Differentiation (Part-38) | S N De

 

In the previous articles , we have discussed  Long and Short Answer Type Questions  and their respective solutions. In this article , we will discuss some more mathematical problems of Very Short Answer Type Questions  of S.N. Dey Math book Exercise from Complex Numbers along with their solutions. So, let's start. 

1. Do you think $\,4\,$ as a complex number ? If so why?

Sol. We notice that $\,4=4+i.0\,\,$ which means $\,4\,$ can be expressed in the form $\,a+ib\,$ and so,  $\,4\,$ can be regarded as a complex number.

2(i). If $\,\,x,y\,$ are real and $\,x+iy=-i(-2+3i),\,$ find $\,x,\,y$ ?

Sol. $\,x+iy=2i-3i^2\\~~~~~~~~~~~~=3+2i\,\,[\text{Putting,}\,\,i^2=-1] \\ \therefore x=3,\,\,y=2.$

2(ii) If $\,x,y\,$ are real and $\,x+iy=\frac{5}{-3+4i},\,$ find $\,x,\,y$?

Sol. $\,x+iy=\frac{5}{-3+4i} \\ \Rightarrow x+iy= \frac{5(-3-4i)}{(-3+4i)(-3-4i)}\\~~~~~~~~~~~~~~~~=\frac{-15-20i}{(-3)^2-(4i)^2}\\~~~~~~~~~~~~~~~~=\frac{-15-20i}{25} \\ \Rightarrow x+iy=\frac{-15}{25}+i\frac{-20}{25} \\ \Rightarrow x= -\frac{15}{25}=-\frac 35,\,\,y=-\frac{-20}{25}=\frac{-4}{5}$ 

3. If $\,\,(1+i)(2+i)(3+i)\cdots(n+i)=a+ib,\,\,$ Show that $\, 2.5.10.\cdots (n^2+1)=a^2+b^2.$

Sol. $\,\,(1+i)(2+i)(3+i)\cdots(n+i)=a+ib \\ \Rightarrow |(1+i)(2+i)(3+i)\cdots(n+i)|=|a+ib| \\ \Rightarrow |1+i|.|2+i|.|3+i|.\cdots |n+i|=|a+ib| \\ \Rightarrow \sqrt{1^2+1^2}. \sqrt{2^2+1^2}.\sqrt{3^2+1^2} \\ \cdots \sqrt{n^2+1^2}=\sqrt{a^2+b^2} \\ \Rightarrow \sqrt{2.5.10.\cdots(n^2+1)}=\sqrt{a^2+b^2}\\ \Rightarrow 2.5.10.\cdots (n^2+1)=a^2+b^2.$

4. Prove that , $\, |\frac{x-iy}{-x+iy}|=1.$

Sol. We have, $\, |\frac{x-iy}{-x+iy}|\\=\frac{|x-iy|}{|-x+iy|}\\=\frac{\sqrt{x^2+(-y)^2}}{\sqrt{(-x)^2+y^2}}\\=\frac{\sqrt{x^2+y^2}}{\sqrt{x^2+y^2}}\\=1.$

5. If $\,z=\frac{2+i}{-2+i},\,\,$ find $\,\bar{z},\,$ the conjugate of $\,z\,$ in the form $\,a+ib.$

Sol. $\,z=\frac{2+i}{-2+i}\\~~=\frac{(2+i)(-2-i)}{(-2+i)(-2-i)}\\~~=\frac{-(2+i)^2}{(2-i)(2+i)}\\~=\frac{-(4+4i+i^2)}{4-i^2}\\~=\frac{-(4-1+4i)}{4+1}\\~=\frac{-3-4i}{5}\\~= -\frac 35- \frac 45 i \\ \therefore \bar{z}=-\frac 35+ \frac 45 i.$

6. Express in the form $\,\,A+iB\,(A,B\,\,\,\text{are real}): \\ (i)(1-i)^3 \,\,(ii)\,\,\frac{i}{1+i}+\frac{1+i}{i} \\ (iii) \frac{x+iy}{y-ix}\,\,(x^2+y^2 \neq 0)\,\, (iv) \frac{i}{2+i}+\frac{3}{1+4i}\\(v)\,\frac{\sqrt3-i\sqrt2}{2\sqrt3-i\sqrt2} \,\,(vi)\,\,\frac{1}{1-\cos\theta-i\sin \theta}$

Sol. $\,(i)\, (1-i)^3\\=1^3-i^3-3.1.i(1-i)\\=1+i-3i-3\,\,[i^3=i^2.i=-i]\\=-2-2i\\=-2+i(-2)$

Sol. $\,(ii)\,\frac{i}{1+i}+\frac{1+i}{i}\\=\frac{i^2+(1+i)^2}{i(1+i)}\\=\frac{-1+1+i^2+2i}{i+i^2}\\=\frac{-1+2i}{i-1}\\=\frac{(-1+2i)(i+1)}{(i-1)(i+1)}\\=\frac{-i-1+2i^2+2i}{i^2-1^2}\\=\frac{i-3}{-1-1}\,\,[i^2=-1]\\=\frac{-3+i}{-2}\\=\frac 32+i(-\frac 12)$

Sol. $\,(iii)\, \frac{x+iy}{y-ix}\\=\frac{(x+iy)(y+ix)}{(y-ix)(y+ix)}\\=\frac{xy+i^2xy+i(x^2+y^2)}{y^2-i^2x^2}\\=\frac{xy-xy+i(x^2+y^2)}{y^2+x^2}\\=\frac{i(x^2+y^2)}{x^2+y^2}\\=i\\=0+i.1$

Sol. $\,(iv)\, \frac{i}{2+i}+\frac{3}{1+4i}\\=\frac{i(2-i)}{(2+i)(2-i)}+\frac{3(1-4i)}{(1+4i)(1-4i)}\\=\frac{2i-i^2}{4-i^2}+\frac{3-12i}{1-16i^2}\\=\frac{2i+1}{4+1}+\frac{3-12i}{1+16}\\=\frac{2i+1}{5}+\frac{3-12i}{17}\\=\frac{34i+17+15-60i}{85}\\=\frac{32-26i}{85}\\=\frac{32}{85}+i\left(-\frac{26}{85}\right)$

Sol. $\,(v)\,\frac{\sqrt3-i\sqrt2}{2\sqrt3-i\sqrt2}\\=\frac{(\sqrt3-i\sqrt2)(2\sqrt3+i\sqrt2)}{(2\sqrt3-i\sqrt2)(2\sqrt3+i\sqrt2)}\\=\frac{6+i\sqrt6-2i\sqrt6+2}{(2\sqrt3)^2+(\sqrt2)^2},\,\,[i^2=-1]\\=\frac{8-i\sqrt6}{12+2}\\=\frac{8-i\sqrt6}{14}\\=\frac{8}{14}+i\left(\frac{-\sqrt6}{14}\right)\\=\frac{4}{7}+i\left(\frac{-\sqrt6}{14}\right)$

Sol. $\,(vi)\,\frac{1}{1-\cos\theta-i\sin \theta}\\=\frac{1(1-\cos\theta+i\sin \theta)}{(1-\cos\theta-i\sin \theta)(1-\cos\theta+i\sin \theta)}\\=\frac{(1-\cos\theta)+i \sin \theta}{(1-\cos \theta)^2+\sin^2\theta}\\=\frac{2 \sin^2{(\theta/2)}+i. 2\sin (\theta/2)\cos (\theta/2)}{1+\cos^2\theta-2\cos \theta+\sin^2\theta}\\=\frac{2 \sin (\theta/2)\left[\sin (\theta/2)+i\cos (\theta/2)\right]}{2(1-\cos \theta)}\\=\frac{2 \sin (\theta/2)\left[\sin (\theta/2)+i\cos (\theta/2)\right]}{2.2\sin^2(\theta/2)}\\=\frac{\sin (\theta/2)+i\cos (\theta/2)}{2 \sin (\theta/2)}\\=\frac 12+ i \left(\frac 12 \cot(\theta/2)\right)$

7. Find the conjugate: 

$\,(i)\, \sqrt{-5}-2\,\,(ii)\,2-i\sqrt3-\sqrt2 \\ (iii)\, \frac{2-i}{(1-2i)^2}$

Sol. (i) To find the conjugate of a complex number $z=\,\,a+ib,\,$ we have to replace $\,i\,$ by $\,-i\,$ which is $\,\,\bar{z}=\,a-ib.$

 Now, let $z=\sqrt{-5}-2=-2+i\sqrt5 \,\,[\text{where,}\,\,i=\sqrt{-1}]\\  \therefore \bar{z}=-2-i\sqrt5$ 

Sol.(ii) To find the conjugate of a complex number $z=\,\,a+ib,\,$ we have to replace $\,i\,$ by $\,-i\,$ which is $\,\,\bar{z}=\,a-ib.$

 Now, let $z=2-i\sqrt3-\sqrt2\\=(2-\sqrt2)+i(-\sqrt3) \,\,[\text{where,}\,\,i=\sqrt{-1}]\\ \therefore \bar{z}=(2-\sqrt2)-i(-\sqrt3)\\=(2-\sqrt2)+i\sqrt3$ 

Sol. (iii) Let  $\,z=\frac{2-i}{(1-2i)^2}\\=\frac{2-i}{1-4-4i}\\=\frac{2-i}{-3-4i}\\=\frac{i-2}{3+4i}\\=\frac{(i-2)(3-4i)}{(3+4i)(3-4i)}\\=\frac{3i-4i^2-6+8i}{3^2-16i^2}\\=\frac{11i+4-6}{9+16}\\=\frac{-2+11i}{25}\\=-\frac{2}{25}+i \frac{11}{25} \\ \therefore \bar{z}=-\frac{2}{25}-i \frac{11}{25}$

8(i). Simplify : $\,\,\,\frac{i+i^2+i^3+i^4}{1+i}\\=\frac{i-1+i^2.i+(i^2)^2}{1+i}\,\,[\text{Using,}\,i^2=-1,i^3=-i]\\=\frac{i-1-i+1}{1+i}\\=\frac{0}{1+i}\\=0.$

8(ii). Simplify : $\,\,(1+i^3)\left(1+\frac 1i\right)^2\left(i^4+\frac{1}{i^4}\right)\\=(1+i^2.i)(1+\frac{i}{i^2})^2\left((i^2)^2+\frac{1}{(i^2)^2}\right)\\=(1-i)(1-i)^2(1+1)\,\,[\text{Using,}\,\,i^2=-1]\\=2(1-i)^3\\=2(1-i)(1-i)^2\\=2(1-i)(1+i^2-2i)\\=2(1-i)(1-1-2i)\\=2(1-i)(-2i)\\=-4i(1-i)\\=-4i+4i^2\\=-4-4i\\=-4(1+i)$

8(iii). Simplify : $\,(1+i)^2+(1-i)^2 \\=2(1^2+i^2) \\ [\text{Since,}\,\,(a+b)^2+(a-b)^2=2(a^2+b^2)]\\=2(1-1)\\=2 \times 0\\=0.$

8(iv). Simplify : $\,\,(1+i)^{-2}-(1-i)^{-2}\\=\frac{1}{(1+i)^2}-\frac{1}{(1-i)^2}\\=\frac{(1-i)^2-(1+i)^2}{(1+i)^2(1-i)^2}\\=-\left[\frac{(1+i)^2-(1-i)^2}{\left((1+i)(1-i)\right)^2}\right]\\=-\left[\frac{4i}{(1^2-i^2)^2}\right]\\=-\left[\frac{4i}{(1+1)^2}\right]\\=-i$ 

8(v). Simplify : $\,\,\left(\frac{1+i}{1-i}\right)^2+\left(\frac{1-i}{1+i}\right)^2 \cdots(1)$

We first compute $\,\left(\frac{1+i}{1-i}\right)^2\\=\left(\frac{(1+i)(1+i)}{(1-i)(1+i)}\right)^2\\=\left(\frac{(1+i)^2}{1^2-i^2}\right)^2\\=\left(\frac{1+i^2+2i}{1+1}\right)^2\\=\left(\frac{2i}{2}\right)^2 \cdots(2)$

Similarly, $\,\,\left(\frac{1-i}{1+i}\right)^2\\=\left(\frac{-2i}{2}\right)^2 \cdots(3)$

Hence, from (1),(2) and (3), we get $\,\,\left(\frac{1+i}{1-i}\right)^2+\left(\frac{1-i}{1+i}\right)^2\\=\left(\frac{2i}{2}\right)^2 +\left(\frac{-2i}{2}\right)^2\\=i^2+i^2\\=-1-1\\=-2.$

9. If $\,\,x,y\,$ are real and $\,(x+3i)\,$ and $\,(-2+iy)\,$ are conjugate of each other , find $\,\,x\,$ and $\,y.$

Sol. By question, we have $\,\overline{x+3i}= -2+iy \\ \Rightarrow x-3i=-2+iy \rightarrow(1)$

Comparing both sides of (1), we have $\,\,x=-2,\,\, y=-3.$

10. Find the least positive integral value of $\,n\,$ so that $\,\left(\frac{1+i}{1-i}\right)^n=1.$

Sol. We first compute $\,\left(\frac{1+i}{1-i}\right)\\=\left(\frac{(1+i)(1+i)}{(1-i)(1+i)}\right)\\=\left(\frac{(1+i)^2}{1^2-i^2}\right)\\=\left(\frac{1+i^2+2i}{1+1}\right)\\=\left(\frac{2i}{2}\right)\\=i \cdots(1)$

Hence, using the result in (1), we get $\,i^n=1 \\ \Rightarrow i^n=(-1)^2\\ \Rightarrow i^n=(i^2)^2 \\ \Rightarrow i^n=i^4 \\ \Rightarrow n=4.$

So, the least positive integral value of $\,n=4.$ 

11(i). Find the modulus : $\,\, \, 2\sqrt2-i\sqrt6$

Sol. $\,\,|2\sqrt2-i\sqrt6|\\=\sqrt{(2 \sqrt2)^2+(-\sqrt6)^2}\\=\sqrt{4 \times 2 +6}\\=\sqrt{8+6}\\=\sqrt{14}$

11(ii). Find the modulus : $\,\,(\sqrt5+i\sqrt3)(-2+i)$

Sol. $\,\,(\sqrt5+i\sqrt3)(-2+i)\\=-2\sqrt5-\sqrt3+i(-2\sqrt3+\sqrt5) \rightarrow(1)$

Hence, $\,|(\sqrt5+i\sqrt3)(-2+i)|\\= \sqrt{(-2\sqrt5-\sqrt3)^2+(-2\sqrt3+\sqrt5)^2} \,\,[\text{By (1)}]\\=\sqrt{(2\sqrt5+\sqrt3)^2+(-2\sqrt3+\sqrt5)^2}\\=\sqrt{(2\sqrt5)^2+(\sqrt3)^2+2.2\sqrt5. \sqrt3\\+(-2\sqrt3)^2+(\sqrt5)^2-2.2\sqrt3.\sqrt5}\\=\sqrt{20+3+4\sqrt{15}+12+5-4\sqrt{15}}\\=\sqrt{40}\\=\sqrt{4 \times 10}\\=2\sqrt{10}$

Alternative Sol. $\,\,|(\sqrt5+i\sqrt3)(-2+i)|\\=|(\sqrt5+i\sqrt3)||(-2+i)|\\=\sqrt{(\sqrt5)^2+(\sqrt3)^2} \times \sqrt{(-2)^2+1^2}\\=\sqrt{5+3} \times \sqrt{4+1}\\=\sqrt{8} \times \sqrt{5} \\=\sqrt{8 \times 5}\\=\sqrt{40}\\=\sqrt{4 \times 10}\\=2\sqrt{10}$

11(iii). Find the modulus : $\,\, \frac{1-i}{3-4i}$

Sol. $\,\,|\frac{1-i}{3-4i}|\\=\\=\frac{|1-i|}{|3-4i|}\\= \frac{\sqrt{1^2+(-1)^2}}{\sqrt{(3)^2+(-4)^2}}\\=\frac{\sqrt2}{\sqrt{9+16}}\\=\frac{\sqrt2}{\sqrt{25}}\\=\frac{\sqrt2}{5}$

11(iv). Find the modulus : $\,\, \frac{2}{1+\cos \theta+i \sin \theta}$

Sol. $\,\left|\frac{2}{1+\cos \theta+i \sin \theta}\right|\\=\frac{|2|}{|1+\cos \theta+i \sin \theta|}\\=\frac{2}{\sqrt{(1+\cos \theta)^2+\sin^2\theta}}\\=\frac{2}{\sqrt{1+\cos^2\theta+2\cos\theta+\sin^2\theta}}\\=\frac{2}{\sqrt{2(1+\cos \theta)}}\\=\frac{2}{\sqrt{2.2\cos^2 (\theta/2)}}\\=\frac{2}{2 \cos (\theta/2)}\\=\sec(\theta/2)$

12. Find the argument of each of the following complex numbers: $\,(i)\,2-2i \, (ii)\, -3-3i\,\\ (iii)\,(1+i)(\sqrt3+i)\,(iv)\, \frac{\sqrt3+i}{-1-i\sqrt3}$

Sol. $\,(i)\, \arg (2-2i) \\=\tan^{-1}(-2/2)\\=\tan^{-1}(-1)\\=-\frac{\pi}{4} \,\, [\text{Since, the point}\,\,(2,-2) \\ \text{lies in 4th quadrant}]$

Sol. $\,(ii)\, \arg(-3-3i)\\=\tan^{-1}\left(\frac{-3}{-3}\right)\\=\tan^{-1}(1)\\=-\pi+\frac{\pi}{4}\\=-\frac{3\pi}{4} \,\, [\text{Since, the point}\,\,(-3,-3) \\ \text{lies in 3rd quadrant}]$

Sol. $\,(iii)\,(1+i)(\sqrt3+i)=(\sqrt3-1)+i(\sqrt3+1) \\ \therefore \arg\left((1+i)(\sqrt3+i)\right)\\= \tan^{-1} \left(\frac{\sqrt3+1}{\sqrt3-1}\right) \\=\tan^{-1} \frac{(\sqrt3+1)^2}{(\sqrt3-1)(\sqrt3+1)}\\=\tan^{-1} \frac{4+2\sqrt3}{2}\\=\tan^{-1}(2+\sqrt3)\\=\frac{5 \pi}{12}\,\, [\text{Since, the point}\,\,(\sqrt3-1,\sqrt3+1) \\ \text{lies in 1st quadrant}]$

Sol. $\,(iv)\, \frac{\sqrt3+i}{-1-i\sqrt3}\\=\frac{(\sqrt3+i)(-1+i\sqrt3)}{(-1-i\sqrt3)(-1+i\sqrt3)}\\=\frac{-\sqrt3+3i-i-\sqrt3}{1+3}\\=\frac{-2\sqrt3+2i}{4}\\=\frac{-\sqrt3+i}{2}\\=-\frac{\sqrt3}{2}+\frac 12i \\ \therefore \arg \left(\frac{\sqrt3+i}{-1-i\sqrt3}\right)\\=\arg \left(-\frac{\sqrt3}{2}+\frac 12i \right)\\=\tan^{-1}\left(\frac{1/2}{-\sqrt3/2}\right)\\=\tan^{-1}\left(-\frac{1}{\sqrt3}\right)\\=\pi-\pi/6 \\=\frac{5 \pi}{6}\,\, [\text{Since, the point}\,\,(-\frac{\sqrt3}{2},\frac 12) \\ \text{lies in 2nd quadrant}]$

13. If $\,\,a=\frac{1+i}{\sqrt2},\,\,$ show that $\,\,a^6+a^4+a^2+1=0.$

Sol. We have , $\,\,a=\frac{1+i}{\sqrt2} \\ \Rightarrow \sqrt2 a=1+i \\ \Rightarrow (\sqrt2a)^2=(1+i)^2\\ \Rightarrow 2a^2=1+2i+i^2 \\ \Rightarrow 2a^2=1+2i-1 \\ \Rightarrow 2a^2=2i \\ \Rightarrow a^2=i \\ \therefore a^6+a^4+a^2+1\\=(a^2)^3+(a^2)^2+a^2+1\\=i^3+i^2+i+1\\=-i-1+i+1\\=0 \,\,\text{(proved)}$

14. Find the cube roots of $\,\,(-1)\,$ and $\,\,27.$

Sol. $\,\,\sqrt[3]{-1}=x \\ \Rightarrow x^3=-1 \\ \Rightarrow x^3+1=0 \\ \Rightarrow (x+1)(x^2-x+1)=0 \\ \Rightarrow x+1=0,\,\,x^2-x+1=0 \cdots(1)$

From (1), $\,\,x+1=0 \Rightarrow x=-1 \\  \text{and,}\,\,\,x^2-x+1=0 \\ \Rightarrow x=\frac{1 \pm \sqrt{1-4}}{2}\\~~~~~~~~=\frac{1 \pm i\sqrt3}{2} \\ \therefore \text{Cube roots of}\,\, \sqrt[3]{-1}=-1, -\omega,\,-\omega^2.$

Let $\,\,\sqrt[3]{27}=y \\ \Rightarrow y^3=27 \\ \Rightarrow\left(\frac y3\right)^3=1 \cdots(2)$

Since the cube roots of $\,1\,$ is $\,1,\omega,\,\omega^2$, hence from (2), we can say cube roots of $\,27\,$ is $\,\,3,3\omega,\,3\omega^2.$ 

15. Factorize : $\,\,a^2+1.$

Sol. $\,\,a^2+1\\=a^2-(-1)\\=a^2-i^2\\=(a+i)(a-i)$

16.  If $\,1, \omega, \omega^2\,\,$ denotes the cube roots of unity, find the roots of $\,\,(x+5)^3+27=0.$

Sol. $\,\,(x+5)^3+27=0 \\ \Rightarrow (x+5)^3=-27 \\ \Rightarrow \left(\frac{x+5}{-3}\right)^3=1 \\ \Rightarrow \frac{x+5}{-3}=\sqrt[3]{1} \\ \therefore \text{The values of}\,\,\,\,\frac{x+5}{-5}\,\, =1, \omega,\omega^2 \\ \implies x=-3-5,\,-3\omega-5,\,-3\omega^2-5 \\ \implies x=-8,\,-3\omega-5,\,-3\omega^2-5.$

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