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Class 11 NCERT Solutions- Chapter 5 Complex Numbers And Quadratic Equations – Exercise 5.1 | Set 2

  • Last Updated : 05 Apr, 2021

Chapter 5 Complex Numbers And Quadratic Equations – Exercise 5.1 | Set 1

For Q.11 to Q.13 find the multiplicative inverse of the given number 

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) = 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)}

Attention reader! Don’t stop learning now. Participate in the Scholorship Test for First-Step-to-DSA Course for Class 9 to 12 students.




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