达式

问题14。 $\frac{(3+√5i)(3-√5i)} {(√3+√2i)-(√3-√2i)}$

解决方案：

Let’s denote the given expression as z,

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

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

z = $\frac{(3+√5i)(3-√5i)} {(2√2i)}$

As we know that (a+b)(a-b) = a² – b²

z = $\frac{(3)^2-(√5i)^2}{2√2i}$

z = $\frac{(9-5i^2)}{(2√2i)}$

z = $\frac{(9+5)}{(2√2i)}$

z = $\frac{14}{(2√2i)}$

z = $\frac{7}{(√2)} . \frac{1}{(i)}$

As we can write $\frac{1}{(i)} = \frac{1}{(i)}. \frac{i}{(i)} = \frac{i}{(i^2)}$ = -i

z = $\frac{-7i}{(√2)}$

z = 0-$\frac{-7i}{(√2)}$