NCERT Exemplar Solutions for Class 11 Maths Chapter 5 Complex Numbers and Quadratic Equations are considered to be very useful when you are preparing for the CBSE Class 11 Maths exams. Here, we bring to you detailed solutions to the exercises of NCERT Exemplar Class 11 Maths Chapter 5. Subject matter experts who created the solutions have compiled these questions for you to revise from the chapter 5 of NCERT Exemplar Textbook. These questions have been devised, as per the NCERT syllabus and the guidelines. We provide to you the accurate solutions to all the questions that are covered in the NCERT Exemplar books. These NCERT Exemplar Solutions will rely on the syllabus and its guidelines. You will get enough practice solving these exercises and it will also help you to score high marks.

This chapter defines the concept of Complex numbers and Quadratic Equations. A complex number can simply be defined as a real number, a pure imaginary number or even the sum of both the numbers. The real number can be represented as â€˜aâ€™ and the imaginary number can be represented as â€˜biâ€™. Quadratic equations are related to imaginary numbers and complex numbers only when the value of the quadratic formula is negative. In Chapter 5 of NCERT Exemplar Solutions for Class 11 Maths, students will learn and solve exemplar problems based on;

- Imaginary numbers
- Integral powers of i
- Complex numbers
- Conjugate of a complex number
- Modulus of a complex number
- Properties of modulus of a complex number
- Develop the algebra of complex numbers
- Addition, Difference, Multiplication and Division of two complex numbers
- Power of imaginary number i
- The square roots of a negative real number
- Argand Plane and Polar Representation of the complex number
- Solutions for quadratic equations

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### Access Answers of NCERT Exemplar Class 10 Maths Chapter 5 – Complex Numbers And Quadratic Equations

Exercise Page No: 91

**Short Answer Type**

1. **For a positive integer n, find the value of (1 â€“ i) ^{n}**

**(1 â€“ 1/i)**

^{n}.**Solution:**

** **According to the question,

We have,

= (1 â€“ i)^{n}Â (1 + i)^{n}

= (1 â€“ i^{2})^{n}

= 2^{n}

Therefore, (1 â€“ i)^{n} (1 â€“ 1/i)^{n} = 2^{n}

**2. Evaluate **

**Â where nÂ ÏµÂ N.**

**Solution:**

** **According to the question,

We have,

** **

**3. IfÂ **

**,Â then find (x, y).**

**Solution:**

According to the question,

We have,

= i^{3}Â â€“ (-i^{3})

= 2i^{3}

= 0 â€“ 2i

Thus, (x, y) = (0, -2)

**4. IfÂ **

**, then find the value of x + y.**

**Solution:**

** **According to the question,

We have,

**5. IfÂ **

**Â then find (a, b).**

**Solution:**

** **According to the question,

We have,

= (i^{4})^{25}

= 1

Hence, (a, b) = (1, 0)

6. **If a = cos Î¸ + i sin Î¸, find the value ofÂ **

**Solution:**

According to the question,

We have,

a = cos Î¸ + i sin Î¸

**7. If (1 + i)z = (1 â€“ i)Â zÌ…**,**Â then show thatÂ z = i** zÌ…**.**

**Solution:**

According to the question,

We have,

= -izÌ…

Hence proved.

**8. If z = x + iy, then show that zzÌ… + 2(z + zÌ…) + b = 0 where bÏµR, representing z in the complex plane is a circle.**

**Solution:**

According to the question,

We have,

z = x + iy

â‡’Â zÌ…Â = x â€“ iy

Now, we also have,

z zÌ…Â + 2 (z + zÌ…) + b = 0

â‡’Â (x + iy) (x â€“ iy) + 2 (x + iy + x â€“ iy) + b = 0

â‡’Â x^{2}Â + y^{2}Â + 4x + b = 0

The equation obtained represents the equation of a circle.

**9. If the real part of ( zÌ… + 2)/ ( zÌ… â€“ 1)Â is 4, then show that the locus of the point representing z in the complex plane is a circle.**

**Solution:**

According to the question,

Let z = x + iy

Now,

â‡’Â x^{2}Â + x â€“ 2 + y^{2}Â = 4 (x^{2}Â â€“ 2x + 1 + y^{2})

â‡’Â 3x^{2}Â + 3y^{2}Â â€“ 9x + 6 = 0

The equation obtained represents the equation of a circle.

Hence, locus of z is a circle.

**10. Show that the complex number z, satisfying the condition arg ((z-1)/(z+1)) = Ï€/4Â lies on a circle.**

**Solution:**

According to the question,

Let z = x + iy

arg ((z-1)/(z+1)) = Ï€/4Â

â‡’Â arg (z â€“ 1) â€“ arg (z + 1) = Ï€/4

â‡’Â arg (x + iy â€“ 1) â€“ arg (x + iy + 1) = Ï€/4

â‡’Â arg (x â€“ 1 + iy) â€“ arg (x + 1 + iy) = Ï€/4

â‡’Â x^{2}Â + y^{2}Â â€“ 1 = 2y

â‡’Â x^{2}Â + y^{2}Â â€“ 2y â€“ 1 = 0

The equation obtained represents the equation of a circle.

11. **Solve that equation |z| = z + 1 + 2i.**

**Solution:**

According to the question,

We have,

|z| = z + 1 + 2i

Substituting z = x + iy, we get,

â‡’Â |x + iy| = x + iy + 1 + 2i

We know that,

|z| = âˆš(x^{2} + y^{2})

âˆš(x^{2} + y^{2}) = (x + 1) + i(y + 2)

Comparing real and imaginary parts,

We get,

âˆš(x^{2} + y^{2}) = (x + 1)

And 0 = y + 2

â‡’Â y = -2

Substituting the value of y inÂ âˆš(x^{2} + y^{2}) = (x + 1),

We get,

â‡’Â x^{2}Â + (-2)^{2}Â = (x + 1)^{2}

â‡’Â x^{2}Â + 4 = x^{2}Â + 2x + 1

Hence, x = 3/2

Hence, z = x + iy

= 3/2 â€“ 2i

**Long Answer Type**

**12. If |z + 1| = z + 2 (1 + i), then find z.**

**Solution:**

** **According to the question,

We have,

|z + 1| = z + 2 (1 + i)

Substituting z = x + iy, we get,

â‡’Â |x + iy + 1| = x + iy + 2 (1 + i)

We know,

|z| = âˆš(x^{2} + y^{2})

âˆš((x + 1)^{2} + y^{2}) = (x + 2) + i(y + 1)

Comparing real and imaginary parts,

â‡’ âˆš((x + 1)^{2} + y^{2}) = x + 2

And 0 = y + 2

â‡’Â y = -2

Substituting the value of y inÂ âˆš((x + 1)^{2} + y^{2}) = x + 2,

â‡’Â (x + 1)^{2}Â + (-2)^{2}Â = (x + 2)^{2}

â‡’Â x^{2}Â + 2x + 1 + 4 = x^{2}Â + 4x + 4

â‡’Â 2x = 1

Hence, x = Â½

Hence, z = x + iy

= Â½ â€“ 2i

**13. If arg (z â€“ 1) = arg (z + 3i), then find x â€“ 1 : y. where z = x + iy**

**Solution:**

** **According to the question,

Let z = x + iy

Given that,

arg (z â€“ 1) = arg (z + 3i)

â‡’Â arg (x + iy â€“ 1) = arg (x + iy + 3i)

â‡’Â arg (x â€“ 1 + iy) = arg (x + I (y) = Ï€/4

â‡’Â xy = xy â€“ y + 3x â€“ 3

â‡’Â 3x â€“ 3 = y

â‡’Â (x â€“ 1)/y = 1/3

Hence, (x â€“ 1): y = 1: 3

**14. Show thatÂ |(z â€“ 2) / (z â€“ 3)| = 2Â represents a circle. Find its centre and radius.**

**Solution:**

** **According to the question,

We have,

|(z â€“ 2) / (z â€“ 3)| = 2Â

Substituting z = x + iy, we get

â‡’ |(x + iy â€“ 2) / (x + iy â€“ 3)| = 2

â‡’Â |x â€“ 2 + iy| = 2 |x â€“ 3 + iy|

â‡’ âˆš((x â€“ 2)^{2} + y^{2}) = 2âˆš((x â€“ 3)^{2} + y^{2})

â‡’Â x^{2}Â â€“ 4x + 4 + y^{2}Â = 4 (x^{2}Â â€“ 6x + 9 + y^{2})

â‡’Â 3x^{2}Â + 3y^{2}Â â€“ 20x + 32 = 0

Therefore, centre of circle is (10/3, 0) and radius is 4/9 or 2/3.

**15. IfÂ (z â€“ 1)/(z + 1) is a purely imaginary number (z â‰ â€“1), then find the value of |z|.**

**Solution:**

** **According to the question,

Let z = x + iy

Now,

â‡’Â x^{2}Â â€“ 1 + y^{2}Â = 0

â‡’Â x^{2}Â + y^{2}Â = 1

â‡’Â âˆš(x^{2}Â + y^{2})Â = 1

Hence, |z| = 1

**16. z _{1}Â and z_{2}Â are two complex numbers such that |z_{1}| = |z_{2}| and arg (z_{1}) + arg (z_{2}) = Ï€, then show thatÂ z_{1} = â€“ zÌ…_{2.}**

**Solution:**

** **According to the question,

Let z_{1}Â = |z_{1}| (cos Î¸_{1}Â + I sin Î¸_{1}) and z_{2}Â = |z_{2}| (cos Î¸_{2}Â + I sin Î¸_{2})

Given that |z_{1}| = |z_{2}|

And arg (z_{1}) + arg (z_{2}) = Ï€

â‡’Â Î¸_{1}Â + Î¸_{2}Â = Ï€

â‡’Â Î¸_{1}Â = Ï€ â€“ Î¸_{2}

Now, z_{1}Â = |z_{2}| (cos (Ï€ – Î¸_{2}) + I sin (Ï€ – Î¸_{2}))

â‡’Â z_{1}Â = |z_{2}| (-cos Î¸_{2}Â + I sin Î¸_{2})

â‡’Â z_{1}Â = -|z_{2}| (cos Î¸_{2}Â â€“ I sin Î¸_{2})

â‡’Â z_{1}Â = – [|z_{2}| (cos Î¸_{2}Â â€“ I sin Î¸_{2})]

Hence, z_{1}Â = -zÌ…_{2}

Hence proved.

**17. If |z _{1}| = 1 (z_{1}Â â‰ â€“1) and z_{2} = (z_{1} â€“ 1) / (z + 1),Â then show that the real part of z_{2}Â is zero.**

**Solution:**

According to the question,

Let z_{1}Â = x + iy

Therefore, the real part of z_{2}Â is zero.

**18. If z _{1}, z_{2}Â and z_{3}, z_{4}Â are two pairs of conjugate complex numbers, then find arg(z_{1}/z_{4}) + arg(z_{2}/z_{3}).**

**Solution:**

According to the question,

We have,

z_{1}Â and z_{2}Â are conjugate complex numbers.

The negative side of the real axis

= r_{1}Â (cos Î¸_{1}Â – i sin Î¸_{1})

= r_{1}Â [cos (-Î¸_{1}) + I sin (-Î¸_{1})]

Similarly, z_{3}Â = r_{2}Â (cos Î¸_{2}Â – i sin Î¸_{2})

â‡’Â z_{4}Â = r_{2}Â [cos (-Î¸_{2}) + I sin (-Î¸_{2})]

= Î¸_{1}Â â€“ (-Î¸_{2}) + (-Î¸_{1}) â€“ Î¸_{2}

= Î¸_{1}Â + Î¸_{2}Â â€“ Î¸_{1}Â â€“ Î¸_{2}

= 0

â‡’ arg(z_{1}/z_{4}) + arg(z_{2}/z_{3}) = 0

**19.** **If |z _{1}| = |z_{2}| = â€¦.. = |z_{n}| = 1, then show that |z_{1}Â + z_{2}Â + z_{3}Â + â€¦. + z_{n}|Â = | 1/z_{1} + 1/z_{2} + 1/z_{3} + â€¦ + 1/z_{n}|**

**Solution:**

According to the question,

We have,

|z_{1}| = |z_{2}| = â€¦ = |z_{n}| = 1

â‡’Â |z_{1}|^{2}Â = |z_{2}|^{2}Â = â€¦ = |z_{n}|^{2}Â = 1

â‡’Â z_{1}Â zÌ…Â _{1}= z_{2}Â zÌ…Â _{2}= z_{3}Â zÌ…Â _{3}= â€¦ = z_{n}Â zÌ…Â _{n}= 1

Hence proved.

**20. If for complex numbers z _{1}Â and z_{2}, arg (z_{1}) â€“ arg (z_{2}) = 0, then show that |z_{1}Â â€“ z_{2}| = |z_{1}| â€“ |z_{2}|.**

**Solution:**

** **According to the question,

Let z_{1}Â = |z_{1}| (cos Î¸_{1}Â + I sin Î¸_{1}) and z_{2}Â = |z_{2}| (cos Î¸_{2}Â + I sin Î¸_{2})

We have,

arg (z_{1}) – arg (z_{2}) = 0

â‡’Â Î¸_{1}Â – Î¸_{2}Â = 0

â‡’Â Î¸_{1}Â = Î¸_{2}

We also have,

z_{2}Â = |z_{2}| (cos Î¸_{1}Â + I sin Î¸_{1})

â‡’Â z_{1}Â â€“ z_{2}Â = ((|z_{1}|cos Î¸_{1}Â – |z_{2}| cos Î¸_{1}) + i (|z_{1}| sin Î¸_{1}Â – |z_{2}| sin Î¸_{1}))

Hence, |z_{1}Â â€“ z_{2}| = |z_{1}| – |z_{2}|

Hence proved.

**21. Solve the system of equations Re (z _{2}) = 0, |z| = 2.**

**Solution:**

** **According to the question,

We have,

Re (z^{2}) = 0, |z| = 2

Let z = x + iy.

Then, |z| = âˆš(x^{2} + y^{2}) Â

Given in the question,

âˆš(x^{2} + y^{2}) Â = 2

â‡’Â x^{2}Â + y^{2}Â = 4 â€¦ (i)

z^{2}Â = x^{2}Â + 2ixy â€“ y^{2}

= (x^{2}Â – y^{2}) + 2ixy

Now, Re (z^{2}) = 0

â‡’Â x^{2}Â â€“ y^{2}Â = 0 â€¦ (ii)

Equating (i) and (ii), we get

â‡’Â x^{2}Â = y^{2}Â = 2

â‡’Â x = y = Â±âˆš2

Hence, z = x + iy

= Â±âˆš2 Â± iâˆš2

= âˆš2 + iâˆš2, âˆš2 â€“ iâˆš2, â€“âˆš2 + iâˆš2 and â€“âˆš2 â€“ iâˆš2

Hence, we have four complex numbers.

**22. Find the complex number satisfying the equation z + âˆš2 |(z + 1)| + i = 0.**

**Solution:**

According to the question,

We have,

z + âˆš2 |(z + 1)| + i = 0 â€¦ (1)

Substituting z = x + iy, we get

â‡’Â x + iy + âˆš2 |x + iy + 1| + i = 0

â‡’Â 2x^{2}Â + 4x + 4 = x^{2}

â‡’Â x^{2}Â + 4x + 4 = 0

â‡’Â (x + 2)^{2}Â = 0

â‡’Â x = -2

Hence, z = x + iy

= â€“ 2 â€“ i

**23. Write the complex number**

**Â Â in polar from.**

**Solution:**

According to the question,

We have,

**24. If z and w are two complex numbers such that |zw| = 1 and arg (z) â€“ arg (w) = Ï€/2, then show thatÂ zÌ…w = â€“ i.**

**Solution:**

** **Let z = |z| (cos Î¸_{1}Â + I sin Î¸_{1}) and w = |w| (cos Î¸_{2}Â + I sin Î¸_{2})

Given |zw| = |z| |w| = 1

Also arg (z) â€“ arg (w) = Ï€/2

â‡’Â Î¸_{1}Â – Î¸_{2}Â = Ï€/2

Now, zÌ…Â w = |z| (cos Î¸_{1}Â – I sin Î¸_{1}) |w| (cos Î¸_{2}Â + I sin Î¸_{2})g | = 1

= |z| |w| (cos (-Î¸_{1}) + I sin (-Î¸_{1})) (cos Î¸_{2}Â + I sin Î¸_{2})

= 1 [cos (Î¸_{2}Â â€“ Î¸_{1}) + I sin (Î¸_{2}Â â€“ Î¸_{1})]

= [cos (-Ï€/2) + I sin (-Ï€/2)]

= 1 [0 â€“ i]

= â€“ i

Hence proved