Differentiation (Part-38) | S N De

 

In the previous article, we have discussed Short Answer Type Questions (from 1 to 10) 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. 

11. If $\,\,a,b,p,q\,$ are real and $\,\,\sqrt[3]{a-ib}=p-iq,\,\,$ prove that $\,\,\sqrt[3]{a+ib}=p+iq.$

Sol. $\,\,\sqrt[3]{a-ib}=p-iq \\ \Rightarrow a-ib=(p-iq)^3\\ \Rightarrow a-ib=p^3-3p^2(iq)+3p(iq)^2-(iq)^3\\~~~~~~~~~~~~~~~=(p^3-3pq^2)-i(3p^2q-q^3) \\ ~~~~~~ [\text{Using,}\,\,i^2=-1]\\ \Rightarrow a=p^3-3pq^2,\,\,b=3p^2q-q^3.$ 

Now, $\,\,a+ib=(p^3-3pq^2)+i(3p^2q-q^3)\\~~~~~~~~~~~~=p^3+3.p^2.(iq)+3p(iq)^2+(iq)^3\\~~~~~~~~~~~~=(p+iq)^3\\ \Rightarrow \sqrt[3]{a+ib}=p+iq.$

12. If $\,\,z=x+iy\,\,\text{and}\,\, |z+6|=|2z+3|,\,\,$prove that $\,\,x^2+y^2=9.$

Sol. $\,\,|z+6|=|2z+3| \\ \Rightarrow |(x+iy)+6|=|2(x+iy)+3| \\ \Rightarrow |(x+6)+iy|=|(2x+3)+i(2y)| \\ \Rightarrow \sqrt{(x+6)^2+y^2}=\sqrt{(2x+3)^2+(2y)^2} \\ \Rightarrow (x+6)^2+y^2=(2x+3)^2+(2y)^2 \\ \Rightarrow x^2+12x+36+y^2=4x^2+12x+9+4y^2 \\ \Rightarrow 3x^2+3y^2=27 \\ \Rightarrow x^2+y^2=9.$

13(i) If $\,z=x+iy,\,\,|2z-1|=|z-2|,\,\,$ prove that, $\,x^2+y^2=1$

Sol. $\,|2z-1|=|z-2|\\ \Rightarrow |2(x+iy)-1|=|(x+iy)-1|\\ \Rightarrow |(2x-1)+i(2y)|=|(x-1)+iy| \\ \Rightarrow \sqrt{(2x-1)^2+(2y)^2}=\sqrt{(x-2)^2+y^2}\\ \Rightarrow 4x^2-4x+1+4y^2=x^2-4x+4+y^2 \\ \Rightarrow 3x^2+3y^2=3 \\ \Rightarrow x^2+y^2=1.$

13(ii) If $\,\,z=x+iy,\,\,|2z+1|=|z-2i|,\,\,$ show that, $\,3(x^2+y^2)+4(x+y)=3,\,\,[x,y \in \mathbb R,\,i=\sqrt{-1}]$ 

Sol. $\,\,|2z+1|=|z-2i|\\ \Rightarrow |2(x+iy)+1|=|(x+iy)-2i| \\ \Rightarrow |(2x+1)+i(2y)|=|x+i(y-2)|\\ \Rightarrow \sqrt{(2x+1)^2+(2y)^2}=\sqrt{x^2+(y-2)^2} \\ \Rightarrow 4x^2+4x+1+4y^2=x^2+y^2-4y+4 \\ \Rightarrow 3(x^2+y^2)+4(x+y)=3$

14. If $\,x=2+3i,\,\,y=2-3i,\,\,$ then find the values of $(i) \frac{x^3-y^3}{x^3+y^3}\,\,(ii)\frac{x^2+xy+y^2}{x^2-xy+y^2}$

Sol. $\,\,x=2+3i,\,\,y=2-3i \\ \therefore x+y=(2+3i)+(2-3i)=4 \\ xy=(2+3i)(2-3i)\\=2^2-(3i)^2\\=4+9\\=13 \\ \text{and}\,\, x-y=(2+3i)-(2-3i)=6i. \\ (i)\,\,\frac{x^3-y^3}{x^3+y^3}\\=\frac{(x-y)^3+3xy(x-y)}{(x+y)^3-3xy(x+y)}\\=\frac{(6i)^3+3 \times 13 \times 6i}{4^3-3 \times 13 \times 4}\\=\frac{-216i+234i}{64-156}\\=\frac{18i}{-92}\\=\frac{-9i}{46} \\ \,(ii)\,\frac{x^2+xy+y^2}{x^2-xy+y^2}\\=\frac{(x+y)^2-xy}{(x+y)^2-3xy}\\=\frac{4^2-13}{4^2-3 \times 13}\\=\frac{16-13}{16-39}\\=-\frac{3}{23}$

15. Find the square roots : $\,(i)\, 7-24i\,\,(ii)\, 16+30i\,\,(iii)\,i \,(iv)\,1-i \\(v)\,1+2\sqrt{-6}\,(vi)\, \frac{1+i}{1-i}\,(vii)\,\frac{7-24i}{3+4i}\,(viii)\,x^2-1+2ix \\ (ix)\,1+i \sqrt{a^4-1}\,\,(x)\,x^4+x^2+1\\(xi)\,x-i\sqrt{x^4+x^2+1} \\ (xii)\,\,a^2+\frac{1}{a^2}+4i\left(a +\frac 1a\right)-2$

Sol. (i) $\,\,7-24i \\=16-9-2\times 12i\\=4^2+(3i)^2-2 \times 4\times 3i\\=(4-3i)^2 \\ \Rightarrow \sqrt{7-24i}=\pm(4-3i) \\ (ii)\, 16+30i\\=25-9+2 \times 15\\=5^2+(3i)^2+2 \times 5 \times 3i\\=(5+3i)^2 \\ \Rightarrow \sqrt{16+30i}=\pm(5+3i) \\(iii)\,i\\=\frac 12 \times (2i)\\=\frac 12\times (1-1+2i)\\=\frac 12\times (1^2+i^2+2 \times 1 \times i)\\=\frac 12 (1+i)^2 \\ \Rightarrow \sqrt{i}=\pm \frac{1}{\sqrt2}(1+i) \\ (iv)\, \text{Let}\,\, \sqrt{1-i}=x-iy \\ \Rightarrow 1-i=(x-iy)^2 \\ \Rightarrow 1-i=(x^2-y^2)-i(2xy) \\ \Rightarrow 1=x^2-y^2,\rightarrow(1)\quad 1=2xy \\ (x^2+y^2)^2=(x^2-y^2)^2+4x^2y^2\\~~~~~~~~~~~~~~~~~~=1^2+1^2=2 \\ \Rightarrow x^2+y^2=\sqrt2 \rightarrow(2)\\ \text{Hence from (1) and (2)},\,\,2x^2=\sqrt2+1 \\ \Rightarrow x=\pm \frac{1}{\sqrt2}\sqrt{(\sqrt2+1)}\rightarrow(3) \\ \text{Similarly,}\,\,y= \pm \frac{1}{\sqrt2}\sqrt{(\sqrt2-1)}\\ \therefore \sqrt{1-i}=\pm \frac{1}{\sqrt2}\sqrt{[(\sqrt2+1)-i(\sqrt2-1)]}$

Sol. 15(v) $\,1+2 \sqrt{-6}\\=1+2i\sqrt6\,\,\,[\text{Using,}\,\,i=\sqrt{-1}]\\=(\sqrt3)^2+(i\sqrt2)^2+2 \times \sqrt3 \times i\sqrt2\,\,\,[i^2=-1]\\=(\sqrt3+i\sqrt2)^2 \\ \Rightarrow \sqrt{1+2 \sqrt{-6}}= \pm(\sqrt3+i\sqrt2) $

Sol. 15(vi) $\,\,\frac{1+i}{1-i}\\=\frac{(1+i)^2}{(1-i)(1+i)}\\=\frac{(1+i)^2}{1+1}\,\,[i^2=-1]\\=\frac 12 (1+i)^2 \\ \Rightarrow \sqrt{\frac{1+i}{1-i}}=\pm \frac{1}{\sqrt2}(1+i) $

Sol. $\,\,15(vii)\,\,\frac{7-24i}{3+4i}\\=\frac{(7-24i)(3-4i)}{(3+4i)(3-4i)}\\=\frac{21-28i-72i-96}{3^2-(4i)^2}\\=\frac{-75-100i}{9+16}\,\,\,[\text{Using,}\,\,\,i^2=-1]\\=-3-4i\\=1^2+(2i)^2-2 \times 1 \times 2i\\=(1-2i)^2 \\ \therefore \sqrt{\frac{7-24i}{3+4i}}=\pm (1-2i)$

Sol. $\,\,15(viii)\,x^2-1+2ix\\=x^2+i^2+2 \times x \times i\\=(x+i)^2 \\ \therefore \sqrt{x^2-1+2ix}=\pm (x+i)$

Sol. $\,\,15(ix)\,\,1+i \sqrt{a^4-1}\\=\frac 12\left(2+2i\sqrt{(a^2+1)(a^2-1)}\right)\\=\frac 12\{(a^2+1)-(a^2-1)\\+2.i\sqrt{a^2+1}.\sqrt{a^2-1}\}\\=\frac 12\{(\sqrt{a^2+1})^2+(i\sqrt{a^2-1})^2\\+2.\sqrt{a^2+1}.i\sqrt{a^2-1}\}\\=\frac 12\left(\sqrt{a^2+1}+i\sqrt{a^2-1}\right)^2\\ \Rightarrow \sqrt{1+i \sqrt{a^4-1}}\\=\pm\frac{1}{\sqrt2}\left(\sqrt{a^2+1}+i\sqrt{a^2-1}\right)$

Sol. $\,\,15(x)\,\,x^4+x^2+1\\=(x^2)^2+2.x^2.1+1^2-x^2\\=(x^2+1)^2-x^2\\=(x^2+1+x)(x^2+1-x)$

Now, $\,\,x-i\sqrt{x^4+x^2+1}\\=\frac 12\left(2x-2i\sqrt{(x^2+1+x)(x^2+1-x)}\right)\\=\frac 12\left[(x^2+1+x)-(x^2+1-x)\\-2i \sqrt{x^2+x+1} \sqrt{x^2-x+1}\right]\\=\frac 12\left[(\sqrt{x^2+1+x})^2+(i\sqrt{x^2+1-x})^2\\-2. \sqrt{x^2+x+1}.i \sqrt{x^2-x+1}\right]\\=\frac 12\left[\sqrt{x^2+1+x}-i\sqrt{x^2+1-x}\right]^2 \\ \Rightarrow \sqrt{x-i\sqrt{x^4+x^2+1}}\\=\pm\frac{1}{\sqrt2}\left[\sqrt{x^2+1+x}-i\sqrt{x^2+1-x}\right]$

Sol. $\,15(xi)\,\, y+\sqrt{y^2-x^2}\\=y+i\sqrt{x^2-y^2}\\=\frac 12 \left[(x+y)-(x-y)+2i \sqrt{x+y}\sqrt{x-y}\right]\\=\frac{1}{2}\left[(\sqrt{x+y})^2+(i\sqrt{x-y})^2\\+2.\sqrt{x+y}.i\sqrt{x-y}\right]\\=\frac 12 \left(\sqrt{x+y}+i\sqrt{x-y}\right)^2 \\ \therefore \sqrt{y+\sqrt{y^2-x^2}} \\=\pm \frac{1}{\sqrt2}\left(\sqrt{x+y}+i\sqrt{x-y}\right) $

Sol. $\,\,15(xii)\,\,a^2+\frac{1}{a^2}+4i\left(a +\frac 1a\right)-2\\=\left(a+\frac 1a\right)^2-2.a.\frac 1a+4.i\left(a+\frac 1a\right)-2\\=\left(a+\frac 1a\right)^2+4.i\left(a+\frac 1a\right)-4\\=\left(a+\frac 1a\right)^2+2.\left(a+\frac 1a\right).2i+(2i)^2\\=\left(a+\frac 1a+2i\right)^2 \\ \therefore \sqrt{a^2+\frac{1}{a^2}+4i\left(a +\frac 1a\right)-2}\\=\pm\left(a+\frac 1a+2i\right)$