Differentiation (Part-38) | S N De

 

In the previous chapter,  we have discussed Long Answer Type Questions [From Quest. No. 1 to 12]. In this article we will solve the rest of the problems of S.N. Dey Math exercise book. 

13. If $\,\,a,b\,$ are real and $\,a^2+b^2=1,\,$ show that the equations $\frac{\sqrt{1+x}-i\sqrt{1-x}}{\sqrt{1+x}+i\sqrt{1-x}}=a-ib\,\,$ is satisfied by a real value of $\,x.$

Sol. $\left|\frac{\sqrt{1+x}-i\sqrt{1-x}}{\sqrt{1+x}+i\sqrt{1-x}}\right|=\frac{\sqrt{1+x+1-x}}{\sqrt{1+x+1-x}}=1. \\ \text{Again,}\,\, |a-ib|=\sqrt{a^2+b^2}=1 .$ 

Hence, the equation is satisfied by a real value of $\,x.$

14. If $\,z=x+iy\,\,$ and $\,|z|=1,\,$ show that , $\,\,\left|\frac{z-1}{z+1}\right|\,(z \neq 1)\,\,$ is a purely imaginary quantity.

Sol. $\,\frac{z-1}{z+1}\\=\frac{x+iy-1}{x+iy+1}\\=\frac{(x-1+iy)(x+1-iy)}{(x+1+iy)(x+1-iy)}\\=\frac{(x^2+y^2-1)+2iy}{(x+1)^2+y^2}\\=i.\frac{2y}{(x+1)^2+y^2}\,\,\,[\text{Since,}\,\,x^2+y^2=1] \rightarrow(1)$

From (1), we notice that $\,\,\left|\frac{z-1}{z+1}\right|\,(z \neq 1)\,\,$ is a purely imaginary quantity.

15. If $\,z_1\,$ and $\,z_2\,$ be any two non-zero complex numbers such that $\,\,|z_1+z_2| =|z_1|+|z_2|\,\,$ then prove that $\,\,\arg z_1 =\arg z_2.$ 

Sol. Let $\,z_1=r_1(\cos \theta_1+i \sin \theta_1),\\z_2=r_2(\cos \theta_2+i \sin \theta_2). $

Now, $\,\,|z_1+z_2| =|z_1|+|z_2| \\ \Rightarrow |(r_1 \cos \theta_1+r_2 \cos \theta_2)+i(r_1\sin\theta_1+r_2\sin\theta_2)|=r_1+r_2 \\ \Rightarrow \sqrt{r_1^2+r_2^2+2r_1r_2\cos(\theta_1-\theta_2)}=r_1+r_2 \\ \Rightarrow r_1^2+r_2^2+2r_1r_2\cos(\theta_1-\theta_2)\\=r_1^2+r_2^2+2r_1r_2\\ \Rightarrow \cos(\theta_1-\theta_2)=1 \\ \Rightarrow \theta_1-\theta_2=0,\,\,[\text{since,}\,\,-\pi<\theta_1,\theta_2 \leq \pi] \\ \Rightarrow \theta_1=\theta_2 \\ \therefore \arg z_1=\arg z_2.$

16. If $\,\,\alpha\,$ be the real positive cube root and $\,\,\beta,\gamma\,$ be the complex cube roots of $m,\,\,$a real positive number, then for any $\,x,y,z\,$ show that, $\,\frac{x\beta+y\gamma+z\alpha}{x\gamma+y\alpha+z\beta}=w^2,\,\,$ where $\,w\,\,\,$is a complex cube root of unity.

Sol. Since $\,\,\alpha\,$ be the real positive cube root and $\,\,\beta,\gamma\,$ be the complex cube roots of $m,\,\,$a real positive number, $\,\alpha=m^{\frac 13},\,\,\beta=m^{\frac 13}w,\,\,\gamma=m^{\frac 13}w^2. \\ \text{Now,}\,\quad \frac{x\beta+y\gamma+z\alpha}{x\gamma+y\alpha+z\beta} \\=\frac{x.m^{\frac 13}w+y.m^{\frac 13}w^2+z.m^{\frac 13}}{x.m^{\frac 13}w^2+y.m^{\frac 13}+z.m^{\frac 13}w}\\=\frac{xw+yw^2+z}{xw^2+y+zw}\\=\frac{w^2(xw^2+y+zw)}{xw^2+y+zw}\\=w^2.$

17. If $\,\, \arg \left(\frac{z-1}{z+1}\right)=\frac{\pi}{4},\,$ show that the locus of $\,\,z\,$ in the complex plane is a circle.

Sol. $\,\, \arg \left(\frac{z-1}{z+1}\right)=\frac{\pi}{4} \\ \Rightarrow \arg \left(\frac{x+iy-1}{x+iy+1}\right)=\frac{\pi}{4}\\ \Rightarrow \arg \left(\frac{(x-1)+iy}{(x+1)+iy} \times \frac{(x+1)-iy}{(x+1)-iy}\right)=\frac{\pi}{4} \\ \Rightarrow \arg \left[\frac{(x^2+y^2-1)+i.(2y)}{(x+1)^2+y^2}\right]=\frac{\pi}{4} \\ \therefore \tan{(\pi/4)}=\frac{2y}{(x+1)^2+y^2} \\ \Rightarrow x^2+y^2-2y-1=0 \\ \Rightarrow x^2+y^2-2y+1-2=0 \\ \Rightarrow x^2+(y-1)^2=2 \cdots(1)$

Clearly, the equation (1) indicates a circle with center  $\,(0,1)\,$ and radius $\,\sqrt2\,$ unit.   

Hence,the locus of $\,\,z\,$ in the complex plane is a circle.

18. In the complex plane, the vertices of an equilateral triangle are represented by the complex numbers $\,\,z_1,z_2\,\,$ and $\,z_3.\,$ Prove that, $\,\,\frac{1}{z_1-z_2}+\frac{1}{z_2-z_3}+\frac{1}{z_3-z_1}=0.$

Sol. Let $\,\,z_2-z_3=p,\\ z_3-z_1=q,\\z_1-z_2=r. \\ \therefore p+q+r=0 \\ \therefore \bar{p}+\bar{q}+\bar{r}=0.$

Since, the triangle is an equilateral triangle, $\,|z_2-z_3|^2=|z_3-z_1|^2=|z_1-z_2|^2 \\ \therefore p.\bar{p}= q.\bar{q}= r.\bar{r}=k(\neq 0) \\ \therefore \bar{p}=\frac kp,\,\,\bar{q}= \frac kq,\,\,\bar{r}=\frac kr. \\ \therefore k(\frac 1p+\frac 1q+\frac 1r)=0 \\ \Rightarrow \frac{1}{z_1-z_2}+\frac{1}{z_2-z_3}+\frac{1}{z_3-z_1}=0.$

19. If $\,\,z_1^2+z_2^2+z_3^2-z_1z_3-z_3z_2-z_1z_2=0,\,\,$ prove that, $\,\,|z_2-z_3|=|z_3-z_1|=|z_1-z_2|,\,\,$ where $\,z_1,z_2,z_3\,\,$ are complex numbers.

Sol. $\,\,z_1^2+z_2^2+z_3^2-z_1z_3-z_3z_2-z_1z_2=0, \\ \Rightarrow (z_1+wz_2+w^2z_3)(z_1+w^2z_2+wz_3)=0.$

Now, $\,z_1+wz_2+w^2z_3=0 \\ \Rightarrow z_1+wz_2+(-1-w)z_3=0 \\ \Rightarrow z_1-z_3=w(z_3-z_2) \\ \Rightarrow |z_1-z_3|=|w||z_3-z_2| \\ \Rightarrow |z_1-z_3|=|z_3-z_2| \rightarrow(1) \\ \text{Next,}\,\,z_1+wz_2+w^2z_3=0 \\ \Rightarrow z_1+(-1-w^2)z_2+w^2z_3=0 \\ \Rightarrow z_1-z_2=w^2(z_2-z_3)\\ \Rightarrow |z_1-z_2|=|w^2||z_2-z_3|\\ \Rightarrow |z_1-z_2|=|z_2-z_3| \rightarrow(2)$

Hence, from (1) and (2), we get $\,\,|z_2-z_3|=|z_3-z_1|=|z_1-z_2|.$

Similarly, with $\,\,z_1+w^2z_2+wz_3=0,\,$ we can prove that $\,\,|z_2-z_3|=|z_3-z_1|=|z_1-z_2|.$

20. Solve : $\,\,\bar{z}=iz^2.$

Sol. Let $\,z=a+ib \\ \therefore \,\bar{z}=iz^2 \\ \Rightarrow \overline{a+ib}=i(a+ib)^2 \\ \Rightarrow a-ib=i(a^2-b^2+i(2ab))\\ \Rightarrow a-ib=i(a^2-b^2)-2ab \\ \Rightarrow a+2ab-ib=i(a^2-b^2) \\ \therefore a+2ab=0 \cdots(1) \\ -b=a^2-b^2 \cdots(2)$

hence, from (1), we get $\,\,a=0,\,\,b=-\frac 12.$

Now, putting $\,\,a=0\,$ in (2), we get $\,\,b=0,\,\,b=1.$

Hence, $\,z=a+ib=0+i.0=0 ,\\ z=0+i.1=i.$

Again, putting $\,\,b=-\frac 12,\,\,$ in (2) , we get $\,a=\pm \frac{\sqrt3}{2}$

So, $\,z=\pm \frac{\sqrt3}{2}-\frac{i}{2}.$

21. Solve : $\,\,|z|+z=2+i,\,\,z\,\,$ being a complex number.

Sol. Let $\,\,z=a+ib \\ \therefore |z|+z=2+i \\ \Rightarrow \sqrt{a^2+b^2}+(a+ib)=2+i\\ \sqrt{a^2+b^2}+a=2,\,\,\text{and}\,\,b=1. \\ \therefore a=\frac 34 \\ \therefore z=\frac 34+i. $ 

22.If $\,\,z=x+iy,\,\,$ then show that $\,\,|x|+|y| \leq \sqrt2|x+iy|.$

Sol. Since we know $\,\,G.M. \leq A.M. \\ \Rightarrow \sqrt{|x|^2|y|^2} \leq \frac{|x|^2+|y|^2}{2} \\ \Rightarrow 2|x||y| \leq |x|^2+|y|^2 \\ \Rightarrow |x|^2+|y|^2 +2|x||y| \leq 2(|x|^2+|y|^2 ) \\ \Rightarrow (|x|+|y|)^2 \leq 2(x^2+y^2 ) \\ \Rightarrow |x|+|y| \leq \sqrt2 \sqrt{x^2+y^2 } \\ \Rightarrow |x|+|y| \leq \sqrt2 |x+iy|.$