Differentiation (Part-38) | S N De

In the previous article, we have discussed Short Answer Type Questions (from 11 to 15(xii)) and their respective solutions. In this article , we will discuss some more mathematical problems of S.N. Dey Math book Exercise from Complex Numbers along with their solutions. So, let's start. 

16. If $\,\,x\sqrt2=1+\sqrt{-1},\,\,$ find the value of $\,\,x^6+x^4+x^2+2.$

Sol. $\,\,x\sqrt2=1+\sqrt{-1} \\ \Rightarrow x=\frac{1+i}{\sqrt2},\,\,[\text{Since,}\,\,i=\sqrt{-1}]\\ \Rightarrow x^2=\frac{1+i^2+2i}{2}\\~~~~~~~~~=\frac 12(1-1+2i)\\~~~~~~~~~=\frac 12 \times 2i=i\\ \Rightarrow x^2=i \rightarrow(1)$

Hence, $\,\,x^6+x^4+x^2+2\\=(x^2)^3+(x^2)^2+x^2+2\\=i^3+i^2+i+2\,\,\,[\text{By (1)}]\\=i^2.i-1+i+2\\=-i-1+i+2\\=1$

17(i) Show that one value of $\,\,\sqrt{i}+\sqrt{-i}\,\,$ is $\,\sqrt{2}.$

Sol. We know, $\,\, \sqrt{i}=\pm \frac{1}{\sqrt2}(1+i)\rightarrow(1)\,\,[\text{Follow}\,15(iii)].$

Similarly, $\,\, \sqrt{-i}=\pm \frac{1}{\sqrt2}(1-i) \rightarrow(2)$

Hence from (1) and (2), we get $\,\,\sqrt{i}+\sqrt{-i}\\=\pm \frac{1}{\sqrt2}(1+i)\pm \frac{1}{\sqrt2}(1-i)\\=\pm \sqrt{2}.$

So, one value of $\,\,\sqrt{i}+\sqrt{-i}\,\,$ is $\,\sqrt{2}.$ 

17(ii) Show that one value of $\,\,\sqrt{1+i}-\sqrt{1-i}\,\,$ is $\,\,\,\,i\sqrt{2(\sqrt2-1)}.$

Sol.  We know, $\,\, \sqrt{1-i}=\pm \frac{1}{\sqrt2}\left(\sqrt{\sqrt2+1}-i\sqrt{\sqrt2-1}\right) \\ \rightarrow(1)\,\,[\text{Follow}\,15(iv)].$

Similarly, $\,\, \sqrt{1+i}=\pm \frac{1}{\sqrt2}\left(\sqrt{\sqrt2+1}+i\sqrt{\sqrt2-1}\right) \\ \rightarrow(2)$

Hence from (1) and (2), we get $\,\,\sqrt{1+i}+\sqrt{1-i}\\=\pm \frac{1}{\sqrt2}\left(\sqrt{\sqrt2+1}+i\sqrt{\sqrt2-1}\right)\\ -\left[\pm \frac{1}{\sqrt2}\left(\sqrt{\sqrt2+1}-i\sqrt{\sqrt2-1}\right)\right]\\=\pm i\sqrt2 \sqrt{\sqrt2-1}\\=\pm i\sqrt{2(\sqrt2-1)}$

So, one value of $\,\,\sqrt{1+i}+\sqrt{1-i}\,\,$ is $\,i\sqrt{2(\sqrt2-1)}.$ 

17(iii) Show that one value of $\,\,(4+3i)^{-1/2}+(4-3i)^{-1/2}\,\,$is $\,\,\frac{3\sqrt2}{5}.$

Sol.  $\,\,4+3i\\=\frac 12(8+6i)\\=\frac 12(9-1+2.3.i)\\=\frac 12 \left[3^2+i^2+2.3.i\right]\\=\frac 12(3+i)^2. \\ \therefore (4+3i)^{-1/2}=\pm \frac{\sqrt2}{3+i}\\=\pm \frac{\sqrt2(3-i)}{(3+i)(3-i)}\\=\pm \frac{\sqrt2(3-i)}{10}$

Similarly, $\,\,(4-3i)^{-1/2}=\pm \frac{\sqrt2(3+i)}{10}$

Hence, $\,\,(4+3i)^{-1/2}+(4-3i)^{-1/2} \\=\pm \left[\frac{\sqrt2(3-i)}{10}+\frac{\sqrt2(3+i)}{10}\right]\\=\pm \frac{3\sqrt2}{5}.$

So, one value of $\,\,(4+3i)^{-1/2}+(4-3i)^{-1/2}\,\,$is $\,\,\frac{3\sqrt2}{5}.$

18. If $\,\omega\,$ be an imaginary cube root of unity, show that $\,\,(i)\,(1-\omega)(1-\omega^2)(1-\omega^4)(1-\omega^5)=9$ 

Sol. Since $\,\omega\,$ be an imaginary cube root of unity, $\,\omega^3=1,\,\,1+\omega+\omega^2=0.$

Now, $\,\,(i)\,(1-\omega)(1-\omega^2)(1-\omega^4)(1-\omega^5)\\=(1-\omega)(1-\omega^2)(1-\omega^3 .\omega)(1-\omega^3. \omega^2)\\=(1-\omega)(1-\omega^2)(1-\omega)(1-\omega^2)\\=\left[(1-\omega)(1-\omega^2)\right]^2\\=\left(1-\omega^2-\omega+\omega^3\right)^2\\=\left[1-(\omega^2+\omega)+1\right]^2\\=\left[1+1+1\right]^2\\=3^2\\=9$ 

18. If $\,\omega\,$ be an imaginary cube root of unity, show that $\,\,(ii)\,(1+\omega-\omega^2)(1-\omega+\omega^2)=4$

Sol. Since $\,\omega\,$ be an imaginary cube root of unity, $\,\omega^3=1,\,\,1+\omega+\omega^2=0.$ 

Now,$\,\,(1+\omega-\omega^2)(1-\omega+\omega^2)\\=(1+\omega+\omega^2-2\omega^2)(1+\omega+\omega^2-2\omega)\\=(0-2\omega^2)(0-2\omega)\\=4\omega^3\\=4\times 1\\=4$

18. If $\,\omega\,$ be an imaginary cube root of unity, show that $\,\,(iii)\,(3+3\omega+5\omega^2)^6\\=(3+5\omega+3\omega^2)^5=64$

Sol. Since $\,\omega\,$ be an imaginary cube root of unity, $\,\omega^3=1,\,\,1+\omega+\omega^2=0.$ 

Now, $(3+3\omega+5\omega^2)^6\\=\left[3(1+\omega+\omega^2)+2\omega^2\right]^6\\=(2\omega^2)^6\\=2^6. (\omega^2)^6\\=64 \times  (\omega^3)^4\\=64 \times 1\\=64 \rightarrow (1)$

Similarly,  $(3+5\omega+3\omega^2)^6\\=\left[3(1+\omega+\omega^2)+2\omega\right]^6\\=(2\omega)^6\\=2^6. (\omega)^6\\=64 \times  (\omega^3)^2\\=64 \times 1\\=64 \rightarrow (2)$

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

18. If $\,\omega\,$ be an imaginary cube root of unity, show that $\,\,(iv)\, \frac{x\omega^2+y\omega+z}{x\omega+y+z\omega^2}=\omega.$

Sol. Since $\,\omega\,$ be an imaginary cube root of unity, $\,\omega^3=1,\,\,1+\omega+\omega^2=0.$ 

Now, $\,\,\, \frac{x\omega^2+y\omega+z}{x\omega+y+z\omega^2}\\=\frac{x\omega^2+y\omega+z \omega^3}{x\omega+y+z\omega^2}\,\,\,\,[\text{Since,}\,\,\omega^3=1]\\=\frac{\omega(x\omega+y+z\omega^2)}{(x\omega+y+z\omega^2)}\\=\omega$

18. If $\,\omega\,$ be an imaginary cube root of unity, show that  $\,\,(x+y)^2+(x \omega+y \omega^2)^2+(x\omega^2+y\omega)^2\\=6xy.$

Sol. Since $\,\omega\,$ be an imaginary cube root of unity, $\,\omega^3=1,\,\,1+\omega+\omega^2=0 \rightarrow(1)$ 

Now, $\,\,(x+y)^2+(x \omega+y \omega^2)^2+(x\omega^2+y\omega)^2\\=x^2+2xy+y^2+x^2\omega^2+2xy \omega^3 \\+y^2\omega^4+x^2\omega^4+2xy\omega^3+y^2\omega^2\\=x^2(1+\omega^2+\omega^3.\omega)+6xy\\+y^2(1+\omega^3.\omega+\omega^2)\\=x^2(1+\omega+\omega^2)+6xy\\+y^2(1+\omega+\omega^2) \,\, [\omega^3=1]\\=6xy\,\,\,[\text{By (1)}]$

18. If $\,\omega\,$ be an imaginary cube root of unity, show that  $ (vi)\,\,(x+y\omega+z\omega^2)^2+(x \omega+y \omega^2+z)^2\\+(x\omega^2+y+z\omega)^2=0.$

Sol. Since $\,\omega\,$ be an imaginary cube root of unity, $\,\omega^3=1,\,\,1+\omega+\omega^2=0 \rightarrow(1)$ 

Now, $\,\,(x+y\omega+z\omega^2)^2+(x \omega+y \omega^2+z)^2\\+(x\omega^2+y+z\omega)^2\\=(x+y\omega+z\omega^2)^2+(x \omega+y \omega^2+z\omega^3)^2\\+(x\omega^2+y\omega^3+z\omega.\omega^3)^2\\=(x+y\omega+z\omega^2)^2+(x+y\omega+z\omega^2)^2 \omega^2 \\+(x+y\omega+z\omega^2)^2 (\omega^2)^2\\=(x+y\omega+z\omega^2)^2(1+\omega^2+\omega.\omega^3)\\=(x+y\omega+z\omega^2)^2(1+\omega^2+\omega)\\=(x+y\omega+z\omega^2)^2 \times 0 \,\,\,[\text{By (1)}]\\=0.$

19. If $\,\,\alpha=\frac{-1-\sqrt{-3}}{2},\,\, \beta=\frac{-1+\sqrt{-3}}{2},\,\,$ show that $\,\alpha^2+\alpha \beta+\beta^2=0.$

Sol. We have, $\,\,\alpha=\frac{-1-\sqrt{-3}}{2},\,\, \beta=\frac{-1+\sqrt{-3}}{2} \\ \therefore \alpha+\beta=\frac{-1-\sqrt{-3}}{2}+\frac{-1+\sqrt{-3}}{2}=-1 \\ \text{and}\,\,\, \alpha . \beta= \frac{-1-\sqrt{-3}}{2} \times \frac{-1+\sqrt{-3}}{2}=\frac{1-(-3)}{4}=1$

Now, $\,\alpha^2+\alpha \beta+\beta^2\\=(\alpha+\beta)^2-\alpha \beta\\=(-1)^2-1\\=0.$

20. Show that $\,\,\left(\frac{-1+\sqrt{-3}}{2}\right)^{19}+\left(\frac{-1-\sqrt{-3}}{2}\right)^{19}=-1.$

Sol. Let $\,\,\frac{-1+\sqrt{-3}}{2}=\omega,\\ \frac{-1-\sqrt{-3}}{2}=\omega^2. \\ \therefore \left(\frac{-1+\sqrt{-3}}{2}\right)^{19}+\left(\frac{-1-\sqrt{-3}}{2}\right)^{19}\\=(\omega)^{19}+(\omega^2)^{19}\\=(\omega^3)^6. \omega+(\omega^3)^{12}.\omega^2\\=1.\omega+1.\omega^2\,\,[\text{Using,}\,\,\omega^3=1]\\=-1\,\,[\text{Using,}\,\,\,1+\omega+\omega^2=0]$

21.  If $\,\,\alpha,\,\beta\,\,$ are the complex cube roots of $\,1,\,$ show that $\,\,\alpha^4+\beta^4+\alpha^{-1}.\beta^{-1}=0.$

Sol. Since $\,\,\alpha,\,\beta\,\,$ are the complex cube roots of $\,1,\,$ let $\,\alpha=\omega,\,\beta=\omega^2. \\ \therefore 1+\omega+\omega^2=0 \\ \Rightarrow 1+\alpha +\beta=0 \\ \Rightarrow \alpha+\beta=-1 \\ \text{and}\,\,\omega. \omega^2=\omega^3=1 \\ \Rightarrow\alpha .\beta=1$

Now, $\,\,\alpha^4+\beta^4+\alpha^{-1}.\beta^{-1}\\=(\alpha^2+\beta^2)^2-2\alpha^2\beta^2+\frac{1}{\alpha \beta}\\=\left[(\alpha+\beta)^2-2\alpha \beta\right]^2-2 \times 1+\frac 11\\=\left[(-1)^2-2 \times 1\right]^2-2+1\\=(1-2)^2-1\\=1-1\\=0$

22. Find the value of $\,\,\sqrt{[-3+\sqrt{\{-3+\sqrt{-3+\cdots \infty}\}}]}$

Sol. Let $\,\,z=\sqrt{[-3+\sqrt{\{-3+\sqrt{-3+\cdots \infty}\}}]} \\ \Rightarrow z^2=-3+\sqrt{[-3+\sqrt{\{-3+\sqrt{-3+\cdots \infty}\}}]} \\ \Rightarrow z^2=-3+z \\ \Rightarrow z^2-z+3=0 \\ \Rightarrow z=\frac{1 \pm \sqrt{(-1)^2-4 \times 1 \times 3}}{2 \times 1}\\~~~~~~~=\frac{1 \pm i \sqrt{11}}{2}$

23(i). If $\,\omega\,$ be a complex cube root of unity and $\,\,x=\alpha+\beta,\,y= \alpha+\beta \omega,\,z=\alpha+\beta \omega^2,\,\,$ show that $\,\,x^3+y^3+z^3=3(\alpha^3+\beta^3).$

Sol.  We notice $\,x^3+y^3+z^3\\=(\alpha+\beta)^3+(\alpha+\beta \omega)^3+(\alpha+\beta \omega^2)^3\\=[\alpha^3+\beta^3+3 \alpha \beta(\alpha+\beta)]\\+[\alpha^3+(\beta \omega)^3+3 \alpha(\beta \omega)(\alpha+\beta\omega)]\\+[\alpha^3+(\beta \omega^2)^3+3 \alpha(\beta \omega^2)(\alpha+\beta\omega^2)]\\=3 \alpha^3+ (\beta^3+\beta^3\omega^3+\beta^3\omega^6)\\+3 \alpha \beta[\alpha+\beta+\omega(\alpha+\beta\omega)\\+\omega^2(\alpha+\beta\omega^2)]\\=3 \alpha^3+3 \beta^3+3\alpha\beta[\alpha(1+\omega+\omega^2)\\+\beta(1+\omega^2+\omega^4)]\\=3 \alpha^3+3 \beta^3+3\alpha\beta[\alpha(1+\omega+\omega^2)\\+\beta(1+\omega^2+\omega^3.\omega)]\\=3 \alpha^3+3 \beta^3+3\alpha\beta[\alpha(1+\omega+\omega^2)\\+\beta(1+\omega^2+\omega)]\\=3(\alpha^3+\beta^3)\,\,[\text{Using,}\,1+\omega+\omega^2=0]$

23(ii) If $\,\,x=a+b,\,y=a \alpha+b\beta,\,z=a \beta+b \alpha,\,$ where $\,\,\alpha,\,\beta\,$ are complex cube root of unity , show that $\,\,xyz=a^3+b^3$

Sol. Since $\,\, \alpha, \beta\,$ are complex cube root of unity , without the loss of generality , we can assume that $\,\,\alpha=\omega,\,\beta=\omega^2.$

Now, $yz=(a \alpha+b\beta)(a \beta+b \alpha)\\~~~~=a^2(\alpha\beta)+b^2(\alpha\beta)+ab(\alpha^2+\beta^2) \rightarrow(1)$

We now compute $\,\alpha \beta,\,\,\alpha^2+\beta^2. \\ \therefore \alpha\beta=\omega.\omega^2=\omega^3=1,\rightarrow(2)\\ \alpha^2+\beta^2=(\omega)^2+(\omega^2)^2\\~~~~~~~~~~~~~=\omega^2+\omega^4\\~~~~~~~~~~~~~=\omega^2+\omega^3.\omega\\~~~~~~~~~~~~~=\omega^2+\omega=-1\rightarrow(3)$

Hence, from (1),(2) and (3), we get, $\,\,yz=a^2+b^2-ab \\ \therefore xyz=(a+b)(a^2-ab+b^2)\\~~~~~~~~~~=a^3+b^3$

24. If $\,\,z=x+iy,\,\,\text{and}\,\,|z-1|^2+|z+1|^2=4,\,\,$determine the position of the point $\,z\,$ in the complex plane.

Sol. $|z-1|^2+|z+1|^2=4 \\ \Rightarrow |x+iy-1|^2+|x+iy+1|^2=4 \\ \Rightarrow |(x-1)+iy|^2+|(x+1)+iy|^2 =4 \\ \Rightarrow [\sqrt{(x-1)^2+y^2}]^2+[\sqrt{(x+1)^2+y^2}]^2=4 \\ \Rightarrow (x-1)^2+y^2+(x+1)^2+y^2=4 \\ \Rightarrow [(x+1)^2+(x-1)^2]+2y^2=4 \\ \Rightarrow 2(x^2+1)+2y^2=4 \\ \Rightarrow x^2+1+y^2=2 \\ \Rightarrow x^2+y^2=1 \cdots(1)$

From (1), we can conclude that it represents a circle with center $\,(0,0)\,$ and radius $=1\,$unit. 

25. Factorize : $\,(i)\,a^2-ab+b^2\,\,\,(ii)\,\,x^3+y^3$

Sol. (i) $\,a^2-ab+b^2\\=a^2+(\omega+\omega^2)ab+\omega^3b^2,\\ [\omega \,\text{being cube root of unity},\,1+\omega+\omega^2=0,\,\omega^3=1]\\=(a+\omega b)(a+\omega^2 b) \cdots(1)$

(ii) $\,x^3+y^3\\=(x+y)(x^2-xy+y^2)\\=(x+y)(x+\omega y)(x+\omega^2 y)\,\,[\text{Using (1)}]$

 

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