COMPLEX NUMBERS (Part-1) | S.N. Dey Math Solution series

 

Complex Numbers, S.N.Dey


Definition of complex number : If an ordered pair $\,(x, y)\,$ of two real numbers $\,x\,$ and $\,y\,$ is represented by the symbol $\,x+iy,\,$ where $\,i=\sqrt{-1},\,$ then the ordered pair $\,(x, y)\,$ is called a complex number. If $\,z=x+iy\,$ then $\,x\,$ is called the real part of the complex number and $\,y\,$ is called its imaginary part.

If $\,x, y\,$ are real and $\,i= \sqrt{-1}\,$ then the complex numbers $\,x+iy\,$ and $\,x-iy\,$ are said to be conjugate to each other; conjugate of a complex number z is denoted by $\,\overline{z}\,$.

If $\,z=x+iy\,$ then the positive square root of $\,(x^2 + y^2)\,$ is called the modulus or absolute value of $\,z\,$ and is denoted by $\,|z|\,$ or mod z. Thus, if $\,z=x+iy\,$ then, $\,|z|=\sqrt{x^2 + y^2}\,$. If $\,\overline{z}$ be the conjugate of $\,z\,$ then $\,|z|=\sqrt{z.\bar{z}}. $

Again, if $\,z= x+iy\,$ then the unique value of $\,\theta\,$ satisfying $\,x=|z| \cos \theta,\,\,\, y = |z|\sin \theta$ and $-\pi<\theta<\pi\,\,$ is called the principal value of argument (or amplitude) of $\,z\,$ and is denoted by $\arg z\,$ or amp $z$. If the point $\,P(z)\,$ in the Argand diagram represents the complex number $\,z=(x, y) = x+iy\,$ and $\,\arg z=\theta\,$ then

$\,(i)\, 0<\theta<\pi/2,\,\,$ when $\,P\,$ lies in the first quadrant

$(ii)\,\pi/2 < \theta<\pi,\,\,$ when $\,P\,$ lies in the second quadrant 

$(iii)\,-\pi < \theta<-\pi/2,\,\,$  when $\,P\,$lies in the third quadrant

$(iv)-\pi/2 < \theta<0,\,\,$  when $\,P\,$ lies in the fourth quadrant.

A. $\,\,z=r(\cos \theta+i\sin \theta)\,\,$ where $\,\,r=|z|\,\,$and $\,\,\theta=\arg z,\, -\pi< \theta \leq \pi,\,\,$ is called the modulus-amplitude form of the complex number $\,z.$

B. If $\,x+iy=0,\,$then $\,x=0,\,y=0.$

C. If $\,x, y, p, q\,$ are real and $\,x+iy=p+iq\,$ then $\,x = p\,$ and $\,y=q\,.$

D. $\,\,i=\sqrt{-1};\,i^2=-1;\,i^3=-i;\,i^4=1.\,\,$Any integral power of $\,i\,$ is $\,i\,$ or $\,(-i)\,$ or $\,(-1)\,$ or $\,1\,$.

E. $|z_1+z_2| \leq |z_1|+|z_2|,\,\,\,|z_1z_2|=|z_1||z_2|.$

F.$\,\,(i)\,\, \arg(z_1z_2)=\arg(z_1)+\arg(z_2)+m, \\ (ii)\,\, \arg(\frac{z_1}{z_2})=\arg(z_1)-\arg(z_2)+m.$

G. Cube roots of $\,1\,$ are $\,1,\omega,\omega^2\,\,\text{where}\,\,\omega=\frac{-1 \pm \sqrt{3}i}{2}.$  

Here, $\,\omega\,$ and $\,\omega^2\,$ are called the imaginary cube roots of $\,1. $

H. If $\,\omega\,$ be an imaginary cube root of unity, then $\,\omega^3=1,\,\,1+\omega+\omega^2=0.$ 


1. If $\,\,z_1=1+i\sqrt3,\,\,z_2=\sqrt3-i,\,\,$Show that 

$(i)\,\arg (z_1z_2)=\arg z_1+\arg z_2,  \\ (ii)\, \arg \frac {z_1}{z_2}=\arg z_1-\arg z_2$

Sol. $\,\, \arg z_1=\tan^{-1}(\sqrt3/1)=\pi/3.$  

Since the point $\,z_2\,$  lies in the fourth quadrant on the complex plane,  

$\, \arg  ​z_2=\tan^{-1} \left( -\frac{1}{\sqrt3}\right)=-\frac{\pi}{6}$ 

Now, $(i)\,\,z_1z_2=(1+i\sqrt3)(\sqrt3-i)=2\sqrt3+2i.$  

Since, the point  $\,(2 \sqrt3,2)\,$ lies in the first quadrant on the complex plane,   $\,\, \arg (z_1z_2)=\tan^{-1}\left( \frac{2}{2\sqrt3}\right)=\frac{\pi}{6}, \\ \arg z_1+\arg z_2= \pi/3-\pi/6=\pi/6 \\ \therefore \arg (z_1z_2)=\arg z_1+\arg z_2$


(ii) $\frac{z_1}{z_2}=\frac{1+i\sqrt3}{\sqrt3-i}\\~~~~=\frac{(1+i\sqrt3)(\sqrt3+i)}{(\sqrt3+i)(\sqrt3-i)}\\~~~~~=\frac{\sqrt3+i+3i-\sqrt3}{4}\\~~~~~=i \\ \therefore \arg \frac {z_1}{z_2}=\tan^{-1} (1/0)=\pi/2 \\ \arg z_1-\arg z_2=\pi/3+\pi/6=\pi/2 \\ \therefore \arg \frac {z_1}{z_2}=\arg z_1-\arg z_2$


 2. If $\,|z_1|=|z_2|\,$ and $\,\,\arg z_1+\arg z_2=0,\,\,$then show that $\,z_1\,$ and $\,z_2\,$ are two complex conjugate numbers.

Sol. Let $\,\,z_1=r(\cos \theta +i \sin \theta),\cdots (1) \\ \therefore |z_1|=r,\,\,\arg z_1=\theta.$

Since $\,|z_1|=|z_2|=r\,$ and $\,\,\arg z_1+\arg z_2=0 \\  \Rightarrow\arg z_2=-\arg z_1=-\theta,\\ z_2=r\left[\cos(-\theta)+i\sin(-\theta)\right] \\~~~~=r(\cos \theta-i\sin \theta) \cdots (2)$

Hence, from (1) and (2), we can conclude that $\,z_1\,$ and $\,z_2\,$ are two complex conjugate numbers.


3. If $\,\,|z_1|=|z_2|=1,\,$ and $\,\,\arg z_1+\arg z_2=0\,$ then show that $\,z_1=\frac{1}{z_2}.$

Sol. Since $\,\,|z_1|=|z_2|=1,\,$ we suppose that   
$\,z_1=1(\cos \theta +i\sin \theta),\,\,\text{where}\,\, \theta=\arg z_1$ 

Again, $\,\,\arg z_1+\arg z_2=0 \\ \Rightarrow \arg z_2=-\arg z_1=-\theta.$

Hence, $\,\,z_2=1.(\cos (-\theta)+i\sin (-\theta))\\~~~~~= \cos \theta-i \sin \theta.$

So, $\,\, \frac{1}{z_2}=\frac{1}{(\cos \theta -i\sin \theta)} \\~~~~~=\frac{(\cos \theta +i\sin \theta)}{(\cos \theta +i\sin \theta)(\cos \theta -i\sin \theta)} \\~~~~~= \frac{(\cos \theta +i\sin \theta)}{\cos^2\theta+\sin^2\theta}\\~~~~~=(\cos \theta +i\sin \theta)\\~~~~~=z_1 \cdots (1)$

Hence, from (1), the result follows.

4. If $\,\,|z_1|=|z_2|\,$ and $\,\,\arg z_1 \sim \arg z_2= \pi,\,$ then show that $\,z_1+ z_2=0.$

Sol. By question, let $\,\,|z_1|=|z_2|=r,\, \,\,\arg z_1= \theta_1,\,\, \arg z_2=\theta_2,\,\,$ so that $\,\,z_1=r(\cos \theta_1+i\sin \theta_1), \\ z_2=r(\cos \theta_2+i\sin \theta_2).$

Without any loss of generality, let $\,\, \theta_1 >\theta_2 \\ \therefore \theta_1=\theta_2+\pi.$

Hence, $\,\,z_1=r(\cos(\theta_2+\pi)+i\sin(\theta_2+\pi))\\~~~~~=r(-\cos \theta_2-i\sin \theta_2)\\~~~~=-r(\cos \theta_2+i\sin \theta_2)\\~~~~=-z_2 \\ \Rightarrow z_1+z_2=0.$ 

In the same way, if   $\,\, \theta_2 >\theta_1\, [\Rightarrow \theta_2=\theta_1+\pi],$ we can prove the same result.

5. If $\,\,z_1=(\sqrt3-1)+(\sqrt3+1)i,\,\,z_2=-\sqrt3+i,\,\,$ find $\,\,\arg z_1,\,\,\arg z_2;$ hence, calculate $\,\,\arg (z_1z_2).$

Sol. $\,\,\arg z_1=\tan^{-1} \left(\frac{\sqrt3+1}{\sqrt3-1}\right)\\ \Rightarrow \arg z_1=\tan^{-1}(2+\sqrt3)\\~~~~~~~~~~~~~~~=\frac{5 \pi}{12}, \\ \text{since} \,z_1\, \text{lies in the first quadrant.}$

Also, $\arg z_2=\tan^{-1} \left(-\frac{1}{\sqrt3}\right)=\frac{5 \pi}{6}, \\ \text{since }\,z_2\,\text{lies in the 2nd quadrant.}$

Now, $\arg(z_1z_2)=\arg z_1+\arg z_2+m,\,\,$where $\,m=0\, \,\,\text{or}\,\, \pm 2\pi.$

Hence, $\arg z_1 +\arg z_2=\frac{5 \pi}{12}+\frac{5 \pi}{6}\\~~~~~~~~~~~~~~~~~~~~~~~~~= \frac{15 \pi}{12}\\~~~~~~~~~~~~~~~~~~~~~~~~~= \frac{5 \pi}{4} >\pi \\ \therefore m=-2\pi \\ \therefore \arg(z_1z_2)=\frac{5 \pi}{4}-2\pi=-\frac{3\pi}{4}$ 
  
6. Find the values of $\,x\,$ and $\,y\,$ (real) for which the following equation is satisfied: $$\frac{(1+i)x-2i}{3+i}+\frac{(2-3i)y+i}{3-i}=i $$

Sol. We have , $\frac{(1+i)x-2i}{3+i}+\frac{(2-3i)y+i}{3-i}=i \\ \implies \frac{\{(1+i)x-2i\}(3-i)+\{(2-3i)y+i\}(3+i)}{3^2-i^2}=i \\ \implies ................................................. \\ \implies ~~~~~~~~~~~~~~~\frac{(4x+9y-3)+i(2x-7y-3)}{10}=i \cdots (1) \\ \Rightarrow \text{Comparing both sides of (1), we get } \\ 4x+9y-3=0, \cdots(2) \\ 2x-7y-3=10 \cdots(3)$

Hence, solving (2) and (3), we get $\,\,x=3,\,\,y=-1.$ 

7. If $\,\,x,y,b\,$ are real, $\,\,z=x+iy,\,\,\frac{z-i}{z-1}=ib,\,\,$ show that, $$\left(x-\frac 12\right)^2+\left(y-\frac 12\right)^2=\frac 12.$$

Sol. $\,\frac{z-i}{z-1}=ib \\ \Rightarrow \frac{x+iy-i}{x+iy-1}=ib \\ \Rightarrow \frac{(x+iy-i)(x-1-iy)}{(x-1+iy)(x-1-iy)}=ib \\ \Rightarrow \frac{x(x-1)-y(1-y)-ixy+i(y-1)(x-1)}{(x-1)^2+y^2}=ib \\ \therefore \frac{x(x-1)-y(1-y)}{(x-1)^2+y^2}=0 \\ \Rightarrow x^2+y^2-x-y=0\\ \Rightarrow \left(x^2-2.x.\frac 12+ (\frac 12)^2\right)\\+\left(y^2-2.y.\frac 12+ (\frac 12)^2\right)=\frac 14+\frac 14 \\ \therefore \left(x-\frac 12\right)^2+\left(y-\frac 12\right)^2=\frac 12$

Hence, follows the result.

8. If $\,\,x+iy=\frac{2}{3+\cos \theta+i\sin \theta},\,$ show that $\,2x^2+2y^2=3x-1$

Sol. $\,\,x+iy=\frac{2}{3+\cos \theta+i\sin \theta} \\ \Rightarrow x+iy=\frac{2(3+\cos \theta-i\sin \theta)}{(3+\cos \theta+i\sin \theta)(3+\cos \theta-i\sin \theta)}\\~~~~~~~~~~~~~~~=\frac{3 +\cos \theta}{5+3\cos \theta}+i\frac{-\sin \theta}{5+3\cos \theta} \cdots(1)$

Comparing both sides of (1), we get $\,\,x=\frac{3 +\cos \theta}{5+3\cos \theta} ,\\ y=\frac{-\sin \theta}{5+3\cos \theta}$

Now, $\,\,2(x^2+y^2)-3x+1\\=2 \left[(\frac{3 +\cos \theta}{5+3\cos \theta})^2 +(\frac{-\sin \theta}{5+3\cos \theta})^2\right]-3(\frac{3 +\cos \theta}{5+3\cos \theta})+1\\=\frac{2}{(5+3\cos \theta)^2} \times [(3+\cos \theta)^2+\sin^2\theta]-\frac{9+3\cos\theta}{5+3\cos \theta}+1\\= \frac{2}{(5+3\cos \theta)^2} \times [9+6\cos\theta+\cos^2\theta+\sin^2\theta]-\frac{9+3\cos\theta}{5+3\cos \theta}+1\\=\frac{2}{(5+3\cos \theta)^2} \times [10+6\cos\theta] -\frac{9+3\cos\theta}{5+3\cos \theta}+1\\=\frac{2}{(5+3\cos \theta)^2} \times 2(5+3\cos\theta)-\frac{9+3\cos\theta}{5+3\cos \theta}+1\\=\frac{4}{5+3\cos\theta}-\frac{9+3\cos\theta}{5+3\cos \theta}+1\\=\frac{4-9-3\cos\theta+5+3\cos\theta}{5+3\cos\theta}\\=0 \cdots (2)$

Hence, from (2), the result follows.

9. If $\,\,y=\sqrt{x^2+6x+8},\,\,$ show that one value of $\, \sqrt{1+iy} +\sqrt{1-iy}\,\,[i=\sqrt{-1}]$ is $\,\, \sqrt{2x+8}.$

Sol. $\,\,y=\sqrt{x^2+6x+8} \\ \Rightarrow 1+iy \\=1+i \sqrt{(x+4)(x+2)} \\ =\frac 12 [2+2i\sqrt{(x+4)(x+2)}] \\=\frac 12[(x+4)-(x+2)+2\sqrt{x+4}\{ i\sqrt{x+2}\}]\\=\frac 12[(\sqrt{x+4})^2+2. \sqrt{x+4}. i\sqrt{x+2}\\+(i\sqrt{x+2})^2]\\=\frac 12[\sqrt{x+4}+i\sqrt{x+2}]^2 \\ \Rightarrow \sqrt{1+iy}= \pm\frac{1}{\sqrt2}[\sqrt{x+4}+i\sqrt{x+2}] \rightarrow(1) \\ \text{Similarly, }\sqrt{1-iy}\\=\pm\frac{1}{\sqrt2}[\sqrt{x+4}-i\sqrt{x+2}]\rightarrow (2) $

Hence , from (1) and (2), we get  one value of $\, \sqrt{1+iy} +\sqrt{1-iy}\,\,[i=\sqrt{-1}]$ is $=\frac{1}{\sqrt2}[\sqrt{x+4}+i\sqrt{x+2}+\sqrt{x+4}-i\sqrt{x+2}]\\=\frac{1}{\sqrt2} \times {2 \sqrt{x+4}}\\=\sqrt2 \times \sqrt{x+4}\\=\sqrt{2x+8}$

10.  Factorize : $\,\,x^2+y^2+z^2-xy-yz-zx$

Sol. $\,\,x^2+y^2+z^2-xy-yz-zx \\= x^2+y^2+z^2+(\omega+\omega^2)xy+(\omega+\omega^2)yz\\+(\omega+\omega^2)zx \\= x^2+y^2+z^2+\omega xy+\omega^2xy+\omega yz\\+\omega^2 yz+\omega zx+\omega^2 zx \\=x(x+\omega y+\omega^2z)+\omega^2y(x+\omega y+\omega^2z)\\+\omega z(x+\omega y+\omega^2 z)\,\,[\text{Using}\,\,\omega^3=1]\\=(x+\omega y+\omega^2 z)(x+\omega^2 y+\omega z)$

11. Prove that $\,\,(a+b\omega+c \omega^2)^3+(a+b\omega^2+c\omega)^3=27abc,\,$ if $\,\,a+b+c=0.$

Sol. Let $\,\,x= a+b\omega+c\omega^2,\,\,y=a+b\omega^2+c\omega.$

We know, $\,\,x^3+y^3=(x+y)(x+y\omega)(x+y\omega^2)\rightarrow(1)$

Now $\,\,x+y \omega=a+b\omega+c\omega^2+a\omega+b\omega^3+c\omega^2\\=a(1+\omega)+b(\omega+1)+2c\omega^2\,\,[\text{Since,}\,\omega^3=1,\,1+\omega+\omega^2=0]\\=-a\omega^2-b\omega^2+2c\omega^2\\=\omega^2(2c-a-b)\\=\omega^2(3c-\overline{a+b+c})\\=3c\omega^2\,\,[\,\,\text{Since,}\,\,a+b+c=0] \rightarrow(2)$

Again $\,\,x+y\omega^2=a+b\omega+c\omega^2+a\omega^2+b\omega^4+c\omega^3\\=a+b\omega+c\omega^2+a\omega^2+b\omega+c\\=a(1+\omega^2)+2b\omega+c(1+\omega^2)\\=-a\omega+2b\omega-c\omega \,\,[\text{Since,}\,\omega^3=1,\,1+\omega+\omega^2=0]\\=\omega(2b-a-c)\\=\omega(3b-\overline{a+b+c})\\=3b\omega\,\,[\,\,\text{Since,}\,\,a+b+c=0]\rightarrow(3)$

and , $x+y= a+b\omega+c\omega^2+a+b\omega^2+c\omega\\=2a+b(\omega+\omega^2)+c(\omega+\omega^2)\\=2a-b-c \,\,[\text{Since,}\,\omega^3=1,\,1+\omega+\omega^2=0]\\=(3a-\overline{a+b+c})\\=3a\,\,[\,\,\text{Since,}\,\,a+b+c=0] \rightarrow(4)$

Hence, from $\,(1),(2),(3)\, \text{and}\, (4)$, we get   
$(a+b\omega+c\omega^2)^3+(a+b\omega^2+c\omega)^3\\=(3a)(3c\omega^2)(3b\omega)\\=27abc\omega^3\\=27abc$

12. Express each of the following expressions as the sum of two squares :

$(i)\,\,(1+x^2)(1+y^2) \\ (ii)\,(a^2+b^2)(c^2+d^2) \\ (iii)\,\,(1+x^2)(1+y^2)(1+z^2) $

Sol. $\,(i)\,(1+x^2)(1+y^2)\\=[(1+ix)(1-ix)] \times [(1+iy)(1-iy)]\\=[(1+ix)(1+iy)]\times [(1-ix)(1-iy)]\\=[(1-xy)+i(x+y)] \\  \times [(1-xy)-i(x+y)]\\=(1-xy)^2+(x+y)^2 \\ (ii)\,(a^2+b^2)(c^2+d^2)\\=(a^2-i^2b^2)(c^2-i^2d^2)\\=[(a+ib)(a-ib)]\times [(c+id)(c-id)] \\=[(a+ib)(c+id)] \times [(a-ib)(c-id)]\\=[(ac-bd)+i(ad+bc)] \\ \times [(ac-bd)-i(ad+bc)]\\=(ac-bd)^2+(ad+bc)^2 \\ (iii)\,\,(1+x^2)(1+y^2)(1+z^2)\\=[(1+ix)(1-ix)]\times[(1+iy)(1-iy)] \\ \times[(1+iz)(1-iz)]\\=[(1+ix)(1+iy)(1+iz)] \\ \times [(1-ix)(1-iy)(1-iz)]\\=[(1-xy-yz-zx)+i(x+y+z-xyz)]\\ \times [(1-xy-yz-zx)-i(x+y+z-xyz)]\\=(1-xy-yz-zx)^2+(x+y+z-xyz)^2$



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