Differentiation (Part-38) | S N De

 

In the previous two articles (Part-1 and Part-2) , we have discussed Long Answer Type Questions of S.N. Dey Math exercise . In this article, Short Answer Type Questions and their respective solutions will be solved. So, let's start.

1. Find all the roots $\,\,x^3-1=0.\,\,$ Show that if $\,w\,$ is a complex root of this equation , the other complex root is $\,w^2\,$ and $\,\,1+w+w^2=0.$

Sol. $\,\,x^3-1=0 \\ \Rightarrow x^3-1^3=0  \\ \Rightarrow (x-1)(x^2+x+1) =0 \\ \Rightarrow x-1=0,\,\,x^2+x+1=0.$

Now, $\,\,x-1=0 \Rightarrow x=1. \\ x^2+x+1=0 \\ \therefore x=\frac{-1 \pm \sqrt{1^2-4 \times 1}}{2}=\frac{-1 \pm \sqrt{-3}}{2}$

Hence, the solution of the aforesaid equation is given by : $\,x=1,\frac{-1 + \sqrt{-3}}{2},\frac{-1 - \sqrt{-3}}{2}. $

Now, let without loss of generality, $\,w=\frac{-1 + \sqrt{-3}}{2} \\ \Rightarrow w^2=\left(\frac{-1 + \sqrt{-3}}{2}\right)^2 \\ \Rightarrow w^2=\frac{1-2\sqrt{-3}+(-3)}{4}=\frac{-1-\sqrt{-3}}{2}$

Again, $\,w=\frac{-1 - \sqrt{-3}}{2} \\ \Rightarrow w^2=\left(\frac{-1 - \sqrt{-3}}{2}\right)^2 \\ \Rightarrow w^2=\frac{1+2\sqrt{-3}+(-3)}{4}=\frac{-1+\sqrt{-3}}{2}$

Now, $\,\,1+w+w^2 \\=1+ \frac{-1+\sqrt{-3}}{2}+\frac{-1-\sqrt{-3}}{2}\\=\frac{2-1+\sqrt{-3}-1-\sqrt{-3}}{2}\\=0.$

2. If $\,\sqrt{x-iy}=a-ib,\,\,$and $\,\,a,b,x,y\,$ are real then show that, $\,\,\sqrt{x+iy}=a+ib.$

Sol. We have $\,\sqrt{x-iy}=a-ib \\ \Rightarrow x-iy=(a-ib)^2 =(a^2-b^2)-i(2ab)\cdots(1)$

From (1), we get by comparing both sides, $\,x=a^2-b^2,\, y=2ab.$  

Hence, $(x+iy)\\=(a^2-b^2)+i(2ab) \\=a^2+2a(ib)+(ib)^2\\=(a+ib)^2 \\ \Rightarrow a+ib=\sqrt{x+iy} \cdots(2)$

Hence by (2), we get the required result.

3.If $\,|z+5| \leq 6,\,$ find the maximum and minimum values of $\,|z+2|,\,z$ being a complex number.

Sol. $\,(z+2)=|(z+5)+(-3)|\\ \leq |z+5|+|-3| \\ \leq |z+5|+3 \\ \leq 6+3 \\=9.$

Hence max. value of $\,|z+2|\,$ is $\,9\,$ whereas since we know $\,|z+2| \geq 0,\,$ min. value of $\,|z+2|\,$ is $\,\,0.$

4. If $\,\,x,y\,$ are real and $\,\,(y^2x-5i)\,$ and $\,\{4+i(x+y^2)\}\,$ are conjugate to each other, find $\,\,x,y.$

Sol. Let $\,z=y^2x-5i  \\ \therefore \bar{z}=y^2x+5i \\ \Rightarrow 4+i(x+y^2)=y^2x+5i.$

Since $\,\,x,y \in \mathbb R,\,y^2x=4 \rightarrow (1),\\x+y^2=5 \rightarrow (2)$

Hence from (1) and (2), we get $\,\,x(5-x)=4 \\ \Rightarrow x^2-5x+4=0\\ \Rightarrow (x-1)(x-4)=0 \\ \Rightarrow x=1,4.$

From (2), we get for $\,\,x=1,\,y= \pm2 \\ x=4,\,\,y=\pm1.$

Hence follows the result.

5. If $\,\,z_1=4-3i,\, z_2=-12+5i,\,$show that $\,(i)\, \overline{z_1+z_2}=\bar{z_1}+\bar{z_2}\\ (ii)\, |z_1+z_2| <|z_1|+|z_2|\\ (iii)\,\overline{z_1.z_2}=\bar{z_1}.\bar{z_2} \\ (iv)|z_1z_2|=|z_1||z_2| \\ (v)\,\left|\frac{z_1}{z_2}\right|=\frac{|z_1|}{|z_2|}$

Sol. (i) By question, we have $\,\,\bar{z_1}=4+3i,\,\bar{z_2}=-12-5i, \\z_1+z_2=(4-3i)+(-12+5i)\\~~~~~~~~~~~~=-8+2i \\ \therefore \overline{z_1+z_2}=-8-2i,\cdots(1)\\ \bar{z_1}+\bar{z_2}=(4+3i)+(-12-5i) \\~~~~~~~~~~~~=-8-2i \cdots(2)$

Hence from (1) and (2), the result follows.

Sol. (ii)By question, $\,|z_1|=\sqrt{4^2+(-3)^2}=5,\\ |z_2|=\sqrt{(-12)^2+5^2}=13,\\ \therefore|z_1|+|z_2|=5+13=18 =\sqrt{324} \cdots(1) \\ z_1+z_2=(4-3i)+(-12+5i)\\~~~~~~~~~~~~=-8+2i\\ \therefore |z_1+z_2|=\sqrt{(-8)^2+2^2}=\sqrt{68} \cdots(2)$

Hence, from (1) and (2), follows the result.

Sol. (iii) By question, $z_1z_2=(4-3i).(-12+5i)\\~~~~~~~~~=-48+20i+36i+15\,\,[\text{Using}\,\,i^2=-1] \\~~~~~~~~~=-33+56i \\ \therefore \overline{z_1z_2}=-33-56i \cdots(1) \\\bar{z_1}.\bar{z_2}=(4+3i).(-12-5i)\\~~~~~~~~~=-48-20i-36i+15\\~~~~~~~~~=-33-56i \cdots(2)$

Hence, from (1) and (2), the result follows.

Sol. (iv) By question, $\,|z_1|=\sqrt{4^2+(-3)^2}=5,\\ |z_2|=\sqrt{(-12)^2+5^2}=13, \\ z_1z_2=(4-3i).(-12+5i)\\~~~~~~~~~=-48+20i+36i+15\,\,[\text{Using}\,\,i^2=-1] \\~~~~~~~~~=-33+56i \\ \therefore |z_1z_2|=\sqrt{(-33)^2+(56)^2}\\~~~~~~~~~~~~~=\sqrt{4225}\\~~~~~~~~~~~~=65\\~~~~~~~~~~~~= 5 \times 13\\~~~~~~~~~~~~=|z_1|.|z_2|$

Sol. (v)By question, $\,|z_1|=\sqrt{4^2+(-3)^2}=5,\\ |z_2|=\sqrt{(-12)^2+5^2}=13, \\ \text{Hence,}\, \left|\frac{z_1}{z_2}\right|=\left|\frac{4-3i}{-12+5i}\right|\\~~~~~~~~~~~~~~~~~~~=\left|\frac{(4-3i)(-12-5i)}{(-12+5i)(-12-5i)}\right|\\~~~~~~~~~~~~~~~~~~~=\left|\frac{-48-20i+36i-15}{(-12)^2-(5i)^2}\right|\\~~~~~~~~~~~~~~~~~~~=\left|\frac{-63+16i}{169}\right|\\~~~~~~~~~~~~~~~~~~~=\left|\frac{-63}{169}+i\frac{16}{169}\right|\\~~~~~~~~~~~~~~~~~~~=\sqrt{\left(\frac{-63}{169}\right)^2+\left(\frac{16}{169}\right)^2}\\~~~~~~~~~~~~~~~~~~~=\sqrt{\frac{4225}{(169)^2}}\\~~~~~~~~~~~~~~~~~~~=\frac{65}{169}\\~~~~~~~~~~~~~~~~~~~=\frac{5}{13}\\~~~~~~~~~~~~~~~~~~~=\frac{|z_1|}{|z_2|}$

6 (i). If $\,z=2-3i,\,$express $\,\frac{1-2z}{z-3}\,$ in the form $\,\,A+iB\,$where $\,A,\,B\,$ are real.

Sol. $\,\frac{1-2z}{z-3} \\=\frac{1-2(2-3i)}{(2-3i)-3}\\=\frac{-3+6i}{-1-3i}\\=\frac{3-6i}{1+3i}\\=\frac{(3-6i)(1-3i)}{(1+3i)(1-3i)}\\=\frac{3-9i-6i-18}{1+9}\,\,[\text{Since,}\,i^2=-1]\\=\frac{-15-15i}{10}\\=\frac{-15}{10}+i \left(\frac{-15}{10}\right)$

6(ii) If $\,\frac{2+i}{2-3i}=A+iB\,(A,B\,$ are real,$\,\,\,i=\sqrt{-1},\,$ find the value of $\,A^2+B^2.$

Sol. $\,\frac{2+i}{2-3i}\\=\frac{(2+i)(2+3i)}{(2-3i)(2+3i)}\\=\frac{4+6i+2i-3}{4+9}\\=\frac{1+8i}{13}\\=\frac{1}{13}i\frac{8}{13} \cdots(1)$

Comparing (1) with $\,A+iB\,$ we get, $\,A=\frac{1}{13},\,\,B=\frac{8}{13}$

Hence, $\,A^2+B^2=(1/13)^2+(8/13)^2=\frac{65}{169}=\frac{5}{13}.$

6(iii) If $\quad \frac{\sqrt3+i\sqrt2}{2\sqrt3+i\sqrt2}=x+iy\,\,$ where $\,x,y \in \mathbb R,\,i=\sqrt{-1},\,$ find $\,x,y.$

Sol. $\frac{\sqrt3+i\sqrt2}{2\sqrt3+i\sqrt2}\\=\frac{(\sqrt3+i\sqrt2)(2\sqrt3-i\sqrt2)}{(2\sqrt3+i\sqrt2)(2\sqrt3-i\sqrt2)}\\=\frac{6-i\sqrt6+i2\sqrt6+2}{12+2}\\=\frac{8+i\sqrt6}{14}\\=\frac{8}{14}+i\frac{\sqrt6}{14}\\ \Rightarrow \frac{8}{14}+i\frac{\sqrt6}{14}=x+iy \cdots(1)$

Hence $\,\,x=\frac{8}{14}=\frac{4}{7},\\ y=\frac{\sqrt6}{14}$

7. Express each of the following complex numbers in modulus-amplitude form:

$\,(i)\, \sqrt3+i \,\, (ii)\, 5-5i \,\,(iii)\,4i  \\(iv)\,(-3+3i)(1-i)\,(v)\,\frac{\sqrt3-i}{1-\sqrt3i}$

Sol. Let $\,(i)\, z=\sqrt3+i.\,$ If $\,|z|=r,\, \text{amp}\,\, z=\theta\,(-\pi<\theta \leq \pi),$ then $\,z\,$ can be written in the form: $\,z=r(\cos \theta+i \sin \theta)\rightarrow (1)$

Now, $|z|=\sqrt{(\sqrt3)^2+(1)^2}=2,\\ \text{Again,}\,z=\sqrt3+i\times 1 \\ \Rightarrow \text{the point}\,\,(\sqrt3,1)\,\,\text{lies in the 1st quadrant.}$ 

So, amp z $\,=\theta \Rightarrow \tan \theta=\frac{1}{\sqrt3},\,\,\text{where}\,\,0 <\theta <\pi/2. $

So, $\,\,\tan \theta=\tan (\frac{\pi}{6}) \Rightarrow \theta =\pi/6.$

Hence, $\,\,z=\sqrt3+i,\,$ can be written in modulus-amplitude form: 

$\,z=2\left(\cos (\frac{\pi}{6})+i \sin (\frac{\pi}{6})\right) \\ [\,\text{Putting the value of}\,\,r, \theta \,\,\,\,\text{in (1)}]$

Sol. 7(ii)Let $\,\, z=5-5i.\,$ If $\,|z|=r,\, \text{amp}\,\, z=\theta\,(-\pi<\theta \leq \pi),$ then $\,z\,$ can be written in the form: $\,z=r(\cos \theta+i \sin \theta)\rightarrow (1)$

Now, $|z|=\sqrt{(5)^2+(-5)^2}=5\sqrt2,\\ \text{Again,}\,z=5+i\times (-5) \\ \Rightarrow \text{the point}\,\,(5,-5)\,\,\text{lies in the 4th quadrant.}$    

So,  amp z $=\theta \\ \Rightarrow \tan \theta=\frac{-5}{5}=-1,\,\,\text{where}\,\,-\frac{\pi}{2} <\theta <0. $

So, $\,\,\tan \theta=\tan (-\frac{\pi}{4}) \Rightarrow \theta =-\pi/4.$

Hence, $\,\,z=5-5i,\,$ can be written in modulus-amplitude form:  

$\,z=5 \sqrt2\left(\cos (\frac{-\pi}{4})+i \sin (\frac{-\pi}{4})\right) \\  [\,\text{Putting the value of}\,\,r, \theta \,\text{in (1)}]$

Sol.7(iii) Let $\,\, z=4i=0+4i.\,$ If $\,|z|=r,\, \text{amp}\,\, z=\theta\,(-\pi<\theta \leq \pi),$ then $\,z\,$ can be written in the form: $\,z=r(\cos \theta+i \sin \theta)\rightarrow (1)$

Now, $|z|=\sqrt{(0)^2+(4)^2}=4,\\ \text{Again,}\,z=0+i\times 4 \\ \Rightarrow \text{the point}\,\,(0,4)\,\,\text{lies on the positive x-axis.}$ 

So,  in this case,  amp z $\,=\theta \Rightarrow \tan \theta=\tan \frac{\pi}{2},\,\,\text{where}\,\,0 <\theta <\pi/2. $

Hence,$\,\,\theta=\pi/2$

Hence, $\,\,z=4i,\,$ can be written in modulus-amplitude form: 

$\,z=4\left(\cos (\frac{\pi}{2})+i \sin (\frac{\pi}{2})\right) \\ [\,\text{Putting the value of}\,\,r, \theta \,\text{in (1)}]$

7(iv) Let $\,\, z=(-3+3i)(1-i)\\=-3+3i+3i+3\\=0+6i.\,$ 

If $\,|z|=r,\, \text{amp}\,\, z=\theta\,(-\pi<\theta \leq \pi),$ then $\,z\,$ can be written in the form: $\,z=r(\cos \theta+i \sin \theta)\rightarrow (1)$

Now, $|z|=\sqrt{(0)^2+(6)^2}=6,\\ \text{Again,}\,z=0+i\times 6 \\ \Rightarrow \text{the point}\,\,(0,6)\,\,\text{lies on the positive x-axis.}$ 

So,  in this case,  amp z $\,=\theta \Rightarrow \tan \theta=\tan \frac{\pi}{2},\,\,\text{where}\,\,0 <\theta <\pi/2. $

So,$\,\,\theta=\pi/2$

Hence, $\,\,z=6i,\,$ can be written in modulus-amplitude form: 

$\,z=6\left(\cos (\frac{\pi}{2})+i \sin (\frac{\pi}{2})\right) \\ [\,\text{Putting the value of}\,\,r, \theta \,\text{in (1)}]$

7(v)Let $\,\, z=\frac{\sqrt3-i}{1-\sqrt3i}\\=\frac{(\sqrt3-i)(1+\sqrt3i)}{(1-\sqrt3i)(1+\sqrt3i)}\\=\frac{\sqrt3+3i-i+\sqrt3}{1+3}\\=\frac{\sqrt3}{2}+i\left(\frac 12\right)$ 

If $\,|z|=r,\, \text{amp}\,\, z=\theta\,(-\pi<\theta \leq \pi),$ then $\,z\,$ can be written in the form: $\,z=r(\cos \theta+i \sin \theta)\rightarrow (1)$

Now, $|z|=\sqrt{\left(\frac{\sqrt3}{2}\right)^2+\left(\frac 12\right)^2}=1,\\ \text{Again,}\,z=\frac{\sqrt3}{2}+i\left(\frac 12\right) \\ \Rightarrow \text{the point}\,\,(\frac{\sqrt3}{2},\frac 12)\,\,\text{lies in the first quadrant.}$ 

So,  in this case,  amp z $\,=\theta \\  \Rightarrow \tan \theta=\tan \frac{1/2}{\sqrt3/2}=\frac{1}{\sqrt3},\,\,\text{where}\,\,0 <\theta <\pi/2. $

So,$\,\,\theta=\pi/6$

Hence, $\,\,z=\frac{\sqrt3-i}{1-\sqrt3i},\,$ can be written in modulus-amplitude form: 

$\,z=1.\left(\cos (\frac{\pi}{6})+i \sin (\frac{\pi}{6})\right) \\ [\,\text{Putting the value of}\,\,r, \theta \,\,\,\text{in (1)}]$

8. If $\,\,a,b,c,d,x,y\,\,$ are real and $\,(a+ib)(c+id)=x+iy,\,\,\,$show that 

$\,(i)\,(a-ib)(c-id)=x-iy  \\ (ii)\,\,(ac-bd)^2+(ad+bc)^2=x^2+y^2$ 

Sol. $\,(a+ib)(c+id)=x+iy \\ \Rightarrow (ac-bd)+i(ad+bc)=x+iy,\\ [\,\text{Since,}\,\,i^2=-1] \\ \Rightarrow x=ac-bd,\,\,y=ad+bc \cdots (1) \\ \,\,[\text{Comparing both sides of the equ.}]$

Now, $\,(i)\,(a-ib)(c-id) \\ = (ac-bd)+i(-ad-bc)\\=(ac-bd)-i(ad+bc)\\=x-iy\,\,[\text{By (1)}]$

Using (1), we can easily prove  2nd part [8(ii)]. 

Alternative way to prove $8(ii)\,\,$:

We have , $\,\,(a+ib)(c+id)=x+iy \\ \Rightarrow |(a+ib)(c+id)|=|x+iy| \\ \Rightarrow |(ac-bd)+i(ad+bc)|=|x+iy| \\ \Rightarrow \sqrt{(ac-bd)^2+(ad+bc)^2}=\sqrt{x^2+y^2} \\ \Rightarrow (ac-bd)^2+(ad+bc)^2=x^2+y^2$

9. If $\,\,\frac{a+ib}{c+id}=x+iy,\,\,$prove that $\,(i)\,\frac{a-ib}{c-id}=x-iy\,\\ (ii)\,\frac{a^2+b^2}{c^2+d^2}=x^2+y^2$

Sol. (i) $\,\,\frac{a+ib}{c+id}=x+iy \\ \Rightarrow \frac{(a+ib)(c-id)}{(c+id)(c-id)}=x+iy \\ \Rightarrow \frac{(ac+bd)+i(bc-ad)}{c^2+d^2}=x+iy \\ \Rightarrow \frac{ac+bd}{c^2+d^2}=x,\,\, \frac{bc-ad}{c^2+d^2}=y \rightarrow (1)$

Now, $\,\,\frac{a-ib}{c-id}\\=\frac{(a-ib)(c+id)}{(c-id)(c+id)}\\=\frac{(ac+bd)-i(bc-ad)}{c^2+d^2}\\=\frac{ac+bd}{c^2+d^2}-i\frac{bc-ad}{c^2+d^2}\\=x-iy \rightarrow(2)$

Sol. (ii) $\,x^2+y^2\\=\left(\frac{ac+bd}{c^2+d^2}\right)^2+\left(\frac{bc-ad}{c^2+d^2}\right)^2\\=\frac{(ac+bd)^2+(bc-ad)^2}{(c^2+d^2)^2}\\=\frac{a^2c^2+2abcd+b^2d^2+b^2c^2-2abcd+a^2d^2}{(c^2+d^2)^2}\\=\frac{c^2(a^2+b^2)+d^2(a^2+b^2)}{(c^2+d^2)^2}\\=\frac{(a^2+b^2)(c^2+d^2)}{(c^2+d^2)^2}\\=\frac{a^2+b^2}{c^2+d^2} \rightarrow(3)$

From (2) and (3), the required result follows.

10. If $\,\,x+iy=\,\sqrt{\frac{a+ib}{c+id}},\,\,$prove that $\,(x^2+y^2)^2=\frac{a^2+b^2}{c^2+d^2}$

Sol. $\,\,x+iy=\,\sqrt{\frac{a+ib}{c+id}}\\ \Rightarrow (x+iy)^2=\frac{a+ib}{c+id} \\ \Rightarrow |(x+iy)^2|=|\frac{a+ib}{c+id}|\\ \Rightarrow |x+iy|^2=\frac{|a+ib|}{|c+id|} \\ \Rightarrow x^2+y^2=\frac{\sqrt{a^2+b^2}}{\sqrt{c^2+d^2}}\\ \Rightarrow (x^2+y^2)^2=\frac{a^2+b^2}{c^2+d^2}$