Exercise 5.2 class 11 maths solutions | class 11 ch 5 exercise 5.2 solutions | class 11 chapter 5 exercise 5.2 solution | class 11 maths ncert solutions chapter 5 | ncert solutions for class 11 maths chapter 5 | ncert exemplar class 11 maths | class 11 complex numbers and quadratic equations
Looking for Exercise 5.2 Class 11 Maths solutions? You’ve come to the right place! This section provides step-by-step and accurate solutions for all the questions from Exercise 5.2 of Chapter 5 – Complex Numbers and Quadratic Equations. These solutions are designed as per the latest NCERT syllabus and focus on solving quadratic equations using complex numbers, including cases where the discriminant is negative. Whether you’re revising concepts from the Class 11 Ch 5 Exercise 5.2 solutions or practicing with the NCERT Exemplar Class 11 Maths, these solutions will help you gain complete clarity. Download or view the Class 11 Maths NCERT Solutions Chapter 5 now and master the topic of complex numbers and quadratic equations with confidence!
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Exercise 5.2
1. Find the modulus and the arguments of each of the complex numbers in
\(\mathrm{z}=-1-i \sqrt{3}\)
Answer
\(\mathrm{Z}=-1-i \sqrt{3}\)
Let \( r \cos \theta=-1 \) and \( r \sin \theta=-\sqrt{3} \)
On squaring and adding, we obtain
\((r \cos \theta)^{2}+(r \sin \theta)^{2}=(-1)^{2}+(-\sqrt{3})^{2}\)
\(r^{2}\left(\cos ^{2} \theta+\sin ^{2} \theta\right)=1+3\)
\(=\text { Modules }=2\)
\(\therefore 2 \cos \theta=-1 \text { and } 2 \sin \theta=-\sqrt{3}\)
\(=\cos \theta=\frac{-1}{2} \text { and } \sin \theta=\frac{-\sqrt{3}}{2}\)
Since both the values of \( \sin \theta \) and \( \cos \theta \) are negative and \( \sin \theta \) and \( \cos \theta \) are negative in III quadrant,
Argument \( =-\left(\pi-\frac{\pi}{3}\right)=\frac{-2 \pi}{3} \)
Thus, the modulus and argument of the complex number \( -1-\sqrt{3} i \) are 2 and \( \frac{-2 \pi}{3} \) respectively.
2. Find the modulus and the arguments of each of the complex numbers in \(z=-\sqrt{3}+i\)
Answer
\( \mathrm{z}=-\sqrt{3}+i \)
Let \( r \cos \theta=-\sqrt{3} \) and \( r \sin \theta=1 \)
On squaring and adding, we obtain
\( r^{2} \cos ^{2} \theta+r^{2} \sin ^{2} \theta=(-\sqrt{3})^{2}+1^{2}\)
\(= r^{2}=3+1=4\)
\(= \mathrm{r}=\sqrt{4}=2\)
\(\therefore \text { Modules }=2\)
\(\therefore 2 \cos \theta=-\sqrt{3} \text { and } 2 \sin \theta=1\)
\( \operatorname{Cos} \theta=\frac{-\sqrt{3}}{2} \text { and } \sin \theta=\frac{1}{2}\)
\( \therefore \theta=\pi-\frac{\pi}{6}=\frac{5 \pi}{6}\)
Thus, the modulus and argument of the complex number \( -\sqrt{3}+\mathrm{i} \) are \(2\) and \( \frac{5 \pi}{6} \) respectively.
3. Convert each of the complex numbers given in Exercises 3 to 8 in the polar form: \( 1-i \)
Answer
\( 1-\mathrm{i} \)
Let, \( r \cos \theta=1 \) and \( \mathrm{r} \sin \theta=-1 \)
On squaring and adding, we obtain
\( r^{2} \cos ^{2} \theta+r^{2} \sin ^{2} \theta=1^{2}+(-1)^{2}\)
\(= r^{2}\left(\cos ^{2} \theta+\sin ^{2} \theta\right)=1+1\)
\(= r^{2}=2\)
\(= \mathrm{r}=\sqrt{2}\)
\(\therefore \sqrt{2} \cos \theta=1 \text { and } \sqrt{2} \sin \theta=-1\)
\( \cos \theta=\frac{1}{\sqrt{2}} \) and \( \sin \theta=\frac{-1}{\sqrt{2}} \)
\(\therefore 1-i=r \cos \theta+r \sin \theta=\sqrt{2} \cos \left(-\frac{\pi}{4}\right)+i \sqrt{2} \sin \left(-\frac{\pi}{4}\right)=\)
\(\sqrt{2}\left[\cos \left(-\frac{\pi}{4}\right)+i \sin \left(-\frac{\pi}{4}\right)\right]\)
This is the required polar form.
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4. Convert each of the complex numbers given in Exercises 3 to 8 in the polar form: \( -1+i \)
Answer
\(-1+i\)
Let, \( r \cos \theta=-1 \) and \( r \sin \theta=1 \)
On squaring and adding, we obtain
\(r^{2} \cos ^{2} \theta+r^{2} \sin ^{2} \theta=(-1)^{2}+1^{2}\)
\(=r^{2}\left(\cos ^{2} \theta+\sin ^{2} \theta\right)=1+1\)
\(\quad=r=\sqrt{2} \quad[\text { Conventionally, } \mathrm{r} > 0]\)
\(\therefore \sqrt{2} \cos \theta=-1 \text { and } \sqrt{2} \sin \theta=1\)
\(=\cos \theta=\frac{-1}{\sqrt{2}} \text { and } \sin \theta=\frac{1}{\sqrt{2}}\)
\( \therefore \theta=\pi-\frac{\pi}{4}=\frac{3 \pi}{4} \quad \) [ As \( \theta \) lies in the II quadrant]
It can be written,
\(\therefore-1+i= r \cos \theta+r \sin \theta=\sqrt{2} \cos \frac{3 \pi}{4}+i \sqrt{2} \sin \frac{3 \pi}{4}\)
\( =\sqrt{2}\left(\cos \frac{3 \pi}{4}+i \sin \frac{3 \pi}{4}\right)\)
This is the required polar form.
5. Convert each of the complex numbers given in Exercises 3 to 8 in the polar form: \( -1-i \)
Answer
\(-1+i\)
Let, \( r \cos \theta=-1 \) and \( \mathrm{r} \sin \theta=-1 \)
On squaring and adding, we obtain
\(r^{2} \cos ^{2} \theta+r^{2} \sin ^{2} \theta=(-1)^{2}+(-1)^{2}\)
\(\quad=r^{2}\left(\cos ^{2} \theta+\sin ^{2} \theta\right)=1+1\)
\(=r=\sqrt{2}\)
\(\quad \therefore \sqrt{2} \cos \theta=-1 \text { and } \sqrt{2} \sin \theta=-1\)
\(=\cos \theta=\frac{-1}{\sqrt{2}} \text { and } \sin \theta=\frac{-1}{\sqrt{2}}\)
\( \therefore \theta=-\left(\pi-\frac{\pi}{4}\right)=-\frac{3 \pi}{4} \quad \) [As \( \theta \) lies in the III quadrant]
\(\therefore-1+i=r \cos \theta+i r \sin \theta=\sqrt{2} \cos \frac{-3 \pi}{4}+i \sqrt{2} \sin \frac{-3 \pi}{4}=\)
\(\sqrt{2}\left(\cos \frac{-3 \pi}{4}+i \sin \frac{-3 \pi}{4}\right)\)
This is the required polar form.
6. Convert each of the complex numbers given in Exercises 3 to 8 in the polar form: \( -3 \)
Answer
Let, \( r \cos \theta=-3 \) and \( r \sin \theta=0 \)
On squaring and adding, we obtain
\( r^{2} \cos ^{2} \theta+r^{2} \sin ^{2} \theta=(-3)^{2}\)
\(=r^{2}\left(\cos ^{2} \theta+\sin ^{2} \theta\right)=9\)
\(=r^{2}=9\)
\(=\mathrm{r}=\sqrt{9}=3\)
\(\therefore 3 \cos \theta=-1 \text { and } 3 \sin \theta=0\)
\(\quad=\theta=\pi\)
\(\therefore-3=r \cos \theta+i r \sin \theta=3 \cos \pi+3 i \sin \pi= 3(\cos \pi+i \sin \pi) \)
This is the required polar form.
7. Convert each of the complex numbers given in Exercises 3 to 8 in the polar form: \(\sqrt{3}+i\)
Answer
\( \sqrt{3}+i \)
Let, \( \mathrm{r} \cos \theta=\sqrt{3} \) and \( \mathrm{r} \sin \theta=1 \)
On squaring and adding, we obtain
\(r^{2} \cos ^{2} \theta+r^{2} \sin ^{2} \theta=(\sqrt{3})^{2}+1^{2}\)
\(=r^{2}\left(\cos ^{2} \theta+\sin ^{2} \theta\right)=3+1\)
\(=r^{2}=4\)
\(=\mathrm{r}=\sqrt{4}\)
\(=\mathrm{r}=2\)
\( \therefore 2 \cos \theta=\sqrt{3} \) and \( 2 \sin \theta=1 \)
\( \cos \theta=\frac{\sqrt{3}}{2} \) and \( \sin \theta=\frac{1}{2} \)
\( \therefore \theta=\frac{\pi}{6}\)
\(\therefore \sqrt{3}+i=r \cos \theta+r \sin \theta=2 \cos \frac{\pi}{6}+i 2 \sin \frac{\pi}{6}=2\left(\cos \frac{\pi}{6}+i \sin \frac{\pi}{6}\right)\)
8. Convert each of the complex numbers given in Exercises 3 to 8 in the polar form: \( i \)
Answer
Let, \( \mathrm{r} \cos \theta=0 \) and \( \mathrm{r} \sin \theta=1 \)
On squaring and adding, we obtain
\(r^{2} \cos ^{2} \theta+r^{2} \sin ^{2} \theta=0^{2}+1^{2}\)
\(= r^{2}\left(\cos ^{2} \theta+\sin ^{2} \theta\right)=1\)
\(= r^{2}=1\)
\(= r=\sqrt{1}\)
\( \therefore \cos \theta=0 \sin \theta=1\)
\(\therefore \theta=\frac{\pi}{2}\)
\(\therefore i=r \cos \theta+i r \sin \theta=\cos \frac{\pi}{2}+i \sin \frac{\pi}{2} \)
This is the required polar form.
