NCERT Solutions Class 11 Chapter 4 Complex Numbers And Quadratic Equations Exercise 4.1 (original) (raw)
Last Updated : 19 Apr, 2024
For Q.1 to Q.10 express each complex number in form of a+ib
**Question 1. (5i)(\frac{-3i}{5})
**Solution:
Let the given number be a,
a= (5i)*(\frac{-3i}{5})
a= \frac{-15i^2}{5}
a= (-3)*i2
a= (-3)*(-1)
a= 3+0i
**Question 2. i 9 +i 19
**Solution:
Let the given number be a,
a = i9 * (1+i10)
a = ((i4)2*i )(1 + (i4)2 (i2))
a = (1*i)(1+i2)
a = (i)*(0)
a = 0+0i
**Question 3. i -39
**Solution:
Let the given number be a and let z = i39 ,
z = (i)*(i2)19
z = (i)*(-1)19
z = -i
a = i-39
a = 1/i39
a = 1/z
a = 1/-i
a = (i4)/-i
a = -i3 = -(i2*i)
a = -1*-i
a = 0+i
**Question 4. 3(7+7i) + i(7+7i)
**Solution:
Let the given number be a,
a = 3*(7+7i)+i*(7+7i)
a = 21+21i+7i+7i2
a = 21+7i2+28i
a = 21-7+28i
a = 14+28i
**Question 5. (1-i)-(-1+i6)
**Solution:
Let the given number be a,
a = (1-i)-(-1+6i)
a = 1-i+1-6i
a = 2-7i
****Question 6. (**\frac{1}{5}+\frac{2}{5}i ****)-(4+**\frac{5i}{2} ****)**
**Solution:
Let the given number be a,
a = (\frac{1}{5}+\frac{2i}{5})-(4+\frac{5i}{2})
a = (\frac{1}{5}-4)+(\frac{2i}{5}-\frac{5i}{2})
a = (\frac{-19}{5})+(\frac{2}{5}-\frac{5}{2})i
a = (\frac{-19}{5} )+(\frac{-21i}{10} )
a =\frac{-38-21i}{10}
****Question 7. [(**\frac{1}{3}+\frac{7i}{3} ****)+(4+**\frac{i}{3} ****]-(**\frac{-4}{3} ****+i)**
**Solution:
Let the given number be a,
a = (\frac{1}{3} +\frac{7i}{3} )+(4+\frac{i}{3} )-(\frac{-4}{3} +i)
a = (\frac{1}{3} +4+\frac{4}{3} )+(\frac{7i}{3}+\frac{i}{3} -i)
a = (\frac{5}{3} +4)+(\frac{8i}{3} -i)
a = \frac{17}{3}+ \frac{5i}{3}
a = \frac{17+5i}{3}
**Question 8. (1-i) 4
**Solution:
Let the given number be a,
a = ((1-i)2)2
As we know , ****(a-b)** 2 = (a 2 +b 2 -2ab)
a = (1+i2-2i)2
a = (1-1-2i)2
a = (-2i)2
a = 4i2
a = -4+0i
****Question 9. (**\frac{1}{3} ****+3i)** 3
**Solution:
Let the given number be a,
a = (\frac{1}{3} +3i)3
As we know, ****(a+b)** 3 = (a 3 +b 3 +3ab(a+b))
a = ((\frac{1}{27} )+(3i)3 +3(\frac{1}{3} )*(3i)(\frac{1}{3} +3i))
a = (\frac{1}{27} +(-27i)+ 3i*(\frac{1}{3} +3i))
a = (\frac{1}{27} +(-27i)+i+9i2)
a = ((\frac{1}{27} )-9+(-27)i+i)
a = ((\frac{-242}{27} )-26i)
****Question 10. (-2-(**\frac{i}{3} ****))** 3
**Solution:
Let the given number be a,
a = (-2-\frac{i}{3} )3
a = -((2+\frac{i}{3} )3)
As we know, ****(a+b)** 3 = (a 3 +b 3 +3ab(a+b))
a = -((8)+(\frac{i}{3} )3 +3(2)*(\frac{i}{3} )(2+\frac{i}{3} ))
a = -(8+(\frac{-i}{27} )+ 2i*(2+\frac{i}{3} ))
a = -(8-\frac{i}{27} +4i+\frac{2i^2}{3} )
a = -(8-\frac{2}{3} +(\frac{-i}{27} )+4i)
a = -(\frac{22}{3} +(\frac{107i}{27} ))
a = \frac{-22}{3}-\frac{107i}{27}
**Question 11. 4-3i
**Solution:
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}
**Question 12. √5+3i
**Solution:
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}
**Question 13. -i
**Solution:
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
Express the following expression in form of a+ib
**Question 14. \frac{(3+√5i)(3-√5i)} {(√3+√2i)-(√3-√2i)}
**Solution:
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) = a** 2 - b 2
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)}