Class 11 NCERT

NCERT

1   Express the given complex number in the form $a+ib:(5i)(-\dfrac{3}{5}i)$

Solution :

$( 5 i ) ( - \dfrac {3} {5} i ) \\ = - 5 * \dfrac {3} {5} * i * i \\ = - 3 i^2 \\ = - 3 (-1) \ \ \ \ [ i^2 = - 1 ] \\ = 3 \\ = 3 + i(0)$

2   Express the given complex number in the form $a+ib:i^9+i^{19}$

Solution :

$i^9+i^{19}\\ =i^{4*2+1}+i^{4*4+3}\\ =(i^4)^2.i+(i^4)^4.i^3\\ =1*i+1*(-i) \ \ \ \ \ \ \ [i^4=1,i^3=-i]\\ =i+(-i)\\ =0$

3   Express the given complex number in the form $a+ib:i^{-39}$

Solution :

$i^{-39}=i^{-4*9-3}\\ =(i^4)^{-9}.i^{-3}\\ =(1)^{-9}.i^{-3} \ \ \ [i^4=1]\\ =\dfrac{1}{i^3}=\dfrac{1}{-i} \ \ \ \ \ \ \ \ [i^3=-i]\\ =\dfrac{-1}{i}*\dfrac{i}{i}\\ =\dfrac{-i}{i^2}\\ =\dfrac{-i}{-1}=i \ \ \ \ \ \ [i^2=-1]$

4   Express the given complex number in the form a $A+ib:3(7+i7)+i(7+i7)$

Solution :

$3 ( 7 + i 7 ) + i ( 7 + i 7 ) \\ = 21 + 21 i + 7 i + 7 i^2 \\ = 21 + 28 i + 7 * ( -1 ) \ \ \ \ \ \ [ \therefore i^2 = - 1 ] \\ = 14 + 28 i$

5   Express the given complex number in the form $a+ib:(1-i)-(-1+i6).$

Solution :

$( 1 - i ) - ( - 1 + i 6 ) \\ = 1 - i + 1 - 6 i \\ = 2 - 7 i$

6   Express the given complex number in the form $a + ib :(\dfrac{1}{5}+i\dfrac{2}{5})-(4+i\dfrac{5}{2})$

Solution :

$(\dfrac{1}{5}+i\dfrac{2}{5})-(4+i\dfrac{5}{2})\\ =\dfrac{1}{5}+\dfrac{2}{5}i-4-\dfrac{5}{2}i\\ =(\dfrac{1}{5-4})+i(\dfrac{2}{5}-\dfrac{5}{2})\\ =\dfrac{-19}{5}+i(\dfrac{-21}{10})\\ =\dfrac{-19}{5}-\dfrac{21}{10}i$

7   Express the given complex number in the form $a + ib : [(\dfrac{1}{3}+i\dfrac{7}{3})+(4+i\dfrac{1}{3})]-(-\dfrac{4}{3}+i)$$\\ Solution : [(\dfrac{1}{3}+i\dfrac{7}{3})+(4+i\dfrac{1}{3})]-(-\dfrac{4}{3}+i)\\ =\dfrac{1}{3}+\dfrac{7}{3}i+4+\dfrac{1}{3}i+\dfrac{4}{3}-i\\ =(\dfrac{1}{3}+4+\dfrac{4}{3})+i(\dfrac{7}{3}+\dfrac{1}{3}-1)\\ =\dfrac{17}{3}+i\dfrac{5}{3} 8 Express the given complex number in the form a + ib : ( 1 - i )^4 Solution : (1-i)^4=[(1-i)^2]^2\\ =[1^2+i^2-2i]^2\\ =[1-1-2i]^2\\ =(2i)^2\\ =(-2i)*(-2i)\\ =4i^2=-4 \quad [i^2=-1]$$\\$

9   Express the given complex number in the form $a + ib : (\dfrac{1}{3}+3i)^3$

Solution :

$(\dfrac{1}{3}+3i)^3=(\dfrac{1}{3})^3+(3i)^3+3(\dfrac{1}{3})(3i)(\dfrac{1}{3}+3i)\\ =\dfrac{1}{27}+27i^3+3i(\dfrac{1}{3}+3i)\\ =\dfrac{1}{27}+27(-i)+i+9i^2 \quad [i^3=-i]\\ =\dfrac{1}{27} -27i+i-9 \quad [i^2=-1]\\ =(\dfrac{1}{27}-9)+i(-27+1)\\ =\dfrac{-242}{27}-26i$

10   Express the given complex number in the form $a + ib :(-2-\dfrac{1}{3}i)^3$

Solution :

$(-2-\dfrac{1}{3}i)^3=(-1)^3(2+\dfrac{1}{3}i)^3\\ =-[2^3+(\dfrac{i}{3})^3+3(2)(\dfrac{i}{3})(2+\dfrac{i}{3})]\\ =-[8+\dfrac{i^3}{27}+2i(2+\dfrac{i}{3})]\\ =-[8-\dfrac{i}{27}+4i+(2i^2){3}]\quad [i^3=-i]\\ =-[8-\dfrac{i}{27}+4i-\dfrac{2}{3}] \quad [i^2=-1]\\ =-[\dfrac{22}{3}+\dfrac{107i}{27}]\\ =-\dfrac{22}{3}-\dfrac{107}{27}i$