Question 12 Marks
Find the number of solutions of $\text{z}^2+|\text{z}|^2=0.$
AnswerLet $\text{z}=\text{x}+\text{iy}$
$\text{z}^2=(\text{x}+\text{iy})^2=\text{x}^2-\text{y}^2+2\text{x}\text{yi}$
$|\text{z}|^2=\text{z}\bar{\text{z}}=(\text{x}+\text{iy})(\text{x}-\text{iy})=\text{x}^2+\text{y}^2$
$\text{z}^2+|\text{z}|^2=0$
$\text{x}^2-\text{y}^2+2\text{x}\text{yi}+\text{x}^2+\text{y}^2=0$
$2\text{x}^2+2\text{x}\text{yi}=0$
$\Rightarrow2\text{x}^2=0$ and $2\text{xy}=0$
$\Rightarrow\text{x}=0$ and $\text{y}\in\text{R}$
$\therefore\text{z}=0+\text{iy}$ where $\text{y}\in\text{R}$
View full question & answer→Question 22 Marks
Find the conjugates of the following complex numbers:
$\frac{(3-2\text{i})(2+3\text{i})}{(1+2\text{i})(2-\text{i})}$
AnswerLet $\text{z}=\frac{(3-2\text{i})(2+3\text{i})}{(1+2\text{i})(2-\text{i})}$
$=\frac{3(2+3\text{i})-2\text{i}(2+3\text{i})}{2-\text{i}+2\text{i}(2-\text{i})}$
$=\frac{6+9\text{i}-4\text{i}+6}{2-\text{i}+4\text{i}+2}$
$=\frac{12+5\text{i}}{4+3\text{i}}$
$=\frac{12+5\text{i}}{4+3\text{i}}\times\frac{4-3\text{i}}{4-3\text{i}}$
$=\frac{12(4-3\text{i})+5\text{i}(4-3\text{i})}{4(4-3\text{i})+3\text{i}(4-3\text{i})}$
$=\frac{48-36\text{i}+20\text{i}+15}{16-12\text{i}+12\text{i}+9}$
$=\frac{63-16\text{i}}{16+9}$
$\Rightarrow\text{z}=\frac{63-16\text{i}}{25}$
$\therefore \bar{\text{z}}=\frac{63+16\text{i}}{25}$
$=\frac{63}{25}+=\frac{16}{25}\text{i}$
View full question & answer→Question 32 Marks
Express in $\sin\frac{\pi}{5}+\text{i}\Big(1-\cos\frac{\pi}{5}\Big)$ polar form.
Answer$\sin\frac{\pi}{5}+\text{i}\Big(1-\cos\frac{\pi}{5}\Big)$
$=2\sin\frac{\pi}{10}\cos\frac{\pi}{10}+\text{i}2\sin^2\frac{\pi}{10}$ $\big[\text{using }\sin2\theta=2\sin\theta\cos\theta\ { \&}\ 1-\cos2\theta=\sin^2\theta\big]$
$=2\sin\frac{\pi}{10}\Big(\cos\frac{\pi}{10}+\text{i}\sin\frac{\pi}{10}\Big)$
View full question & answer→Question 42 Marks
Express the following complex numbers in the standard form a + ib:
$\frac{(1-\text{i})^3}{1-\text{i}^3}$
Answer$\frac{(1-\text{i})^3}{1-\text{i}^3}=\frac{1^3-\text{i}^3-3\times1\times\text{i}(1+\text{i})}{1-(-\text{i})} \ \begin{bmatrix}\because \ (\text{a}-\text{b})^3=\text{a}^3-\text{b}^3-3\text{ab}(\text{a}+\text{b}) \\ \text{and} \ \text{i}^3=-\text{i} \end{bmatrix}$
$=\frac{1-(-\text{i})-3\text{i}(1-\text{i})}{1+\text{i}}$
$=\frac{1+\text{i}-3\text{i}-3}{1+\text{i}}$
$=\frac{-2-2\text{i}}{1+\text{i}}$
$=\frac{-2(1+\text{i})}{1+\text{i}}$
$=-2$
$=-2+0\text{i}$
$\therefore \ \frac{(1-\text{i})^3}{1-\text{i}^3}=-2+0\text{i}$
View full question & answer→Question 52 Marks
Express the following complex numbers in the standard form a + ib:
$\frac{(1-\text{i})(1+\sqrt{3}\text{i})}{1-\text{i}}$
Answer$\frac{(1+\text{i})(1+\sqrt{3}\text{i})}{1-\text{i}}=\frac{1(1+\sqrt{3}\text{i})+\text{i}(1+\sqrt{3}\text{i})}{1-\text{i}}$
$=\frac{(1+\sqrt{3}\text{i}+\text{i}-\sqrt{3})}{1-\text{i}} \ \big(\therefore \ \text{i}^2=-1\big)$
$=\frac{(1-\sqrt{3})+\text{i}(1+\sqrt{3})}{1-\text{i}}\times\frac{(1+\text{i})}{1+\text{i}}$ (Rationalising the denominator)
$=\frac{(1-\sqrt{3})(1+\text{i})+\text{i}(1+\sqrt{3})(1 +\text{i})}{1^2+1^2}$
$=\frac{1+\text{i}-\sqrt{3}(1+\text{i})+\text{i}(1+\text{i}+\sqrt{3}(1+\text{i}))}{2}$
$=\frac{1+\text{i}-\sqrt{3}-\sqrt{3}\text{i}+\text{i}(1+\text{i}+\sqrt{3}+\sqrt{3}\text{i})}{2}$
$=\frac{1-\sqrt{3}+\text{i}-\sqrt{3}\text{i}+\text{i}-1+\sqrt{3}\text{i}+\sqrt{3}}{2}$
$=\frac{-2\sqrt{3}+2\text{i}}{2}$
$=\sqrt{3}+\text{i}$
$\therefore \ \frac{(1+\text{i})(1+\sqrt{3}\text{i})}{1-\text{i}}=-\sqrt{3}+\text{i}$
View full question & answer→Question 62 Marks
Express the following complex numbers in the standard form a + ib:
$\frac{(2+\text{i})^3}{2+3\text{i}}$
Answer$\frac{(2+\text{i})^3}{2+3\text{i}}=\frac{2^3+\text{i}^3+3\times2\times\text{i}(2+\text{i})}{2+3\text{i}} \ \big[\therefore \ (\text{a}+\text{b})^3=\text{a}^3+\text{b}^3+3\text{ab}(\text{a}+\text{b})\big]$
$=\frac{(8-\text{i}+6\text{i}(2+\text{i}))}{2+3\text{i}}\times\frac{(2-3\text{i})}{2-3\text{i}}$ (On rationalising the denominator)
$=\frac{(8-\text{i}+12\text{i}+6\text{i}^2)(2-3\text{i})}{2^2+3^3}$
$=\frac{(8-6+11\text{i})(2-3\text{i})}{4+9} \ \big(\because \ \text{i}^2=-1\big)$
$=\frac{(2+11\text{i})(2-3\text{i})}{13}$
$=\frac{4-6\text{i}+22\text{i}+33}{13}$
$=\frac{37+16\text{i}}{13}$
$=\frac{37}{13}+\frac{16}{13}\text{i}$
$\therefore\frac{(2+\text{i})^3}{2+3\text{i}}=\frac{37}{13}+\frac{16}{13}\text{i}$
View full question & answer→Question 72 Marks
Express the following complex numbers in the standard form a + ib:
$(1-2\text{i})^{-3}$
Answer$(1-2\text{i})^{-3}=\frac{1}{(1+2\text{i}^3)} \ \Big(\therefore \ \text{z}^{-3}=\frac{1}{\text{z}^3}\Big)$
$=\frac{1}{1^3+(2\text{i})^3+3\times1\times2\text{i}(1+2\text{i})}$
$=\frac{1}{1^3+2^3\times\text{i}^3+6\text{i}(1+2\text{i})}$
$=\frac{1}{1-8\text{i}+6\text{i}-12} \ \big(\therefore \ \text{i}^3=-\text{i} \ \text{and} \ \text{i}^2=-1\big)$
$=\frac{1}{-11-2\text{i}}$
$=\frac{1}{-11-2\text{i}}\times\frac{(-11+2\text{i})}{(-11+2\text{i})}$
$=\frac{-11+2\text{i}}{121+4}$
$=\frac{-11}{125}+\frac{2}{125}\text{i}$
$\therefore \ (1-2\text{i})^{-3}=\frac{-11}{125}+\frac{2}{125}\text{i}$
View full question & answer→Question 82 Marks
Find the modulus and argument of the following complex numbers and hence express each of them in the polar form:
$1-\text{i}$
AnswerModulus, $|1-\text{i}|=\sqrt{1^2+1^2}=\sqrt{2}$
Argument, $\text{arg}(1-\text{i})=\tan^{-1}\Big(\frac{-1}{1}\Big)=\tan^{-1}(-1)=-\frac{\pi}{4}$
Polar form, $\sqrt{2}\Big(\cos\frac{\pi}{4}-\text{i}\sin\frac{\pi}{4}\Big)$
View full question & answer→Question 92 Marks
Express the following complex numbers in the standard form a + ib:
$\frac{2+3\text{i}}{4+5\text{i}}$
Answer$\frac{2+3\text{i}}{4+5\text{i}}=\frac{2+3\text{i}}{4+5\text{i}}\times\frac{(4-5\text{i})}{(4-5\text{i})}$ (On rationalising the denominator)
$=\frac{2(4-5\text{i})+3\text{i}(4-5\text{i})}{4^2+5^2}$
$=\frac{8-10\text{i}+12\text{i}+15}{16+25} \ \big(\because\ \text{i}^2=-1\big)$
$=\frac{23+2\text{i}}{41}$
$=\frac{23}{41}+\frac{2}{41}\text{i}$
$\therefore \ \frac{2+3\text{i}}{4+5\text{i}}=\frac{23}{41}+\frac{2}{41}\text{i}$
View full question & answer→Question 102 Marks
Find the real values of x and y, if
$(\text{x}+\text{iy})(2-3\text{i})=4+\text{i}$
AnswerWe have $(\text{x}+\text{iy})(2-3\text{i})=4+\text{i}$
$\Rightarrow\text{x}(2-3\text{i})+\text{iy}(2-3\text{i})=4+\text{i}$
$\Rightarrow2\text{x}-3\text{xi}+2\text{iy}+3\text{y}=4+\text{i}$
$\Rightarrow2\text{x}+3\text{y}+\text{i}(-3\text{x}+2\text{y})=4+\text{i}$
Equating the real and imaginary parts we get
$2\text{x}+3\text{y}=4 \ ...(\text{i})$
$-3\text{x}+2\text{y}=1 \ ...(\text{ii})$
Multiplying (i) by 3 and (ii) by 2 and adding
$6\text{x}-6\text{x}-9\text{y}+4\text{y}=12+2$
$\Rightarrow13\text{y}=14$
$\Rightarrow\text{y}=\frac{14}{13}$
Substituting the value of y in (i), we get
$2\text{x}+3\times\frac{14}{13}=4$
$\Rightarrow2\text{x}+\frac{42}{13}=4$
$\Rightarrow2\text{x}=4-\frac{42}{13}$
$\Rightarrow2\text{x}=\frac{52-42}{13}$
$\Rightarrow2\text{x}=\frac{10}{13}$
$\Rightarrow\text{x}=\frac{5}{13}$
Hence
$\text{x}=\frac{5}{13}$ and $\text{y}=\frac{14}{13}$
View full question & answer→Question 112 Marks
Find the square root of the following complex numbers:
$-5+12\text{i}$
AnswerLet $\text{z}=-5+12\text{i}$
$\Rightarrow|\text{z}|=\sqrt{(-5)^2+12^2}$
$=\sqrt{25+144}$
$=\sqrt{169}$
$=13$
$\therefore\sqrt{-5+12\text{i}}=\pm\Bigg\{\sqrt{\frac{13+(-5)}{2}}+\text{i}\sqrt{\frac{13-(-5)}{2}}\Bigg\} \ (\because\text{y}>0)$
$=\pm\Bigg\{\sqrt{\frac{8}{2}}+\text{i}\sqrt{\frac{18}{2}}\Bigg\}$
$\pm\{2+3\text{i}\}$
View full question & answer→Question 122 Marks
Evaluate the following:
$\frac{1}{\text{i}^{58}}$
AnswerWe know that
$\text{i}=\sqrt{-1}$
$\text{i}^2 = -1$
$\text{i}^3 = -\text{i}$
$\text{i}^4 = 1$
In order to find $ i^n $ Where $n > 4,$ we divide $n$ by $4$ to get quotient $p$ and remainder $q,$ So that $\text{n} = 4\text{p} + \text{q}, \ 0\leq\text{q}<4 $
Then $\text{i}^\text{n} =\text{i}^{4\text{p}+\text{q}}$
$=\text{i}^{4\text{p}}\times\text{i}^\text{q}$
$=\big(\text{i}^{4}\big)^{\text{p}}\times\text{i}^\text{q}$
$=\text{i}^{\text{p}}\times\text{i}^\text{q}$
$=\text{i}^\text{q} \ \big[\therefore \ 1^{\text{p}-1}\big]$
Hence $\text{i}^\text{n} =\text{i}^\text{q},$ where $0\leq\text{q}<4 $
$\therefore \ \frac{1}{\text{i}^{58}}=\frac{1}{\text{i}^{4\times14}\times\text{i}^2}$
$=\frac{1}{1\times\text{i}^2}$
$=\frac{1}{-1} \ \big[\therefore \ \text{i}^2=-1\big]$
$=-1$
View full question & answer→Question 132 Marks
If $Z_1, z_2, z_3$ are complex numbers such that $|\text{z}_1|=|\text{z}_2|=|\text{z}_3|=\Big|\frac{1}{\text{z}_1}+\frac{1}{\text{z}_2}+\frac{1}{\text{z}_3}\Big|=1,$ then find the value of $|\text{z}_1+\text{z}_2+\text{z}_3|.$
Answer$|\text{z}_1+\text{z}_2+\text{z}_3|=\Big|\frac{\text{z}_1\bar{\text{z}_1}}{\text{z}_1}+\frac{\text{z}_2\bar{\text{z}_2}}{\text{z}_2}+\frac{\text{z}_3\bar{\text{z}_3}}{\text{z}_3}\Big|$
$=\Big|\frac{|\text{z}_1|^2}{\text{z}_1}+\frac{|\text{z}_2|^2}{\text{z}_2}+\frac{|\text{z}_3|^2}{\text{z}_3}\Big|$
$=\frac{\Bigg|\frac{1}{\text{z}_1}+\frac{1}{\text{z}_2}+\frac{1}{\text{z}_3}\Bigg|}{\Bigg|\frac{1}{\text{z}_1}+\frac{1}{\text{z}_2}+\frac{1}{\text{z}_3}\Bigg|} \ ...\big[\because|\text{z}_1|=|\text{z}_2|=|\text{z}_3|=1\big]$
$=\Bigg|\frac{1}{\text{z}_1}+\frac{1}{\text{z}_2}+\frac{1}{\text{z}_3}\Bigg|$
$=1$
View full question & answer→Question 142 Marks
Find the square root of the following complex numbers:
$8-15\text{i}$
AnswerLet $\text{z}=8-15\text{i}$
Then, $|\text{z}|=\sqrt{(8)^2+(-15)^2}$
$=\sqrt{64+225}$
$=\sqrt{289}$
$=17$
$\therefore\sqrt{8-15\text{i}}=\pm\Bigg\{\sqrt{\frac{17+8}{2}}-\text{i}\sqrt{\frac{17-8}{2}}\Bigg\} \ (\because\text{y}<0)$
$=\pm\Bigg\{\frac{5}{\sqrt{2}}-\text{i}\frac{3}{\sqrt{2}}\Bigg\}$
$=\pm\frac{1}{\sqrt{2}}\{5-3\text{i}\}$
View full question & answer→Question 152 Marks
Express the following complex numbers in the standard form a + ib:
$\frac{5+\sqrt{2}\text{i}}{1-\sqrt{2}\text{i}}$
AnswerWe have,
$\frac{5+\sqrt{2}\text{i}}{1-\sqrt{2}\text{i}}=\frac{5+\sqrt{2}\text{i}}{1-\sqrt{2}\text{i}}\times\frac{1+\sqrt{2}\text{i}}{1+\sqrt{2}\text{i}}$
$=\frac{5(1+\sqrt{2}\text{i})+\sqrt{2}\text{i}(1+\sqrt{2}\text{i})}{1+2}$
$=\frac{5+5\sqrt{2}\text{i}+\sqrt{2}\text{i}-2}{3}$
$=\frac{3+6\sqrt{2}\text{i}}{3}$
$=1+2\sqrt{2}\text{i}$
$\therefore \ \frac{5+\sqrt{2}\text{i}}{1-\sqrt{2}\text{i}}=1+2\sqrt{2}\text{i}$
View full question & answer→Question 162 Marks
Evaluate the following:
$\text{i}^{457}$
AnswerWe know that
$\text{i}=\sqrt{-1}$
$\text{i}^2 = -1$
$\text{i}^3 = -\text{i}$
$\text{i}^4 = 1$
In order to find $i^n$ Where $n > 4,$ we divide $n$ by $4$ to get quotient $p$ and remainder $q,$ So that $\text{n} = 4\text{p} + \text{q}, \ 0\leq\text{q}<4 $
Then $\text{i}^\text{n} =\text{i}^{4\text{p}+\text{q}}$
$=\text{i}^{4\text{p}}\times\text{i}^\text{q}$
$=\big(\text{i}^{4}\big)^{\text{p}}\times\text{i}^\text{q}$
$=\text{i}^{\text{p}}\times\text{i}^\text{q}$
$=\text{i}^\text{q} \ \big[\therefore \ 1^{\text{p}-1}\big]$
Hence $\text{i}^\text{n} =\text{i}^\text{q},$ where $0\leq\text{q}<4 $
$\therefore \ \text{i}^{457}=\text{i}^{4\times114}\times\text{i}^1$
$=\text{i}^1$
$=\text{i}$
View full question & answer→Question 172 Marks
Find the modulus and argument of the following complex numbers and hence express each of them in the polar form:
$\sqrt{3}+\text{i}$
AnswerThe polar form of a complex number $\text{z}=\text{x}+\text{iy},$ is given by
$\text{z}=|\text{z}|(\cos\theta+\text{i}\sin\theta)$
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 $\text{z}=\sqrt{3}+\text{i}$
$|\text{z}|=\sqrt{(\sqrt{3}^2)+(1)^2}$
$=\sqrt{3+1}$
$=\sqrt{4}$
$=2$
$\because\text{x}=\sqrt{3}>0$ & $\text{y}=1>0,$
$\therefore\theta$ lies in first quadrant
Hence
$\theta=\text{arg(z)}=\tan^{-1}\big(\frac{\text{y}}{\text{x}}\big)$
$=\tan^{-1}\Big(\frac{1}{\sqrt{3}}\Big)$
$=\tan^{-1}\Big(\frac{\tan\pi}{6}\Big)$
$=\tan^{-1}\big(\therefore \tan^{-1}(\tan\text{x})=\text{x}\big)$
Polar form of is given by $\text{z}=|\text{z}|(\cos\theta+\text{i}\sin\theta)$
i.e., $\text{z}=2\Big(\cos\frac{\pi}{6}+\text{i}\sin\theta\frac{\pi}{6}\Big)$
View full question & answer→Question 182 Marks
Find the modulus and argument of the following complex numbers and hence express each of them in the polar form:
$1+\text{i}$
AnswerThe polar form of a complex number $\text{z}=\text{x}+\text{iy},$ is given by $\text{z}=|\text{z}|(\cos\theta+\text{i}\sin\theta)$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 $\text{z}=1+\text{i}$
$|\text{z}|=\sqrt{1^2+1^2}$
$=\sqrt{2}$
$\because\text{x},\text{y}>0,$ So $\theta$ lies in first quadrant
Now,
$\theta=\tan^{-1}\big(\frac{\text{b}}{\text{a}}\big)$
$=\tan^{-1}\big(\frac{\text{1}}{\text{1}}\big) \ \big[\because\text{a}=1 \ \text{and} \ \text{b}=1\big]$
$=\tan^{-1}(1)$
$=\tan^{-1}\big(\frac{\tan\pi}{4}\big) \ \Big(\because \ \frac{\tan\pi}{4}=1\Big)$
$=\frac{\pi}{4} \ \big(\because\tan^{-1}(\tan\text{x})=\text{x}\big)$
$\Rightarrow\text{arg(z)}=\frac{\pi}{4}$
Polar form of $1+\text{i}$ is given by $\text{z}=\sqrt{2}\big(\cos\frac{\pi}{4}+\text{i}\sin\frac{\pi}{4}\big)$
View full question & answer→Question 192 Marks
Express the following complex numbers in the standard form a + ib:
$\frac{3-4\text{i}}{(4-2\text{i})(1+\text{i})}$
Answer$\frac{3-4\text{i}}{(4-2\text{i})(1+\text{i})}=\frac{3-4\text{i}}{4(1+\text{i})-2\text{i}(1+\text{i})}$
$=\frac{3-4\text{i}}{4+4\text{i}-2\text{i}+2}$
$=\frac{3-4\text{i}}{6+2\text{i}}$
$=\frac{3-4\text{i}}{6+2\text{i}}\times\frac{6-2\text{i}}{6-2\text{i}}$
$=\frac{3(6-2\text{i})-4\text{i}(6-2\text{i})}{6^2+2^2}$
$=\frac{18-6\text{i}-24\text{i}-8}{36+4}$
$=\frac{10(1-3\text{i})}{40}$
$=\frac{1-3\text{i}}{4}$
$=\frac{1}{4}-\frac{3}{4}\text{i}$
$\therefore \ \frac{3-4\text{i}}{(4-2\text{i})(1+\text{i})}=\frac{1}{4}-\frac{3}{4}\text{i}$
View full question & answer→Question 202 Marks
Evaluate the following:
$\text{i}^{37}+\frac{1}{\text{i}^{67}}$
AnswerWe know that
$\text{i}=\sqrt{-1}$
$\text{i}^2 = -1$
$\text{i}^3 = -\text{i}$
$\text{i}^4 = 1$
In order to find $i^n$ Where $n > 4,$ we divide n by $4$ to get quotient $p$ and remainder $q,$ So that $\text{n} = 4\text{p} + \text{q}, \ 0\leq\text{q}<4 $
Then $\text{i}^\text{n} =\text{i}^{4\text{p}+\text{q}}$
$=\text{i}^{4\text{p}}\times\text{i}^\text{q}$
$=\big(\text{i}^{4}\big)^{\text{p}}\times\text{i}^\text{q}$
$=\text{i}^{\text{p}}\times\text{i}^\text{q}$
$=\text{i}^\text{q} \ \big[\therefore \ 1^{\text{p}-1}\big]$
Hence $\text{i}^\text{n} =\text{i}^\text{q},$ where $0\leq\text{q}<4 $
$\therefore \ \text{i}^{37}+\frac{1}{\text{i}^{67}}=\text{i}^{4\times9}\times\text{i}^1+\frac{1}{\text{i}^{4\times16}\times\text{i}^3}$
$=1\times\text{i}^1+\frac{1}{1\times\text{i}^3}$
$=\text{i}+\frac{1}{\text{i}^3\times\text{i}}\times\text{i}$
$=\text{i}+\frac{\text{i}}{\text{i}^4}$
$=\text{i}+\frac{\text{i}}{1} \ \big[\therefore \ \text{i}^4=1\big]$
$=2\text{i}$
$\therefore\text{i}^{37}+\frac{1}{\text{i}^{67}}=2\text{i}$
View full question & answer→Question 212 Marks
Find the square root of the following complex numbers:
$8-6\text{i}$
AnswerLet $\text{z}=8-6\text{i}$
Then, $|\text{z}|=\sqrt{(8)^2+(-6)^2}$
$=\sqrt{64+36}$
$=\sqrt{100}$
$=10$
$\therefore\sqrt{8-6\text{i}}=\pm\Bigg\{\sqrt{\frac{10-8}{2}}-\text{i}\sqrt{\frac{10+8}{2}}\Bigg\} \ (\because\text{y}<0)$
$=\pm\Bigg\{\sqrt{\frac{2}{2}}-\text{i}\sqrt{\frac{18}{2}}\Bigg\}$
$=\pm\{\sqrt{1}-\text{i}\sqrt{9}\}$
$=\pm\{1-3\text{i}\}$
View full question & answer→Question 222 Marks
Express the following complex numbers in the standard form a + ib:
$\frac{1}{(2+\text{i})^2}$
Answer$\frac{1}{(2+\text{i})^2}=\frac{1}{2^2+(\text{i})^2+2\times2\times\text{i}}$
$=\frac{1}{4-1+4\text{i}}$
$=\frac{1}{3+4\text{i}}$
$\frac{1}{(3+4\text{i})}\times\frac{(3-4\text{i})}{(3-4\text{i})}$ [On rationalising the denominator]
$=\frac{3-4\text{i}}{3^2+4^2} \ \big[\because \ (\text{a}+\text{ib})(\text{a}-\text{ib})=\text{a}^2+\text{b}^2\big]$
$=\frac{3-4\text{i}}{25}$
$=\frac{3}{25}-\frac{4}{25}\text{i}$
$\therefore \ \frac{1}{(2+\text{i})^2}=\frac{3}{25}-\frac{4}{25}\text{i}$
View full question & answer→Question 232 Marks
Express the following complex numbers in the standard form a + ib:
$\frac{1-\text{i}}{1+\text{i}}$
Answer$\frac{1-\text{i}}{1+\text{i}}=\frac{(1-\text{i})}{(1+\text{i})}\times\frac{(1-\text{i})}{(1-\text{i})}$ [On rationalising the denominator]
$=\frac{(1-\text{i})^2}{1^2+1^2} \ \big[\therefore \ (\text{a}+\text{ib})(\text{a}-\text{ib})=\text{a}^2+\text{b}^2\big]$
$=\frac{1^2+\text{i}^2-2\times\text{i}\times1}{2}$
$=\frac{-2\text{i}}{2}$
$=\text{-i}$
$=0-\text{i}$
$\therefore \ \frac{1-\text{i}}{1+\text{i}}=0-\text{i}$
View full question & answer→Question 242 Marks
Find the square root of the following complex numbers:
$1-\text{i}$
AnswerLet $\text{z}=1-\text{i}$ Then, $|\text{z}|=\sqrt{1^2+(-1)^2}$ $=\sqrt{1+1}$ $=\sqrt{2}$ $=25$ $\therefore\sqrt{1-\text{i}}=\pm\Bigg\{\sqrt{\frac{\sqrt{2}+1}{2}}-\text{i}\sqrt{\frac{\sqrt{2}-1}{2}}\Bigg\} \ (\because\text{y}<0)$$=\pm\Bigg\{\sqrt{\frac{\sqrt{2}+1}{2}}-\text{i}\sqrt{\frac{\sqrt{2}-1}{2}}\Bigg\}$
View full question & answer→Question 252 Marks
Express the following complex numbers in the standard form a + ib:
$(1+\text{i})(1+2\text{i})$
Answer$(1+\text{i})(1+2\text{i})=1\times(1+2\text{i})+\text{i}(1+2\text{i})$
$=1+2\text{i}+\text{i}+2\text{i}^2$
$=1+3\text{i}-2$
$=-1+3\text{i}$
$\therefore \ (1+\text{i})(1+2\text{i})=-1+3\text{i}$
View full question & answer→Question 262 Marks
Express the following complex numbers in the standard form a + ib:
$\Big(\frac{1}{1-4\text{i}}-\frac{2}{1+\text{i}}\Big)\Big(\frac{3-4\text{i}}{5+\text{i}}\Big)$
Answer$\Big(\frac{1}{1-4\text{i}}-\frac{2}{1+\text{i}}\Big)\Big(\frac{3-4\text{i}}{5+\text{i}}\Big)=\frac{(1+\text{i}-2(1-4\text{i}))}{(1-4\text{i})(1+\text{i})}\times\frac{3-4\text{i}}{5+\text{i}}$
$=\frac{(1+\text{i}-2+8\text{i})}{1(1+\text{i})-4\text{i}(1+\text{i})}\times\frac{3-4\text{i}}{5+\text{i}}$
$=\frac{(-1+9\text{i})}{(1+\text{i}-4\text{i}+4)}\times\frac{3-4\text{i}}{5+\text{i}}$
$=\frac{-1(3-4\text{i})+9\text{i}(3-4\text{i})}{(5-3\text{i})(5+\text{i})}$
$=\frac{-3+4\text{i}+27\text{i}+36}{5(5+\text{i})-3\text{i}(5+\text{i})}$
$=\frac{33+31\text{i}}{25+5\text{i}-15\text{i}+3}$
$=\frac{33+31\text{i}}{28-10\text{i}}$
$=\frac{(33+31\text{i})}{28-10\text{i}}\times\frac{(28+10\text{i})}{28+10\text{i}}$
$=\frac{33\times28+33\times10\text{i}+31\text{i}\times28+31\text{i}\times10\text{i}}{28^2+10^2}$
$=\frac{924+330\text{i}+868\text{i}-310}{784+100}$
$=\frac{614+1198\text{i}}{884}$
$=\frac{614}{884}+\frac{1198}{884}\text{i}$
$=\frac{30 7}{442}+\frac{599}{442}\text{i}$
$\therefore\Big(\frac{1}{1-4\text{i}}-\frac{2}{1+\text{i}}\Big)\Big(\frac{3-4\text{i}}{5+\text{i}}\Big)=\frac{30 7}{442}+\frac{599}{442}\text{i}$
View full question & answer→Question 272 Marks
Find the square root of the following complex numbers:
$-\text{i}$
AnswerLet $\text{z}=-\text{i}$
Then $|\text{z}|=|-\text{i}|$
$=|-1|\times|\text{i}| \ (\because|\text{z}_1\text{z}_2|=|\text{z}_1|\times|\text{z}_2|)$
$=1\times\text{i} \ (\because|\text{i}|=1)$
$=1$
$\therefore\sqrt{-\text{i}}=\pm\Bigg\{\sqrt{\frac{1+0}{2}}-\text{i}\sqrt{\frac{1-0}{2}}\Bigg\} \ (\because\text{y}<0)$
$=\pm\Big\{\frac{1}{\sqrt{2}}-\frac{\text{i}}{\sqrt{2}}\Big\}$
$=\pm\frac{1}{\sqrt{2}}(1-\text{i})$
View full question & answer→Question 282 Marks
Evaluate the following:
$\text{i}^{49}+\text{i}^{68}+\text{i}^{89}+\text{i}^{110}$
View full question & answer→Question 292 Marks
Evaluate the following:
$\text{i}^{30}+\text{i}^{40}+\text{i}^{60}$
AnswerWe know that
$\text{i}=\sqrt{-1}$
$\text{i}^2 = -1$
$\text{i}^3 = -\text{i}$
$\text{i}^4 = 1$
In order to find $i^n$ Where $n > 4,$ we divide $n$ by $4$ to get quotient $p$ and remainder $q,$ So that $\text{n} = 4\text{p} + \text{q}, \ 0\leq\text{q}<4 $
Then $\text{i}^\text{n} =\text{i}^{4\text{p}+\text{q}}$
$=\text{i}^{4\text{p}}\times\text{i}^\text{q}$
$=\big(\text{i}^{4}\big)^{\text{p}}\times\text{i}^\text{q}$
$=\text{i}^{\text{p}}\times\text{i}^\text{q}$
$=\text{i}^\text{q} \ \big[\therefore \ 1^{\text{p}-1}\big]$
Hence $\text{i}^\text{n} =\text{i}^\text{q},$ where $0\leq\text{q}<4 $
$\therefore \ \text{i}^{30}+\text{i}^{40}+\text{i}^{60}=\text{i}^{4\times7}\times\text{i}^{2}+\text{i}^{4\times10}+\text{i}^{4\times15}$
$=1\times\text{i}^2+1+1$
$=-1+1+1$
$=1$
$\text{i}^{30}+\text{i}^{40}+\text{i}^{60}1$
View full question & answer→Question 302 Marks
Find the square root of the following complex numbers:
$4\text{i}$
AnswerLet $\text{z}=4\text{i}$
Then $|\text{z}|=|4\text{i}|$
$=|4\text{i}| \ (\because|\text{z}_1\text{z}_2|=|\text{z}_1|\times|\text{z}_2|)$
$=4 \ (\because|\text{i}|=1)$
$\therefore\sqrt{4\text{i}}=\pm\Bigg\{\sqrt{\frac{4+0}{2}}+\text{i}\sqrt{\frac{4-0}{2}}\Bigg\} \ (\because\text{y}>0)$
$=\pm\{\sqrt{2}+\text{i}\sqrt{2}\}$
$=\pm\sqrt{2}(1+\text{i})$
View full question & answer→Question 312 Marks
Evaluate the following:
$\text{i}^{528}$
AnswerWe know that
$\text{i}=\sqrt{-1}$
$\text{i}^2 = -1$
$\text{i}^3 = -\text{i}$
$\text{i}^4 = 1$
In order to find $i^n$ Where $n > 4,$ we divide $n$ by $4$ to get quotient $p$ and remainder $q,$ So that $\text{n} = 4\text{p} + \text{q}, \ 0\leq\text{q}<4 $
Then $\text{i}^\text{n} =\text{i}^{4\text{p}+\text{q}}$
$=\text{i}^{4\text{p}}\times\text{i}^\text{q}$
$=\big(\text{i}^{4}\big)^{\text{p}}\times\text{i}^\text{q}$
$=\text{i}^{\text{p}}\times\text{i}^\text{q}$
$=\text{i}^\text{q} \ \big[\therefore \ 1^{\text{p}-1}\big]$
Hence $\text{i}^\text{n} =\text{i}^\text{q},$ where $0\leq\text{q}<4 $
$\therefore \ \text{i}^{528}=\text{i}^{4\times132}$
$=\big(\text{i}^4\big)^{132}$
$=1^{132}$
$=1$
$\therefore \ \big(\text{i}^{528}\big)=1$
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