Question 15 Marks
Evaluate the following:
$\text{x}^4-4\text{x}^3+4\text{x}^2+8\text{x}+44,$ when $\text{x}=3+2\text{i}$
AnswerWe have,
$\text{x}=3+2\text{i}$
$\Rightarrow\text{x}-3=2\text{i}$
$\Rightarrow(\text{x}-3)^2=(2\text{i})^2$
$\Rightarrow\text{x}^2+3^2-2\times3\times\text{x}=-4$
$\Rightarrow\text{x}^2+9-6\text{x}+4=0$
$\Rightarrow\text{x}^2-6\text{x}+13=0 \ ...\text{(i)}$
Now,
$\text{x}^4-4\text{x}^3+4\text{x}^2+8\text{x}+44$
$=\text{x}^2(\text{x}^2-6\text{x}+13)+6\text{x}^2-13\text{x}^2-4\text{x}^3+4\text{x}^2+8\text{x}+44$ (adding and subtracting $6x^3$ and $13x^2$)
$=\text{x}^2\times0+2\text{x}^3-9\text{x}^2+8\text{x}^2+44$ (using (i))
$2\text{x}(\text{x}^2-6\text{x}+13)+12\text{x}^2-26\text{x}-9\text{x}^2+8\text{x}+44$ (adding and subtracting $12x^3 $and $26x^2$)
$2\text{x}\times0+3\text{x}^2-18\text{x}+44$ (using (i))
$=3(\text{x}^2-6\text{x}+13)+5$
$=3\times0+5$ (using (i))
$=5$
View full question & answer→Question 25 Marks
Express the following complex numbers in the form $\text{r}(\cos\theta+\text{i}\sin\theta):$
$1+\text{i}\tan\alpha$
AnswerLet $\text{z}=1+\text{i}\tan\alpha$$\tan\alpha$ is periodic function with period $\pi$
So, let us take $\alpha$ lying in the interval $\Big[0,\frac{\pi}{2}\Big)\cup\Big(\frac{\pi}{2},\pi\Big].$
Case - 1: when $\alpha\in\Big[0,\frac{\pi}{2}\Big)$
$|\text{z}|=\sqrt{1+\tan^2\alpha}=\sqrt{\sec^2\alpha}=|\sec\alpha|=\sec\alpha$
Let $\beta$ be acute angle given by $\tan\beta=\frac{|\text{Im(z)}|}{|\text{Re(z)|}}.$
$\tan\beta=|\tan\alpha|=\tan\alpha$
$\Rightarrow\beta=\alpha$
As z represented by a point in first quadrant.
$\therefore \ \text{arg(z)}=\beta=\alpha$
So polar form of z is $\sec\alpha\big(\cos\alpha+\text{i}\sin\alpha\big)$
Case - 2: when $\alpha\in\Big(\frac{\pi}{2},\pi\Big]$
$|\text{z}|=\sqrt{1+\tan^2\alpha}=\sqrt{\sec^2\alpha}=|\sec\alpha|=-\sec\alpha$
Let $\beta$ be acute angle given by $\tan\beta=\frac{|\text{Im(z)}|}{|\text{Re(z)|}}.$
$\tan\beta=|\tan\alpha|=-\tan\alpha=\tan(\pi-\alpha)$
$\Rightarrow\beta=\pi-\alpha$
As z represented by a point in fourth quadarnt.
$\therefore \ \text{arg(z)}=-\beta=\alpha-\pi$
So polar form of z is $-\sec\alpha\big(\cos(\alpha-\pi)+\text{i}\sin(\alpha-\pi)\big).$
View full question & answer→Question 35 Marks
If $|\text{z}+1|=\text{z}+2(1+\text{i}),$ find z.
AnswerLet $\text{z}=\text{x}+\text{iy}$
$|\text{z}+1|=\text{z}+2(1+\text{i})$
$\Rightarrow|\text{x}+\text{iy}+1|=\text{x}+\text{iy}+2+2\text{i}$
$\Rightarrow\sqrt{(\text{x}+1)^2+\text{y}^2}=(\text{x}+2)\text{i}(\text{y}+2)$
Comparing, real and imaginary parts, we get
$\text{x}+2=\sqrt{\text{x}^2+2\text{x}+1+\text{y}^2}$ and $\text{y}+2=0$
$\text{y}+2=0$
$\Rightarrow\text{y}=-2$
& $(\text{x}+2)^2=\text{x}^2+2\text{x}+1+\text{y}^2$
$\Rightarrow\text{x}^2+4\text{x}+4=\text{x}^2+2\text{x}+1+\text{y}^2$
$\Rightarrow2\text{x}+3=\text{y}^2$
$\Rightarrow2\text{x}+3=(-2)^2$
$\Rightarrow2\text{x}+3=4$
$\Rightarrow2\text{x}=1$
$\Rightarrow\text{x}=\frac{1}{2}$
$\therefore\text{z}=\text{x}+\text{iy}=\frac{1}{2}-2\text{i}$
View full question & answer→Question 45 Marks
Find the smallest positive integer value of n for which $\frac{(1+\text{i})^\text{n}}{(1-\text{i})^{\text{n}-2}}$ is a real number.
AnswerLet $\text{z}=\frac{(1+\text{i})^\text{n}}{(1-\text{i})^{\text{n}-2}}$
$=\frac{(1+\text{i})^\text{n}}{(1-\text{i})^{\text{n}}}(1-\text{i})^2$
$=\Big(\frac{1+\text{i}}{1-\text{i}}\Big)^\text{n}\times(1-\text{i})^2$
$=\text{i}^\text{n}\big(1+\text{i}^2-2\times1\times\text{i}\big) \ \Big(\because\frac{1+\text{i}}{1-\text{i}}=\text{i},\text{using problem 10}\Big)$
$=\text{i}^\text{n}(1-1-2\text{i})$
$=-2\text{i}\times\text{i}^\text{n}$
$=-2\text{i}^{\text{n}+1}$
$\therefore$ For n = 1
$\text{z}=-2\text{i}^{1+1}$
$=-2\text{i}^2$
$=2,$ which is real number
$\therefore$ The smallest positive integer value of n is 1.
View full question & answer→Question 55 Marks
If $\text{x}+\text{iy}=\frac{\text{a}+\text{ib}}{\text{a}-\text{ib}},$ Prove that $\text{x}^2+\text{y}^2=1$
Answer$\text{x}+\text{iy}=\frac{\text{a}+\text{ib}}{\text{a}-\text{ib}}$
$\Rightarrow \ (\overline{\text{x}+\text{iy}})=\overline{\Big(\frac{\text{a}+\text{ib}}{\text{a}-\text{ib}}\Big)}$ (On taking conjugate both sides)
$\Rightarrow \ ({\text{x}+\text{iy}})={\frac{\overline{\big(\text{a}+\text{ib}\big)}}{\overline{\big(\text{a}-\text{ib}\big)}}} \ \Bigg(\because \ \overline{\Big(\frac{\text{z}_1}{\text{z}_2}\Big)}=\overline{\frac{\text{z}_1}{\text{z}_2}}\Bigg)$
$=\frac{\text{a}-\text{ib}}{\text{a}+\text{ib}}$
$\therefore \ (\text{x}+\text{iy})(\text{x}-\text{iy})=\frac{\text{a}+\text{ib}}{\text{a}-\text{ib}}\times\frac{\text{a}-\text{ib}}{\text{a}+\text{ib}}$
$\Rightarrow\text{x}^2+\text{y}^2=1$
View full question & answer→Question 65 Marks
If $\Big(\frac{1+\text{i}}{1-\text{i}}\Big)^3-\Big(\frac{1-\text{i}}{1+\text{i}}\Big)^3=\text{x}+\text{iy,}$ find (x, y)
Answer$\Big(\frac{1+\text{i}}{1-\text{i}}\Big)^3-\Big(\frac{1-\text{i}}{1+\text{i}}\Big)^3=\text{x}+\text{iy,}$
$\Rightarrow\Big(\frac{(1+\text{i})(1+\text{i})}{(1-\text{i})(1+\text{i})}\Big)^3-\Big(\frac{(1-\text{i})(1-\text{i})}{(1+\text{i})(1-\text{i})}\Big)^3=\text{x}+\text{iy}$ [Rationalizing the denominator]
$\Rightarrow\Big(\frac{1+2\text{i}-1}{1+1}\Big)^3-\Big(\frac{1-2\text{i}-1}{1+1}\Big)^3=\text{x}+\text{iy}$
$\Rightarrow\Big(\frac{2\text{i}}{2}\Big)^3-\Big(\frac{-2\text{i}}{2}\Big)^3=\text{x}+\text{iy}$
$\Rightarrow\text{i}^3-(-\text{i})^3=\text{x}+\text{iy}$
$\Rightarrow-\text{i}-\text{i}=\text{x}+\text{iy}$
$\Rightarrow-2\text{i}=\text{x}+\text{iy}$
Comparing the real and imaginary parts,
$(\text{x},\text{y})=(0,2)$
View full question & answer→Question 75 Marks
Find the conjugates of the following complex numbers:
$\frac{(3-\text{i})^2}{2+\text{i}}$
AnswerLet $\text{z}=\frac{(3-\text{i})^2}{2+\text{i}}$
$=\frac{3^2+\text{i}^2-2\times3\times\text{i}}{2+\text{i}}$
$=\frac{9-1-6\text{i}}{2+\text{i}}$
$=\frac{8-6\text{i}}{2+\text{i}}$
$=\frac{8-6\text{i}}{2+1}\times\frac{2-\text{i}}{2-\text{i}}$
$=\frac{8(2-\text{i})-6\text{i}(2-\text{i})}{2^2+1^2}$
$=\frac{16-8\text{i}-12\text{i}-6}{4+1}$
$=\frac{10-20\text{i}}{5}$
$\Rightarrow\text{z}=2-4\text{i}$
Hence
$\bar{\text{z}}=2+4\text{i}$
View full question & answer→Question 85 Marks
If $z_1, z_2 $and $z_3, z_4$ are two pairs of conjugate complex numbers, prove that $\text{arg}\Big(\frac{\text{z}_1}{\text{z}_4}\Big)+\text{arg}\Big(\frac{\text{z}_2}{\text{z}_3}\Big)=0.$
Answer$z_1, z_2$ are conjugates implies $\text{z}_2=\bar{\text{z}_1}$
$z_3, z_4$ are conjugates implies $\text{z}_4=\bar{\text{z}_3}$
Also we know that $\text{arg(z}_1)+\text{arg}\bar{{(\text{z}_1)}}=0$
$\text{arg}\Big(\frac{\text{z}_1}{\text{z}_4}\Big)+\text{arg}\Big(\frac{\text{z}_2}{\text{z}_3}\Big)=0$
$=\text{arg(z}_1)-\text{arg}{(\text{z}_4)}+\text{arg(z}_2)-\text{arg}{(\text{z}_3)} \ \Big[\therefore\text{arg}\Big(\frac{\text{z}_1}{\text{z}_2}\Big)=\text{arg(z}_1)-\text{arg}{(\text{z}_2)}\Big]$
$=\text{arg(z}_1)-\text{arg}\bar{{(\text{z}_3)}}+\text{arg}\bar{{(\text{z}_1)}}-\text{arg(z}_3)$
$=\text{arg(z}_1)+\text{arg}\bar{{(\text{z}_1)}}-\text{arg}\bar{{(\text{z}_3)}}-\text{arg(z}_3)$
$=\text{arg(z}_1)+\text{arg}\bar{{(\text{z}_1)}}-[\text{arg}\bar{{(\text{z}_3)}}+\text{arg(z}_3)] \ \big[\because\text{arg(z}_1)+\text{arg}\bar{{(\text{z}_1)}}=0\big]$
$=0+0+0$
View full question & answer→Question 95 Marks
If $\text{z}_1=2-\text{i}, \ \text{z}_2=-2+\text{i},$ Find
$\text{Re}\Big(\frac{\text{z}_1\text{z}_2}{\text{z}_1}\Big)$
Answer$\frac{\text{z}_1\text{z}_2}{\text{z}_1}=\frac{\text{z}_1\text{z}_2}{\text{z}_1}\times\frac{\text{z}_1}{\text{z}_1}$ (rationalising the denominator)
$=\frac{(\text{z}_1)^2\text{z}_2}{\text{z}_1\text{z}_1}$
$=\frac{(2-\text{i})^2(-2-\text{i})}{|\text{z}_1|^2} \ \big(\therefore \ \text{z}\bar{\text{z}=|\text{z}|^2}\big)$
$=\frac{(2^2+\text{i}^2-2\times2\times\text{i})(-2+\text{i})}{|2-\text{i}|^2}$
$=\frac{(4-1-4\text{i}^2)(-2+\text{i})}{2^2+(-1)^2}$
$=\frac{(3-4\text{i})(-2+\text{i})}{4+\text{i}}$
$=3(-2+\text{i})-4\text{i}(-2+\text{i})$
$=\frac{-6+3\text{i}+8\text{i}+4}{5}$
$=\frac{-2+11\text{i}}{5}$
$\therefore \ \text{Re}\Big(\frac{\text{z}_1\text{z}_2}{\text{z}_1}\Big)=\text{Re}\Big(\frac{-2}{5}+\frac{11}{5}\text{i}\Big)$
$=\frac{-2}{5}$
View full question & answer→Question 105 Marks
If $\text{z}_1=2-\text{i}, \ \text{z}_2=1+\text{i},$ Find $\Big|\frac{\text{z}_1+\text{z}_2+1}{\text{z}_1-\text{z}_2+\text{i}}\Big|.$
AnswerIf $\text{z}=\text{x}+\text{iy}$ then $|\text{z}|=\sqrt{\text{x}^2+\text{y}^2}$
We have,
$\text{z}_1=2-\text{i}, \ \text{z}_2=1+\text{i}$
$\text{z}_1+\text{z}_2=2-\text{i}+1+\text{i}$
$=3$
And $\text{z}_1-\text{z}_2=2-\text{i}-1-\text{i}$
$=1-2\text{i}$
$\Big|\frac{\text{z}_1+\text{z}_2+1}{\text{z}_1-\text{z}_2+\text{i}}\Big|=\frac{3+1}{1-2\text{i}+\text{i}}$
$=\frac{4}{1-\text{i}}$
$=\frac{4}{1-\text{i}}\times\frac{1+\text{i}}{1+\text{i}}$
$=\frac{4(1+\text{i})}{1^2+1^2}$
$=\frac{4(1+\text{i})}{2}$
$=2{(1+\text{i})}$
$\therefore\Big|\frac{\text{z}_1+\text{z}_2+1}{\text{z}_1-\text{z}_2+\text{i}}\Big|=|2(1+\text{i})| $
$=|2||1+\text{i}| \ \big(\therefore|\text{z}_1\text{z}_2|=|\text{z}_1|\times|\text{z}_2|\big)$
$=2\times\sqrt{1^2+1^2}$
$=2\times\sqrt{2}$
$=2\sqrt{2}$
View full question & answer→Question 115 Marks
For a positive integer n, find the value of $(1-\text{i})^\text{n}\big(1-\frac{1}{\text{i}}\big)^\text{i}.$
Answer$(1-\text{i})^\text{n}\big(1-\frac{1}{\text{i}}\big)^\text{n}$
$=(1-\text{i})^\text{n}\big(1-\frac{1}{\text{i}}\big)^\text{n}$
$=\Big\{\frac{(1-\text{i})(\text{i}-1)}{\text{i}}\Big\}^\text{n}$
$=\Big\{\frac{(1-\text{i})(1-\text{i})}{-\text{i}}\Big\}^\text{n}$
$=\Big\{\frac{(1-\text{i})^2}{-\text{i}}\Big\}^\text{n}$
$=\Big\{\frac{(1-2\text{i}-1)}{-\text{i}}\Big\}^\text{n}$
$=\Big\{\frac{-2\text{i}}{-\text{i}}\Big\}^\text{n}=2^\text{n}$
View full question & answer→Question 125 Marks
Express the following complex numbers in the form $\text{r}(\cos\theta+\text{i}\sin\theta):$
$\frac{1-\text{i}}{\cos\frac{\pi}{3}+\text{i}\sin\frac{\pi}{3}}$
AnswerLet $\text{z}=\frac{1-\text{i}}{\cos\frac{\pi}{3}+\text{i}\sin\frac{\pi}{3}}=\frac{1-\text{i}}{\frac{1}{2}+\text{i}\frac{\sqrt{3}}{2}}=\frac{2-2\text{i}}{1+\text{i}\sqrt{3}}=\frac{(2-2\text{i})(1-\text{i}\sqrt{3})}{(1+\text{i}\sqrt{3})(1-\text{i}\sqrt{3})}=\\\frac{(2-2\sqrt{3})-\text{i}(2\sqrt{3}+2)}{4}=\frac{(1-\sqrt{3})}{2}-\text{i}\frac{({\sqrt{3}+1)}}{2}$
$|\text{z}|=\sqrt{\frac{(1-\sqrt{3})^2}{4}+\frac{(1+\sqrt{3})^2}{4}}=\sqrt{\frac{8}{4}}=\sqrt{2}$
Let $\beta$ be acute angle by $\tan\beta=\frac{|\text{Im(z)}|}{|\text{Re(z)|}}.$
$\tan\beta=\frac{\Big|-\frac{(\sqrt{3}+1)}{2}\Big|}{\Big|\frac{1-\sqrt{3}}{2}\Big|}=\Big|\frac{-(\sqrt{3}+1)}{(1-\sqrt{3})}\Big|=\big|2+\sqrt{3}\big|=\tan\Big(\frac{7\pi}{12}\Big)$
$\Rightarrow\beta=\frac{7\pi}{12}$
Z is represented by a poiny in second quadrant.
So polar form of z is $\sqrt{2}\Big(\cos\frac{7\pi}{12}-\text{i}\sin\frac{7\pi}{12}\Big).$
View full question & answer→Question 135 Marks
If $\text{a}=\cos\theta+\text{i}\sin\theta,$ find the value of $\frac{1+\text{a}}{1-\text{a}}.$
Answer$\text{a}=\cos\theta+\text{i}\sin\theta, \ \frac{1+\text{a}}{1-\text{a}}$
$=\frac{1+\cos\theta+\text{i}\sin\theta}{1-\cos\theta-\text{i}\sin\theta}$
$=\frac{(1+\cos\theta+\text{i}\sin\theta)(1+\cos\theta+\text{i}\sin\theta)}{(1-\cos\theta-\text{i}\sin\theta)(1+\cos\theta+\text{i}\sin\theta)}$ [Rationalizing the denominator]
$=\frac{(1+\cos\theta+\text{i}\sin\theta)(1-\cos\theta+\text{i}\sin\theta)}{(1-\cos\theta)^2-(\text{i}\sin\theta)^2}$
$=\frac{(1+\text{i}\sin\theta)^2-\cos^2\theta}{1-2\cos\theta+\cos^2\theta+\sin^2\theta}$
$=\frac{1+2\text{i}\sin\theta-\sin^2\theta-\cos^2\theta}{1-2\cos\theta+\cos^2\theta+\sin^2\theta}$
$=\frac{1+2\text{i}\sin\theta-1}{1-2\cos\theta+\cos^2\theta+\sin^2\theta} \ \big[\therefore \cos^2\theta+\sin^2\theta=1\big]$
$=\frac{2\text{i}\sin\theta}{1-2\cos\theta+1}$
$=\frac{2\text{i}\sin\theta}{2-2\cos\theta}$
$=\frac{\text{i}\sin\theta}{1-\cos\theta}$
$=\frac{\text{i}2\sin\frac{\theta}{2}\cos\frac{\theta}{2}}{2\sin^2\frac{\theta}{2}}$
$=\frac{\text{i}\cos\frac{\theta}{2}}{\sin^2\frac{\theta}{2}}=\text{i}\cot\frac{\theta}{2}$
View full question & answer→Question 145 Marks
Find the multiplicative inverse of the following complex numbers:
$(1-\text{i}\sqrt{3})^2$
AnswerLet $\text{z}=(1-\text{i}\sqrt{3})^2$
$=1^2+(\text{i}\sqrt{3})^2+2\times1\times\text{i}\sqrt{3}$
$=1-3+2\sqrt{3}\text{i}$
$=-2+2\sqrt{3}\text{i}$
$\therefore \ \text{z}^{-1}=\frac{-2}{(-2)^2+(2\sqrt{3})^2}-\frac{2\sqrt{3}\text{i}}{(-2)^2+(2\sqrt{3})^2}$
$=\frac{-2}{4+12}-\frac{2\sqrt{3}\text{i}}{4+12}$
$=\frac{-2}{16}-\frac{2\sqrt{3}\text{i}}{16}$
$=\frac{-1}{8}-\frac{\sqrt{3}\text{i}}{8}$
View full question & answer→Question 155 Marks
If $z_1, z_2$ are two complex numbers such that $|\text{z}_1|=|\text{z}_2|$ and $\text{arg(z}_1)+\text{arg(z}_2)=\pi,$ then show that $\text{z}_1=-\bar{\text{z}}_2.$
Answer$|\text{z}_1|=|\text{z}_2|$
Let $\text{arg(z}_1)=\theta$
$\therefore\text{arg(z}_2)=\pi-\theta$
In polar form, $\text{z}_1=|\text{z}_1|(\cos\theta+\text{i}\sin\theta) \ ...(\text{i})$
$\text{z}_2=|\text{z}_2|\big(\cos(\pi-\theta)+\text{i}\sin(\pi-\theta\big)$
$=|\text{z}_2|\big(-\cos\theta+\text{i}\sin\theta\big)$
$=-|\text{z}_2|\big(\cos\theta-\text{i}\sin\theta\big)$
Finding conjugate of
$\bar{\text{z}}_2=-|\text{z}_2|\big(\cos\theta-\text{i}\sin\theta\big) \ ...(\text{ii})$
(i), (ii) is equal to
$\frac{\text{z}_1}{\text{z}_2}=-\frac{|\text{z}_1|(\cos\theta+\text{i}\sin\theta)}{|\text{z}_2|(\cos\theta+\text{i}\sin\theta)}$
$\frac{\text{z}_1}{\text{z}_2}=-\frac{|\text{z}_1|}{|\text{z}_2|} \ [\because|\text{z}_1|=|\text{z}_2|]$
$\frac{\text{z}_1}{\text{z}_2}=-1$
$\text{z}_1=\bar{-\text{z}}_2$
Hence proved.
View full question & answer→Question 165 Marks
Find the modulus and argument of the following complex numbers and hence express each of them in the polar form:
$\frac{-16}{1+\text{i}\sqrt{3}}$
AnswerThe polar form of a complex number $\text{z}=\text{x}+\text{iy},$ is give by
$\text{z}=|\text{z}|\big(\cos\theta+\text{i}\sin\theta\big)$
Where,
$|\text{z}|=\sqrt{\text{x}^2+\text{y}^2}$ and
$\text{arg(z)}=\theta=\tan^{-1}\Big(\frac{\text{b}}{\text{a}}\Big)$
Let $\text{z}=\frac{-16}{1+\text{i}\sqrt{3}}$
$=\frac{-16}{1+\text{i}\sqrt{3}}\times\frac{1-\text{i}\sqrt{3}}{1-\text{i}\sqrt{3}}$
$=\frac{-16(1-\text{i}\sqrt{3})}{(1)^2+(\sqrt{3})^2}$
$=\frac{-16(1-\text{i}\sqrt{3})}{1+3}$
$=\frac{-16}{4}(1-\text{i}\sqrt{3})$
$=-4(1-\text{i}\sqrt{3})$
$=-4+4\sqrt{3}\text{i}$
$\therefore \ |\text{z}|=\sqrt{(-4)^2+(4\sqrt{3})^2}$
$=\sqrt{16+48}$
$=\sqrt{64}$
$=8$
Here $\text{x}=-4<0$ & $\text{y}=4\text{x}3>0, \ \therefore\theta$ lies in quadrantII
$=\theta=\text{arg(z)}=\tan^{-1}\Big(\frac{4\sqrt{3}}{-4}\Big)$
$=\tan^{-1}(\sqrt{3})$
$=\tan^{-1}\Big(\tan\big(\pi-\frac{\pi}{3}\big)\Big) \ \big(\therefore\tan(\pi-\theta)=-\tan\theta\big)$
$=\pi-\frac{\pi}{3}$
$=\frac{2\pi}{3}$
The polar form is given by $\text{z}=8\Big(\cos\frac{2\pi}{3}+\text{i}\sin\frac{2\pi}{3}\Big)$
View full question & answer→Question 175 Marks
Find the multiplicative inverse of the following complex numbers:
$\sqrt{5}-3\text{i}$
AnswerLet $\text{z}=5+\sqrt{3}\text{i}$
Then $\text{z}^{-1}=\frac{\sqrt{5}}{(\sqrt{5}^2)+(3)^2}-\frac{(-3)}{(\sqrt{5})^2+(3)^2}\text{i}$
$=\frac{\sqrt{5}}{5+9}-\frac{3}{5+9}\text{i}$
$=\frac{\sqrt{5}}{14}+\frac{3}{14}\text{i}$
View full question & answer→Question 185 Marks
$2\text{x}^4+5\text{x}^3+7\text{x}^2-\text{x}+41,$ when $\text{x}=-2-\sqrt{3}\text{i}$
Answer$\text{x}=-2-\sqrt{3}\text{i}$
$\text{x}^2=(-2-\sqrt{3}\text{i})^2=4+4\sqrt{3}\text{i}+3\text{i}^2=1+4\sqrt{3}\text{i}$
$\text{x}^3=(1+4\sqrt{3}\text{i})(-2-\sqrt{3}\text{i})=-2-8\sqrt{3}\text{i}-\sqrt{3}\text{i}-12\text{i}^2=10+9\sqrt{3}\text{i}$
$\text{x}^4=(1-4\sqrt{3}\text{i})^2=1+8\sqrt{3}\text{i}+48\text{i}^2=-47+8\sqrt{3}\text{i}$
$2\text{x}^4+5\text{x}^3+7\text{x}^2-\text{x}+41$
$=2(-47)+8\sqrt{3}\text{i})+5(10-9\sqrt{3}\text{i})+7(1+4\sqrt{3}\text{i})-(-2-\sqrt{3}\text{i})+41$
$=-94+16\sqrt{3}\text{i}+50-45\sqrt{3}\text{i}+7+28\sqrt{3}\text{i}+2+\sqrt{3}\text{i}+41$
$=(-94+50+7+2+41)+(16\sqrt{3}\text{i}-45\sqrt{3}\text{i}+28\sqrt{3}\text{i}+\sqrt{3}\text{i})$
$=6+0$
$=6$
View full question & answer→Question 195 Marks
Express the following complex numbers in the form $\text{r}(\cos\theta+\text{i}\sin\theta):$
$\tan\alpha-\text{i}$
AnswerLet $\text{z}=\tan\alpha-\text{i}$$\tan\alpha$ is periodic function with period $\pi$
So, let us take $\alpha$ lying in the interval $\Big[0,\frac{\pi}{2}\Big)\cup\Big(\frac{\pi}{2},\pi\Big].$
Case I: when $\alpha\in\Big[0,\frac{\pi}{2}\Big)$
$|\text{z}|=\sqrt{\tan^2\alpha+1}=\sqrt{\sec^2\alpha}=|\sec\alpha|=\sec\alpha$
Let $\beta$ be acute angle given by $\tan\beta=\frac{|\text{Im(z)}|}{|\text{Re(z)|}}.$
$\tan\beta=\frac{1}{|\tan\alpha|}=|\cot\alpha|=\cot\alpha=\tan\Big(\frac{\pi}{2}-\alpha\Big)$
$\Rightarrow\beta={\frac{\pi}{2}}-\alpha$
As z represented by a point in first quadrant.
$\therefore \ \text{arg(z)}=\beta=\alpha-\frac{\pi}{2}.$
So polar form of z is $\sec\alpha\Big(\cos\Big(\alpha-\frac{\pi}{2}\Big)+\text{i}\sin\Big(\alpha-\frac{\pi}{2}\Big)\Big)$
Case II: when $\alpha\in\Big(\frac{\pi}{2},\pi\Big]$
$|\text{z}|=\sqrt{1+\tan^2\alpha+1}=\sqrt{\sec^2\alpha}=|\sec\alpha|=-\sec\alpha$
Let $\beta$ be acute angle given by $\tan\beta=\frac{|\text{Im(z)}|}{|\text{Re(z)|}}.$
$\tan\beta=|\tan\alpha|=-\tan\alpha=\tan(\pi-\alpha)$
$\Rightarrow\beta=\alpha-\frac{\pi}{2}$
As z represented by a point in fourth quadrant.
$\therefore \ \text{arg(z)}=\pi+\beta=\frac{\pi}{2}+\alpha.$
So polar form of z is $-\sec\alpha\Big(\cos\Big(\frac{\pi}{2}+\alpha\Big)+\text{i}\sin\Big(\frac{\pi}{2}+\alpha\Big)\Big).$
View full question & answer→Question 205 Marks
$\text{x}^4+4\text{x}^3+6\text{x}^2+4\text{x}+9,$ when $\text{x}=-1+\text{i}\sqrt{2}$
AnswerWe have
$\text{x}=-1+\text{i}\sqrt{2}$
$\Rightarrow\text{x}+1+\text{i}\sqrt{2}$
$\Rightarrow(\text{x}+1)^2=(\text{i}\sqrt{2})^2$ (squaring both sides)
$\Rightarrow\text{x}^2+1+2\text{x}=-2$
$\Rightarrow\text{x}^2+2\text{x}+3=0 \ ...(\text{i})$
Now,
$\text{x}^4+4\text{x}^3+6\text{x}^2+4\text{x}+9$
$=\text{x}^2(\text{x}^2+2\text{x}+3)+2\text{x}^3+3\text{x}^2+4\text{x}+9$
$=\text{x}^2\times0+2\text{x}(\text{x}^2+2\text{x}+3)-\text{x}^2-2\text{x}+9$ (using(i))
$=2\text{x}\times0-({\text{x}^2+2\text{x}+3})+3+9$ (using(i) and adding and subtracting 3)
$=-0+3+9$ (using(i))
$=12$
View full question & answer→Question 215 Marks
Evaluate the following:
$2\text{x}^3+2\text{x}^2-7\text{x}+72,$ when $\text{x}=\frac{3-5\text{i}}{2}$
Answer$\text{x}=\frac{3-5\text{i}}{2}$
$\Rightarrow2\text{x}=3-5\text{i}$
$\Rightarrow2\text{x}-3=-5\text{i}$
$\Rightarrow(2\text{x}-3)^2=(-5\text{i})^2$
$\Rightarrow4\text{x}^2+9-12\text{x}=-25$
$\Rightarrow4\text{x}^2-12\text{x}+34=0$
$\Rightarrow2(2\text{x}^2-6\text{x}+17)=0$
$\Rightarrow2\text{x}^2-6\text{x}+17=0 \ ...(\text{i})$
$\therefore2\text{x}^3+2\text{x}^2-7\text{x}+72$
$=\text{x}(2\text{x}^2-6\text{x}+17)+6\text{x}^2-17\text{x}+2\text{x}^2-7\text{x}+72$ (Adding and subtracting $6x^2$ and 17x)
$\text{x}\times0+8\text{x}^2-24\text{x}+72$ (Using(i))
$=4(2\text{x}^2-6\text{x}+17)+4$
$=4\times0+4$ (Using(i))
$=4$
View full question & answer→Question 225 Marks
Find the least positive integral value of n for which $\Big(\frac{1+\text{i}}{1-\text{i}}\Big)^{\text{n}}$ is real.
AnswerFor n = 1, we have,
$\Big(\frac{1+\text{i}}{1-\text{i}}\Big)^1=\frac{1+\text{i}}{1-\text{i}}$
$=\frac{1+\text{i}}{1-\text{i}}\times\frac{1+\text{i}}{1+\text{i}}$
$=\frac{(1+\text{i})^2}{1^2+1^2}$
$=\frac{1^2+\text{i}^2+2\times1\times\text{i}}{2}$
$=\frac{2\text{i}}{2} \ \big(\therefore \ \text{i}^2=-1\big)$
$=\text{i},$ which is not real
For n = 2, we have
$\Big(\frac{1+\text{i}}{1-\text{i}}\Big)^2=\text{i}^2 \ \big(\because \ \frac{1+\text{i}}{1-\text{i}}=1 \ \text{from above}\big)$
$=-1,$ which is real
Hence the least positive integral value of n is 2.
View full question & answer→Question 235 Marks
If $z_1$ is a complex number other than -1 such that $|\text{z}|=1$ and $\text{z}_2=\frac{\text{z}_1-1}{\text{z}_1+1},$ then show that the real parts of $z_2$ is zero.
AnswerLet $\text{z}_1=\text{x}+\text{iy}_1,\text{z}_2=\text{x}_2+\text{iy}_2$
$|\text{z}_1|=1\Rightarrow\text{x}_1^2+\text{y}_1^2=1$
$\text{z}_2=\frac{\text{z}_1-1}{\text{z}_1+1}$
$\text{x}_2+\text{iy}_2=\frac{\text{x+iy}_1-1}{\text{x+iy}_1+1}$
$\Rightarrow\text{x}_2+\text{iy}_2=\frac{\text{x}_1-1+\text{iy}_1}{\text{x}_1+1+\text{iy}_1}$
$\Rightarrow\text{x}_2+\text{iy}_2=\frac{(\text{x}_1-1+\text{iy}_1)(\text{x}_1+1-\text{iy}_1)}{(\text{x}_1+1+\text{iy}_1)(\text{x}_1+1-\text{iy}_1)}$ [Rationalizing the denominator]
$\Rightarrow\text{x}_2+\text{iy}_2=\frac{(\text{x}_1-1)(\text{x}_1+1)-\text{iy}_1(\text{x}_1-1)+\text{iy}_1(\text{x}_1+1)+\text{y}_1^2}{{{(\text{x}_1+1)}^2-{(\text{iy}_1)^2}}}$
$\Rightarrow\text{x}_2+\text{iy}_2=\frac{\text{x}_1^2-1+\text{y}_1^2-\text{ix}_1\text{y}_1+\text{iy}_1+\text{ix}_1\text{y}_1+\text{iy}_1}{{(\text{x}_1+1)}^2-{(\text{iy}_1)^2}}$
$\Rightarrow\text{x}_2+\text{iy}_2=\frac{\text{x}_1^2+\text{y}_1^2-1+2\text{iy}_1}{{(\text{x}_1+1)}^2-{(\text{iy}_1)^2}}$
$\Rightarrow\text{x}_2+\text{iy}_2=\frac{1-1+2\text{iy}_1}{{(\text{x}_1+1)}^2-{(\text{iy}_1)^2}} \ \big[\because \ \text{x}_1^2+\text{y}_1^2=1\big]$
$\Rightarrow\text{x}_2+\text{iy}_2=\frac{2\text{iy}_1}{{(\text{x}_1+1)}^2-{(\text{iy}_1)^2}} \ \big[\because \ \text{x}_1^2+\text{y}_1^2=1\big]$
Since there is no real part in the RHS, therefore $x_2 = 0$.
The real part of the $z_2 = 0.$
View full question & answer→Question 245 Marks
If $\frac{(1+\text{i})^2}{2-\text{i}}=\text{x}+\text{iy,}$ find x, y.
Answer$\frac{(1+\text{i})^2}{2-\text{i}}=\text{x}+\text{iy}$
$\Rightarrow\frac{(1+2\text{i}-1)}{2-\text{i}}=\text{x}+\text{iy}$
$\Rightarrow\frac{2\text{i}}{2-\text{i}}=\text{x}+\text{iy}$
$\Rightarrow\frac{2\text{i}(2+\text{i})}{(2-\text{i})(2+\text{i})}=\text{x}+\text{iy}$ [Rationalizing the denominator]
$\Rightarrow\frac{2(2\text{i}-1)}{4+1}=\text{x}+\text{iy}$
$\Rightarrow\frac{4\text{i}-2}{5}=\text{x}+\text{iy}$
$\Rightarrow-\frac{2}{5}+\text{i}\frac{4}{5}=\text{x}+\text{iy}$
Comparing the real and imaginary parts, we get
$\text{x}=-\frac{2}{5},\text{y}=\frac{4}{5}$
$\text{x}+\text{y}=\frac{2}{5}$
View full question & answer→Question 255 Marks
What is the smallest positive integer n for which $(1+\text{i})^{2\text{n}}=(1-\text{i})^{2\text{n}} \ ?$
Answer$(1+\text{i})^{2\text{n}}=(1-\text{i})^{2\text{n}}$
$\Rightarrow\Big(\frac{1+\text{i}}{1-\text{i}}\Big)^{2\text{n}}=1$
$\Rightarrow\Big(\frac{(1+\text{i})(1+\text{i})}{(1-\text{i})(1+\text{i})}\Big)^{2\text{n}}=1$ [Rationalizing the denominator]
$\Rightarrow\Big(\frac{1+2\text{i}-1}{1+1}\Big)^{2\text{n}}=1$
$\Rightarrow\Big(\frac{2\text{i}}{2}\Big)^{2\text{n}}=1$
$\Rightarrow\text{i}^{2\text{n}}=1$
$\therefore\text{n}=2$
View full question & answer→Question 265 Marks
If $\Big(\frac{1-\text{i}}{1+\text{i}}\Big)^{100}=\text{a}+\text{ib},$ find (a, b).
Answer$\Big(\frac{1-\text{i}}{1+\text{i}}\Big)^{100}=\text{a}+\text{ib}$
$\Rightarrow\Big(\frac{(1-\text{i})(1-\text{i})}{(1+\text{i})(1-\text{i})}\Big)^{100}=\text{a}+\text{ib}$ [Rationalizing the denominator]
$\Rightarrow\Big(\frac{(1-2\text{i}-1)}{(1+1)}\Big)^{100}=\text{a}+\text{ib}$
$\Rightarrow\Big(\frac{-2\text{i}}{2}\Big)^{100}=\text{a}+\text{ib}$
$\Rightarrow(-\text{i})^{100}=\text{a}+\text{ib}$
$\Rightarrow1=\text{a}+\text{ib}$
Comparing, we get
$(\text{a},\text{b})=(1,0)$
View full question & answer→Question 275 Marks
Solve the equation $|\text{z}|=\text{z}+1+2\text{i}.$
AnswerLet $\text{z}=\text{x}+\text{iy}.$
Then,
$|\text{z}|=\sqrt{\text{x}^2+\text{y}^2}$
$\therefore|\text{z}|=\text{z}+1+2\text{i}$
$\Rightarrow\sqrt{\text{x}^2+\text{y}^2}=(\text{x}+\text{iy})+1+2\text{i}$
$\Rightarrow\sqrt{\text{x}^2+\text{y}^2}=(\text{x}+1)+\text{i}(\text{y}+2)$
$\Rightarrow\sqrt{\text{x}^2+\text{y}^2}=(\text{x}+1)$ and $\text{y}+2=0$
${\text{x}^2+\text{y}^2}=(\text{x}+1)^2$ and $\text{y}=-2$
${\text{x}^2+\text{y}^2}=\text{x}^2+1+2\text{x}$ and $\text{y}=-2$
$\Rightarrow\text{y}^2=2\text{x}+1$ and $\text{y}=-2$
$\Rightarrow4=2\text{x}+1$ and $\text{y}=-2$
$\Rightarrow2\text{x}=3$ and $\text{y}=-2$
$\Rightarrow\text{x}=\frac{3}{2}$ and $\text{y}=-2$
$\therefore\text{z}=\text{x}+\text{iy}=\frac{3}{2}-2\text{i}$
View full question & answer→Question 285 Marks
Express the following complex numbers in the form $\text{r}(\cos\theta+\text{i}\sin\theta):$
$1-\sin\alpha+\text{i}\cos\alpha$
AnswerLet $\text{z}=(1-\sin\alpha)+\text{i}\cos\alpha$
Since sine and cosine are periodic functions with periodic with peroid $2\pi.$
So, let us take $\alpha$ lying in the interval $[0,2\pi]$
Now, $\text{z}=(1-\sin\alpha)+\text{i}\cos\alpha$
$\Rightarrow|\text{z}|=\sqrt{(1-\sin\alpha)^2+\cos^2\alpha}\\=\sqrt{2-2\sin\alpha}=\sqrt{2}\sqrt{1-\sin\alpha}$
$\Rightarrow|\text{z}|=\sqrt{2}\sqrt{\Big(\cos\frac{\alpha}{2}-\sin\frac{\alpha}{2}\Big)^2}=\sqrt{2}\Big|\cos\frac{\alpha}{2}-\sin\frac{\alpha}{2}\Big|$
Let $\beta$ be acute angle given by $\tan\beta=\frac{|\text{Im(z)}|}{|\text{Re(z)|}}.$
$\tan\beta=\frac{|\cos\alpha|}{|1-\sin\alpha|}=\frac{|\cos\alpha|}{|1-\sin\alpha|}=\Bigg|\frac{\cos^2\frac{\alpha}{2}-\sin^2\frac{\alpha}{2}}{\Big(\cos\frac{\alpha}{2}-\sin\frac{\alpha}{2}\Big)}\Bigg|=\Bigg|\frac{\cos\frac{\alpha}{2}+\sin\frac{\alpha}{2}}{\cos\frac{\alpha}{2}-\sin\frac{\alpha}{2}}\Bigg|$
$\Rightarrow\tan\beta=\Bigg|\frac{1+\tan\frac{\alpha}{2}}{1-\tan\frac{\alpha}{2}}\Bigg|=\Big|\tan\big(\frac{\pi}{4}+\frac{\alpha}{2}\big)\Big|$
Following cases arise:
Case I: when $0\leq\alpha<\frac{\pi}{2}$
$\cos\frac{\alpha}{2}>\sin\frac{\alpha}{2}$ and $\frac{\pi}{4}+\frac{\alpha}{2}\in\Big[\frac{\pi}{4},\frac{\pi}{2}\Big)$
$\therefore\text{arg(z)}=\frac{\pi}{2}+\frac{\alpha}{2}$
So polar form of z is
$\sqrt{2}\Big(\cos\frac{\alpha}{2}-\sin\frac{\alpha}{2}\Big)\Big(\cos\Big(\frac{\pi}{4}+\frac{\alpha}{2}\Big)+\text{i}\sin\Big(\frac{\pi}{4}+\frac{\alpha}{2}\Big)\Big)$
Case II: when $\frac{\pi}{2}<\alpha<\frac{3\pi}{2}$
$\cos\frac{\alpha}{2}<\sin\frac{\alpha}{2}$ and $\frac{\pi}{4}+\frac{\alpha}{2}\in\Big(\frac{\pi}{2},\pi\Big)$
$\therefore|\text{z}|=\sqrt{2}\Big|\cos\frac{\alpha}{2}-\sin\frac{\alpha}{2}\Big|=\sqrt{2}\Big(\cos\frac{\alpha}{2}-\sin\frac{\alpha}{2}\Big)$
and, $\tan\beta=\Bigg|\tan\Big(\frac{\pi}{4}+\frac{\alpha}{2}\Big)\Bigg|=-\tan\Big(\frac{\pi}{4}+\frac{\alpha}{2}\Big)\\=\tan\Big\{\pi\Big(\frac{\pi}{4}+\frac{\alpha}{2}\Big)\Big\}=\tan\Big(\frac{3\pi}{4}-\frac{\alpha}{2}\Big)$
$\Rightarrow\beta=\frac{3\pi}{4}-\frac{\alpha}{2}$
Since $1-\sin\alpha>0$ and $\cos\alpha<0.$
Clearly, z lies in the fourth quadrant.
$\therefore\text{arg(z)}=-\beta=\frac{\alpha}{2}-\frac{3\pi}{4}$
So polar form of z is $\sqrt{2}\Big(\cos\frac{\alpha}{2}-\sin\frac{\alpha}{2}\Big)\Big(\cos\Big(\frac{\alpha}{2}-\frac{3\pi}{4}\Big)+\text{i}\sin\Big(\frac{\alpha}{2}-\frac{3\pi}{4}\Big)\Big)$
Case III: when $\frac{3\pi}{2}<\alpha<2\pi$
$\cos\frac{\alpha}{2}<\sin\frac{\alpha}{2}$ and $\frac{\pi}{4}+\frac{\alpha}{2}\in\Big(\pi,\frac{5\pi}{4}\Big)$
$\therefore|\text{z}|=\sqrt{2}\Big|\cos\frac{\alpha}{2}-\sin\frac{\alpha}{2}\Big|=-\sqrt{2}\Big(\cos\frac{\alpha}{2}-\sin\frac{\alpha}{2}\Big)$
and, $\tan\beta=\Bigg|\tan\Big(\frac{\pi}{4}+\frac{\alpha}{2}\Big)\Bigg|=\tan\Big(\frac{\pi}{4}+\frac{\alpha}{2}\Big)\\=-\tan\Big\{\pi-\Big(\frac{\pi}{4}+\frac{\alpha}{2}\Big)\Big\}=\tan\Big(\frac{\alpha}{2}-\frac{3\pi}{4}\Big)$
$\Rightarrow\beta=\frac{\alpha}{2}-\frac{3\pi}{4}$
Clearly, $\text{Re(z)}<0$ and $\text{Im(z)}>0.$
$\therefore\text{arg(z)}=\beta=\frac{\alpha}{2}-\frac{3\pi}{4}$
So polar form of z is $-\sqrt{2}\Big(\cos\frac{\alpha}{2}-\sin\frac{\alpha}{2}\Big)\Big(\cos\Big(\frac{\alpha}{2}-\frac{3\pi}{4}\Big)+\text{i}\sin\Big(\frac{\alpha}{2}-\frac{3\pi}{4}\Big)\Big).$
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