Question 13 Marks
Write in $(\text{i}^{25})^3$ polar form.
Answer$(\text{i}^{25} )^3=\text{i}^{75}$ $=\text{i}^4\times 18+3$ $=(\text{ i}^4)^{ 18}.\text{ i}^3$ $=\text{i}^3\ \big[\because \text{i}^4=1\big]$ $=\text{-i}\ \big[\because \text{i}^3=-\text{i}\big]$ Let $\text{z} = 0 - \text{i}$ Then, $\text{|z|}=\sqrt{0^2+(-1)^2}=1$ Let $\theta$ be the argument of z and a be the acute angle given by $\tan \alpha=\frac{|\text{Im}\text{(z)|}}{|\text{Re}\text{(z)}|}$
Then, $\tan\alpha=\frac{1}{0}=\infty$ $\Rightarrow \alpha =\frac{\pi}{2}$ Clearly, z lies in fourth quadrant. so $\text{arg(z)}=-\alpha =-\frac{\pi}{2}$ $\therefore$ the polar from of z is $\text{|z|}(\cos\theta+\text{i }\sin\theta)=\cos(-\frac{\pi}{2})+\text{i }\sin(-\frac{\pi}{2})$ Thus, the polar from of $(\text{i}^{25})^3$ is $\cos(\frac{\pi}{2})-\text{i }\sin(\frac{\pi}{2})$
View full question & answer→Question 23 Marks
Find the conjugates of the following complex numbers:
$\frac{1}{3+5\text{i}}$
AnswerLet $\text{z}=\frac{1}{3+5\text{i}}$
$=\frac{1}{3+5\text{i}}\times\frac{(3+5\text{i})}{3+5\text{i}}$ (On rationalising the denominator)
$=\frac{3+5\text{i}}{3^2+5^2}$
$\Rightarrow\text{z}=\frac{3-5\text{i}}{9+25}$
So $\bar{\text{z}}=\frac{3-5\text{i}}{34}$
$=\frac{3}{34}+\frac{5}{34}\text{i}$
View full question & answer→Question 33 Marks
Find the values of the following expressions:
$\text{i}^{30}+\text{i}^{80}+\text{i}^{120}$
Answer$\text{i}^{30}+\text{i}^{80}+\text{i}^{120}=\text{i}^{4\times7}\times\text{i}^2+\text{i}^{4\times20}+\text{i}^{4\times30}$ $=1\times\text{i}^2+1+1$ $=-1+1+1$ $=1$$\therefore\text{i}^{30}+\text{i}^{80}+\text{i}^{120}=1$
View full question & answer→Question 43 Marks
Find the conjugates of the following complex numbers:
$\frac{(1+\text{i})(2+\text{i})}{3+\text{i}}$
AnswerLet $\text{Z}=\frac{(1+\text{i})(2+\text{i})}{3+\text{i}}$
$=\frac{2+\text{i}+\text{i}(2+\text{i})}{3+\text{i}}$
$=\frac{2+\text{i}+2\text{i}+\text{i}}{3+\text{i}}$
$=\frac{1+3\text{i}}{3+\text{i}}$
$=\frac{(1+3\text{i})}{(3+\text{i})}\times\frac{(3-\text{i})}{(3-\text{i})}$
$=\frac{3-\text{i}+3\text{i}(3-\text{i})}{3^2+1^2}$
$=\frac{3-\text{i}+9\text{i}+3}{9+1}$
$=\frac{6+8\text{i}}{10}$
$=\frac{2(3+4\text{i})}{10}$
$\Rightarrow\text{z}=\frac{3+4\text{i}}{5}$
Hence
$\bar{\text{z}}=\frac{3-4\text{i}}{5}$
$=\frac{3}{5}-\frac{4}{5}\text{i}$
View full question & answer→Question 53 Marks
$(1+\text{i})^6+(1-\text{i})^3$
Answer$(1+\text{i})^6+(1-\text{i})^3=\Big[(1+\text{i})^2\Big]^3+(1-\text{i})^3$
$=\big(1+\text{i}^2+2\text{i}\big)^3+\big(1-3\text{i}+3\text{i}^2-\text{i}^3\big)$
$=\big(1-1+2\text{i}\big)^3+\big(1-3\text{i}-3+\text{i}\big)$
$=8\text{i}^3-2-2\text{i}$
$=-2-10\text{i}$
View full question & answer→Question 63 Marks
Find the multiplicative inverse of the following complex numbers:
$1-\text{i}$
AnswerIf $\text{z}=\text{x}+\text{iy}$ is a complex number, then the multiplicative inverse of z, denoted by $\text{z}^{-1}$ or $\frac{1}{\text{z}}$ is defined as $\text{z}^{-1}=\frac{1}{\text{z}}$
$=\frac{1}{\text{x}+\text{iy}}$
$=\frac{1}{\text{x}+\text{iy}}\times\frac{\text{x}-\text{iy}}{\text{x}-\text{iy}}$
$=\frac{\text{x}-\text{iy}}{\text{x}^2+\text{y}^2}$
$=\frac{\text{x}}{\text{x}^2+\text{y}^2}-\frac{\text{y}}{\text{x}^2+\text{y}^2}\text{i}$
Given
$\text{z}=1-\text{i}$
$\therefore \ \text{z}^{-1}=\frac{1}{1^2+1^2}-\frac{(-1)}{1^2+1^2}\times\text{i}$
View full question & answer→Question 73 Marks
Find the conjugates of the following complex numbers:
$\frac{1}{1+\text{i}}$
AnswerLet $\text{z}=\frac{1}{1+\text{i}}$
$=\frac{1}{1+\text{i}}\times\frac{(1-\text{i})}{1-\text{i}}$
$=\frac{1+\text{i}}{1^2+1^2}$
$=\frac{1-\text{i}}{2}$
$\therefore\bar{\text{z}}=\frac{1-\text{i}}{2}$
$=\frac{1}{2}+\frac{1}{2}\text{i}$
View full question & answer→Question 83 Marks
Evaluate the following:
$\big(\text{i}^{77}+\text{i}^{70}+\text{i}^{87}+\text{i}^{414}\big)^3$
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 $
$\big(\text{i}^{77}+\text{i}^{70}+\text{i}^{87}+\text{i}^{414}\big)^3=\big(\text{i}^{4\times9}\times\text{i}^{1}+\text{i}^{4\times17}\times\text{i}^{2}+\text{i}^{4\times21}\times\text{i}^3+\text{i}^{4\times103}\times\text{i}^2\big)^3$
$=\big(1\times\text{i}+1\times\text{i}^2+1\times\text{i}^3+1\times\text{i}^2\big)^3$
$=(\text{i}-1-\text{i}-1)^3$
$=(-2)^3$
$=-8$
$\big(\therefore(\text{i}^{77}+\text{i}^{70}+\text{i}^{87}+\text{i}^{414}\big)^3=-8$
View full question & answer→Question 93 Marks
Find the multiplicative inverse of the following complex numbers:
$4-3\text{i}$
AnswerLet $\text{z}=4-3\text{i}$
$\text{z}^{-1}=\frac{4}{4^2+(-3)^2}-\frac{(-3)}{4^2+(-3)^2}$
$=\frac{4}{16+9}+\frac{3}{16+9}\text{i}$
$=\frac{4}{25}+\frac{3}{25}\text{i}$
View full question & answer→Question 103 Marks
Evaluate the following:
$\Big(\text{i}^{41}+\frac{1}{\text{i}^{257}}\Big)^9$
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 $
$\Big(\text{i}^{41}+\frac{1}{\text{i}^{257}}\Big)^9=\Bigg(\text{i}^{4\times10}\times\text{i}^1+\frac{1}{\text{i}^{4\times64}\times\text{i}^1}\Bigg)^9$
$=\Big(1\times\text{i}+\frac{1}{1\times\text{i}}\Big)^9$
$=\Big(1+\frac{1}{\text{i}}\Big)^9$
$=\Big(\text{i}+\frac{1}{\text{i}\times\text{i}}\times\text{i}\Big)^9$
$=\Big(\text{i}+\frac{\text{i}}{-1}\Big)^9$
$=(\text{i}-\text{i})^9$
$=0$
View full question & answer→Question 113 Marks
If $\text{z}_1=2-\text{i}, \ \text{z}_2=-2+\text{i},$ Find
$\text{Im}\Big(\frac{1}{\text{z}_1\bar{\text{z}}_1}\Big)$
Answer$\frac{1}{\text{z}_1\bar{\text{z}}_1}=\frac{1}{|\text{z}|^2}$
$=\frac{1}{|2-\text{i}|^2}$
$=\frac{1}{2^2+(-1)^2}$
$=\frac{1}{4+1}$
$=\frac{1}{5},$ which is purely real
$\therefore \ \text{Im}\Big(\frac{1}{\text{z}_1\bar{\text{z}}_1}\Big)=0$
View full question & answer→Question 123 Marks
Find the real values of x and y, if
$\frac{(1+\text{i})\text{x}-2\text{i}}{3+\text{i}}+\frac{(2-3\text{i})\text{y}+\text{i}}{3-\text{i}}=\text{i}$
Answer$\frac{(1+\text{i})\text{x}-2\text{i}}{3+\text{i}}+\frac{(2-3\text{i})\text{y}+\text{i}}{3-\text{i}}=\text{i}$
$\Rightarrow\frac{(3-\text{i})((1+\text{i})\text{x}-2\text{i})+(3+\text{i})((2-3\text{i})\text{y}+\text{i})}{(3+\text{i})(3-\text{i})}=\text{i}$
$\Rightarrow\frac{(3-\text{i})(1+\text{i})\text{x}-2\text{i}(3-\text{i})+(3+\text{i})(2-3\text{i})\text{y}+\text{i}(3+\text{i})}{3^2+1^2}=\text{i}$
$\Rightarrow\frac{(3+3\text{i}-\text{i}+1)\text{x}-6\text{i}-2+(6-9\text{i}+2\text{i}+3)\text{y}+3\text{i}}{9+1}=\text{i}$
$\Rightarrow\frac{(4+2\text{i})\text{x}-6\text{i}-2+(9-7\text{i})\text{y}+3\text{i}-1}{10}=1$
$\Rightarrow4\text{x}+2\text{ix}-6\text{i}-2+9\text{y}-7\text{iy}+3\text{i}-1=10\text{i}$
$\Rightarrow4\text{x}+9\text{y}-3+\text{i}(2\text{x}-7\text{y}-3)=10\text{i}$
Equating the real and imaginary parts we get
$4\text{x}+9\text{y}-3=0 \ ...(\text{i})$
and $2\text{x}-7\text{y}-3=10$
i.e., $2\text{x}-7\text{y}=13 \ ...(\text{ii})$
Multiplying (i) by 7 and (ii) by 9 and adding we get
$28\text{x}+18\text{x}+63\text{y}-63\text{y}=117+21$
$\Rightarrow46\text{x}=117+21$
$\Rightarrow46\text{x}=138$
$\Rightarrow\text{x}=\frac{138}{46}$
$=3$
Substituting the value of x = 3 in (i), we get
$4\times3+9\text{y}=3$
$\Rightarrow9\text{y}=-9$
$\Rightarrow\text{y}=\frac{-9}{9}$
$\Rightarrow\text{y}=-1$
Hence
$\text{x}=3, \ \text{y}=-1$
View full question & answer→Question 133 Marks
Find the real values of x and y, if
$(1+\text{i})(\text{x}+\text{iy})=2-5\text{i}$
Answer$(1+\text{i})(\text{x}+\text{iy})=2-5\text{i}$
$\Rightarrow1(\text{x}+\text{iy})+\text{i}(\text{x}+\text{iy})=2-5\text{i}$
$\Rightarrow\text{x}+\text{iy}+\text{i}\text{x}-\text{y}=2-5\text{i}$
$\Rightarrow\text{x}-\text{y}+\text{i}(\text{x}+\text{y})=2-5\text{i}$
Equating the real and imaginary parts we get
$\text{x}-\text{y}=2 \ ...(\text{i})$
$\text{x}+\text{y}=-5 \ ...(\text{ii})$
Adding (i) and (ii) we get
$2\text{x}=2-5$
$\Rightarrow2\text{x}=-3$
$\Rightarrow\text{x}=\frac{-3}{2}$
Substituting the value of x in (i), we get
$\frac{-3}{2}-\text{y}=2$
$\Rightarrow\frac{-3}{2}-2=\text{y}$
$\Rightarrow\text{y}=\frac{-3-4}{2}$
$\Rightarrow\text{y}=\frac{-7}{2}$
Hence
$\text{x}=\frac{-3}{2},\text{y}=\frac{-7}{2}$
View full question & answer→Question 143 Marks
If $\frac{\text{z}-1}{\text{z}+1}$ is purely imaginary number $(\text{z}\neq-1),$ find the value of $|\text{z}|.$
AnswerLet $\text{z}=\text{x}+\text{iy},$
$\frac{\text{z}-1}{\text{z}+1}$
$=\frac{\text{x+iy}-1}{\text{x+iy}+1}$
$=\frac{\text{x-1+iy}}{\text{x+1+iy}}$
$=\frac{(\text{x-1+iy)}(\text{x+1-iy)}}{(\text{x+1+iy)}(\text{x+1-iy)}}$ [Rationalizing the denominator]
$=\frac{(\text{x-1+iy)}(\text{x+1-iy)}}{{(\text{x}+1)}^2-{(\text{iy)}^2}}$
$=\frac{\text{x}^2+\text{x}-\text{ixy}-\text{x}-1+\text{iy}+\text{ixy}+\text{iy}+\text{y}^2}{\text{x}^2+2\text{x}+1+\text{y}^2}$
$=\frac{\text{x}^2-1+2\text{iy}+\text{y}^2}{\text{x}^2+2\text{x}+1\text{y}^2}$
$=\frac{\text{x}^2+\text{y}^2-1}{\text{x}^2+2\text{x}+1+\text{y}^2}+\text{i}\frac{2\text{y}}{\text{x}^2+2\text{x}+1+\text{y}^2}$
$\because$ It is a purely imaginary no. therefore real part = 0
$=\frac{\text{x}^2+\text{y}^2-1}{\text{x}^2+2\text{x}+1+\text{y}^2}=0$
$\Rightarrow\text{x}^2+\text{y}^2-1=0$
$\Rightarrow\text{x}^2+\text{y}^2=1$
$\Rightarrow\sqrt{\text{x}^2+\text{y}^2}=1$
$\Rightarrow|\text{z}|=1$
View full question & answer→Question 153 Marks
Find the values of the following expressions:
$\text{i}^{5}+\text{i}^{10}+\text{i}^{15}$
Answer$\text{i}^{5}+\text{i}^{10}+\text{i}^{15}=\text{i}^{4\times1}\times\text{i}^1+\text{i}^{4\times2}\times\text{i}^2+\text{i}^{4\times3}\times\text{i}^3$
$=1\times\text{i}+1\times\text{i}^2+1\times\text{i}^3$
$=\text{i}-1-\text{i}$
$=-1$
$\therefore\text{i}^{5}+\text{i}^{10}+\text{i}^{15}=-1$
View full question & answer→Question 163 Marks
Find the values of the following expressions:
$\text{i}+\text{i}^{2}+\text{i}^{3}+\text{i}^{4}$
Answer$\text{i}+\text{i}^{2}+\text{i}^{3}+\text{i}^{4}=1+(-1)+(-\text{i})+1$ $=0$$\therefore\text{i}+\text{i}^{2}+\text{i}^{3}+\text{i}^{4}=0$
View full question & answer→Question 173 Marks
Find the modulus of $\frac{1+\text{i}}{1-\text{i}}-\frac{1-\text{i}}{1+\text{i}}$
AnswerLet $\text{z}=\frac{1+\text{i}}{1-\text{i}}-\frac{1-\text{i}}{1+\text{i}}$
$=\frac{(1+\text{i})^2-(1-\text{i})^2}{(1-\text{i})(1+\text{i})}$
$=\frac{1^2+\text{i}^2+2\times1\times\text{i}-(1^2+\text{i}^2-2\times1\times\text{i})}{1^2+1^2}$
$=\frac{1-1+2\text{i}-(1-1-2\text{i})}{2}$
$=\frac{2\text{i}+2\text{i}}{2}$
$=\frac{4\text{i}}{2}$
$\Rightarrow\text{z}=2\text{i}$
$\therefore \ |\text{z}|=|2\text{i}|$
$=2|\text{i}| \ \Big(\because \ |\text{z}_1\text{z}_2|=|\text{z}_1|\times|\text{z}_2|\Big)$
$=2\times\text{i} \ \Big(\because \ |\text{i}|=1\Big)$
$=2$
View full question & answer→Question 183 Marks
Find the square root of the following complex numbers:
$1+4\sqrt{-3}$
AnswerLet $\text{z}=1+4\sqrt{-3}$
$=1+4\sqrt{3}\times\sqrt{-1} \ (\therefore \ \sqrt{-3}=\sqrt{3}\times\sqrt{-1})$
$\Rightarrow\text{z}=1+4\sqrt{ 3\text{i}}$
$\therefore|\text{z}|=\sqrt{(1)^2+(4\sqrt{3})^2}$
$=\sqrt{1+48}$
$=\sqrt{49}$
$=7$
Hence $\therefore\sqrt{1+4\sqrt{-3}}=\pm\Bigg\{\sqrt{\frac{7+1}{2}}+\text{i}\sqrt{\frac{7-1}{2}}\Bigg\} \ (\because\text{y}>0)$
$=\pm\Bigg\{\sqrt{\frac{8}{2}}-\text{i}\sqrt{\frac{6}{2}}\Bigg\}$
$=\pm\{\sqrt{4}+\text{i}\sqrt{3}\}$
$=\pm\{2+\text{i}\sqrt{3}\}$
View full question & answer→Question 193 Marks
Find the real values of x and y, if
$(3\text{x}-2\text{iy})(2+\text{i})^2=10(1+\text{i})$
Answer$(3\text{x}-2\text{iy})(2+\text{i})^2=10(1+\text{i})$
$\Rightarrow(3\text{x}-2\text{iy})(2^2+\text{i}^2+2\times2\times\text{i})=10+10\text{i}$
$\Rightarrow(3\text{x}-2\text{iy})(4-1\times4\text{i})=10+10\text{i}$
$\Rightarrow3\text{x}(3+4\text{i})-2\text{iy}(3+4\text{i})=10+10\text{i}$
$\Rightarrow9\text{x}+12\text{xi}-6\text{yi}+8\text{y}=10+10\text{i}$
$\Rightarrow9\text{x}+8\text{y}+\text{i}(12\text{x}-6\text{y})=10+10\text{i}$
Equating the real and imaginary parts we get
$9\text{x}+8\text{y}=10 \ ...(\text{i})$
$-12\text{x}-6\text{y}=10 \ ...(\text{ii})$
Multiplying (i) by 6 and (ii) by 8 and adding
$54\text{x}+96\text{x}+48\text{y}-48\text{y}=60+80$
$\Rightarrow150\text{x}=140$
$\Rightarrow\text{x}=\frac{140}{150}$
$\Rightarrow\text{x}=\frac{14}{15}$
Substituting the value of x in (i), we get
$9\times\frac{14}{13}+8\text{y}=10$
$\Rightarrow\frac{42}{5}+8\text{y}=10$
$\Rightarrow8\text{y}=10-\frac{42}{5}$
$\Rightarrow8\text{y}=\frac{50-42}{5}$
$\Rightarrow8\text{y}=\frac{8}{5}$
$\Rightarrow\text{y}=\frac{1}{5}$
Hence
$\text{x}=\frac{14}{5}$ and $\text{y}=\frac{1}{5}$
View full question & answer→Question 203 Marks
$1+\text{i}^2+\text{i}^4+\text{i}^6+\text{i}^8+ ...+\text{i}^{20}$
Answer$1+\text{i}^2+\text{i}^4+\text{i}^6+\text{i}^8+ ...+\text{i}^{20}$
$=1+\text{i}^2+\text{i}^4+\text{i}^{4\times1}\times\text{i}^2+\text{i}^{4\times2}+\text{i}^{4\times2}\times\text{i}^{2}+\text{i}^{4\times3}+\text{i}^{4\times3}\times\text{i}^{2}+\text{i}^{4\times4}+\text{i}^{4\times4}\times\text{i}^2+\text{i}^{4\times5}$
$=1-1+1+1\times\text{i}^2+1+1\times\text{i}^2+1+1\times\text{i}^2+1+1\times\text{i}^2+1$
$=1-1+1-1+1-1+1-1+1-1+1$
$=1$
View full question & answer→Question 213 Marks
Solve the system of equations $\text{Re}(\text{z}^2)=0,|\text{z}|=2.$
Answer$\text{Re}(\text{z}^2)=0,|\text{z}|=2$
Let $\text{z}=\text{x}+\text{iy}$
$\text{z}^2=0$
$\Rightarrow(\text{x}+\text{iy})^2=0$
$\Rightarrow\text{x}^2+2\text{ixy}-\text{y}^2=0$
$\Rightarrow\text{x}^2-\text{y}^2=0 \ ...(\text{i}),$ which is the real part of $(\text{x}+\text{iy})^2.$
$|\text{z}|=2$
$\Rightarrow\sqrt{\text{x}^2+\text{y}^2}=2$
$\Rightarrow\text{x}^2+\text{y}^2=4 \ ...(\text{ii})$
Adding (i) and (ii), we get
$2\text{x}^2=4$
$\Rightarrow\text{x}^2=2$
$\Rightarrow\text{x}=\pm\sqrt{2},\text{y}=\pm\sqrt{2}$
$\text{x}+\text{iy}=\sqrt{2}+\text{i}\sqrt{2}$
$=\sqrt{2}-\text{i}\sqrt{2}$
$=\sqrt{2}+\text{i}\sqrt{2}$
View full question & answer→Question 223 Marks
Find the square root of the following complex numbers:
$-7+24\text{i}$
AnswerLet $\text{z}=-7+24\text{i}$
$\Rightarrow|\text{z}|=\sqrt{(-7)^2+24^2}$
$=\sqrt{49+576}$
$=\sqrt{625}$
$=25$
$\therefore\sqrt{-7-24\text{i}}=\pm\Bigg\{\sqrt{\frac{25-7}{2}}+\text{i}\sqrt{\frac{25+7}{2}}\Bigg\} \ (\because\text{y}<0)$
$=\pm\Bigg\{\sqrt{\frac{18}{2}}+\text{i}\sqrt{\frac{32}{2}}\Bigg\}$
$=\pm\{\sqrt{9}+\text{i}\sqrt{16}\}$
$=\pm\{3-4\text{i}\}$
View full question & answer→Question 233 Marks
Find the conjugates of the following complex numbers:
$4-5\text{i}$
AnswerIf $\text{z}=\text{x}+\text{iy}$ is a complex number, then the conjugate of z denoted by $\bar{\text{z}}$ is defined as $\bar{\text{z}}=\text{x}-\text{iy}$
Let $\text{z}=4-5\text{i}$
$\Rightarrow\bar{\text{z}}=4+5\text{i}$
View full question & answer→Question 243 Marks
Find the modulus and argument of the following complex numbers and hence express each of them in the polar form:
$\frac{1+2\text{i}}{1-3\text{i}}$
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 $\text{z}=\frac{1+2\text{i}}{1-3\text{i}}$
$=\frac{1+2\text{i}}{1-3\text{i}}\times\frac{1+3\text{i}}{1+3\text{i}}$
$=\frac{1(1+3\text{i})+2\text{i}(1+3\text{i})}{1^2+3^2}$
$=\frac{1+3\text{i}+2\text{i}-6}{1+9}$
$=\frac{-5+5\text{i}}{10}$
$=\frac{-5}{10}+\frac{5}{10}\text{i}$
$=\frac{-1}{2}+\frac{1}{2}\text{i}$
$\therefore \ |\text{z}|=\sqrt{\Big(\frac{-1}{2}\Big)^2+\Big(\frac{1}{2}\Big)^2}$
$=\sqrt{\frac{1}{4}+\frac{1}{4}}$
$=\sqrt{\frac{2}{4}}$
$=\frac{1}{\sqrt{2}}$
Here $\text{x}=\frac{-1}{2}<0$ & $\text{y}=\frac{1}{2}>0, \ \therefore\theta$ lies i quadrantII
$\theta=\text{arg(z)}=\tan^{-1}\frac{\frac{1}{2}}{\frac{-1}{2}}$
$=\tan^{-1}(-1)$
$=\tan^{-1}\Big(-\tan\frac{\pi}{4}\Big)$
$=\tan^{-1}\Big(\tan\big(\pi-\frac{\pi}{4}\big)\Big) \ \big(\therefore\tan(\pi-\theta)=-\tan\theta\big)$
$=\pi-\frac{\pi}{4}$
$=\frac{3\pi}{4}$
The polar form is given by $\text{z}=\frac{1}{\sqrt{2}}\Big(\cos\frac{3\pi}{4}+\text{i}\sin\frac{3\pi}{4}\Big)$
View full question & answer→Question 253 Marks
Find the modulus and argument of the following complex numbers and hence express each of them in the polar form:
$\frac{1-\text{i}}{1+\text{i}}$
Answer$\frac{1-\text{i}}{1+\text{i}}=\frac{(1-\text{i})(1-\text{i})}{(1+\text{i})(1-\text{i})}=\frac{(1-\text{i})^2}{1^2-\text{i}^2}=\frac{1-2\text{i}-1}{1+1}=\frac{-2\text{i}}{2}=\text{-i}$
Modulus, $\Big|\frac{1-\text{i}}{1+\text{i}}\Big|=|\text{-i}|=1$
Argument, $\tan^{-1}\Big(\frac{-1}{0}\Big)=-\frac{\pi}{2}$
Polar form, $\text{z}=\text{r}(\cos\theta+\text{i}\sin\theta)$
$\text{z}=\Big(\cos\frac{\pi}{2}-\text{i}\sin\frac{\pi}{2}\Big)$
View full question & answer→Question 263 Marks
If $(1+\text{i})\text{z}=(1-\text{i})\bar{\text{z}},$ then show that $\text{z}=-\text{i}\bar{\text{z}}.$
Answer$(1+\text{i})\text{z}=(1-\text{i})\bar{\text{z}}$
$\Rightarrow\text{z}=\frac{(1-\text{i})}{(1+\text{i})}\bar{\text{z}}$
$\Rightarrow\text{z}=\frac{(1-\text{i})(1-\text{i})}{(1+\text{i})(1-\text{i})}\bar{\text{z}}$ [Rationalizing the denominator]
$\Rightarrow\text{z}=\frac{(1-2\text{i}-1)}{(1+1)}\bar{\text{z}}$
$\Rightarrow\text{z}=\frac{-2\text{i}}{2}\bar{\text{z}}$
$\Rightarrow\text{z}=-\text{i}\bar{\text{z}}$
View full question & answer→Question 273 Marks
Find the modulus and argument of the following complex numbers and hence express each of them in the polar form:
$\frac{1}{1+\text{i}}$
Answer$\frac{1}{1+\text{i}}$
Rationalising the denominator:
$\frac{1}{1+\text{i}}\times \frac{1-\text{i}}{1-\text{i}}$
$\Rightarrow \frac{1-\text{i}}{1-\text{i}^2}$
$\Rightarrow \frac{1-\text{i}}{2}\ (\because\ \text{i}^2=-1)$
$\Rightarrow \frac{1}{2}-\frac{\text{i}}{2}$
$\text{r}=|\text{z}|$
$= \sqrt{\frac{1}{4}+\frac{1}{4}}$
$=\frac{1}{\sqrt{2}}$
Let $\tan\alpha=\bigg|\frac{\text{Im}\text{(z)}}{\text{Re}\text{(z)}}\bigg|$
$\therefore\tan\alpha=\Bigg|\frac{\frac{1}{2}}{\frac{-1}{2}}\Bigg|$
$=1$
$\Rightarrow \alpha =\frac{\pi}{4}$
Since ponit $\Big(\frac{1}{2},-\frac{1}{2}\Big)$ lies in the fourth quadrant, the argurment is given by $\theta=-\alpha =\frac{\pi}{4}$
Polar form $=\text{r}(\cos \theta + \text{i } \sin\theta )$
$=\frac{1}{\sqrt{2}}\big\{\cos \big(\frac{-\pi}{4}\big)+\text{i }\sin\big(\frac{-\pi }{4}\big)\big\}$
$=\frac{1}{\sqrt{2}}\big(\cos \frac{\pi}{4}-\text{i }\sin\frac{\pi }{4}\big)$
View full question & answer→Question 283 Marks
Show that $1+\text{i}^{10}+\text{i}^{20}+\text{i}^{30}$ is a real number.
Answer$1+\text{i}^{10}+\text{i}^{20}+\text{i}^{30}=1+\text{i}^{4\times2}\times\text{i}^2+\text{i}^{4\times5}+\text{i}^{4\times7}\times\text{i}^2$ $=1+1\times\text{i}^2+1+1\times\text{i}^2$ $=1-1+1-1$$=0,$ which is real number.
View full question & answer→Question 293 Marks
Find the modulus and argument of the following complex numbers and hence express each of them in the polar form:
$\sin120^\circ-\text{i}\cos120^\circ$
Answer$\sin120^\circ-\text{i}\cos120^\circ$
$\frac{\sqrt{3}}{2}+\frac{\text{i}}{2}$
$\text{r}=|\text{z}|$
$=\sqrt{\frac{3}{4}+\frac{1}{4}}$
$=1$
Let $\tan\alpha=\bigg|\frac{\text{Im}\text{(z)}}{\text{Re}\text{(z)}}\bigg|$
Then, $\tan\alpha=\Bigg|\frac{\frac{1}{2}}{\frac{\sqrt{3}}{2}}\Bigg|$
$=\frac{1}{\sqrt{3}}$
$\Rightarrow \alpha =\frac{\pi}{6}$
Since point $\Big(\frac{\sqrt{3}}{2},\frac{1}{2}\Big)$ lies in the fourth quadrant, the argurment s given by
$\theta=\alpha =\frac{\pi}{6}$
Polar form $=\text{r}(\cos \theta + \text{i } \sin\theta )$
$=\cos\frac{\pi}{6}+\text{i }\sin\frac{\pi}{6}$
View full question & answer→Question 303 Marks
Find the square root of the following complex numbers:
$-11-60\sqrt{-1}$
AnswerLet $\text{z}=11-60\sqrt{-1}$
$\Rightarrow\text{z}=-11-60\text{i} \ (\therefore \ \sqrt{-1}=\text{i})$
Then, $|\text{z}|=\sqrt{(-11)^2+(-60)^2}$
$=\sqrt{121+3600}$
$=\sqrt{3721}$
$=61$
$\therefore\sqrt{-11-60\text{i}}=\pm\Bigg\{\sqrt{\frac{61-11}{2}}-\text{i}\sqrt{\frac{61+11}{2}}\Bigg\} \ (\because\text{y}<0)$
$=\pm\Bigg\{\sqrt{\frac{50}{2}}-\text{i}\sqrt{\frac{72}{2}}\Bigg\}$
$=\pm\{\sqrt{5}-6\text{i}\}$
View full question & answer→Question 313 Marks
Find the values of the following expressions:
$\frac{\text{i}^{592}+\text{i}^{590}+\text{i}^{588}+\text{i}^{586}+\text{i}^{584}}{\text{i}^{582}+\text{i}^{580}+\text{i}^{578}+\text{i}^{576}+\text{i}^{574}}$
Answer$\frac{\text{i}^{592}+\text{i}^{590}+\text{i}^{588}+\text{i}^{586}+\text{i}^{584}}{\text{i}^{582}+\text{i}^{580}+\text{i}^{578}+\text{i}^{576}+\text{i}^{574}}$
$=\frac{\text{i}^{4\times48}+\text{i}^{147}\times\text{i}^{2}+\text{i}^{4\times147}+\text{i}^{4\times146}\times\text{i}^2+\text{i}^{4\times146}}{\text{i}^{4\times145}+\text{i}^{2}\times\text{i}^{4\times145}+\text{i}^{4\times144}+\text{i}^{4\times144} +\text{i}^{4\times143}\times\text{i}^2}$
$=\frac{1+1\times\text{i}^2+1+1\times\text{i}^2+1}{1\times\text{i}^2+1+1\times\text{i}^2+1+1\times\text{i}^2}$
$=\frac{1-1+1-1+1}{-1+1-1+1-1}$
$=\frac{1}{-1}$
$=-1$
$\therefore \ \frac{\text{i}^{592}+\text{i}^{590}+\text{i}^{588}+\text{i}^{586}+\text{i}^{584}}{\text{i}^{582}+\text{i}^{580}+\text{i}^{578}+\text{i}^{576}+\text{i}^{574}}=-1$
View full question & answer→Question 323 Marks
Find the values of the following expressions:
$\text{i}^{49}+\text{i}^{68}+\text{i}^{89}+\text{i}^{110}$
Answer$\text{i}^{49}+\text{i}^{68}+\text{i}^{89}+\text{i}^{110}=\text{i}^{4\times12}\times\text{i}^1+\text{i}^{4\times17}+\text{i}^{4\times22}\times\text{i}^1+\text{i}^{4\times7}\times\text{i}^2$ $=1\times\text{i}+1+1\times\text{i}+1\times\text{i}^2$ $=\text{i}+1+\text{i}-1$ $=2\text{i}$$\therefore\text{i}^{49}+\text{i}^{68}+\text{i}^{89}+\text{i}^{110}=2\text{i}$
View full question & answer→Question 333 Marks
Find the real values of $\theta$ for which the complex number $\frac{1+\text{i}\cos\theta}{1-2\text{i}\cos\theta}$ is purely real.
AnswerLet $\text{z}=\frac{1+\text{i}\cos\theta}{1-2\text{i}\cos\theta}$
$=\frac{1+\text{i}\cos\theta}{1-2\text{i}\cos\theta}\times\frac{1+2\text{i}\cos\theta}{1+2\text{i}\cos\theta}$
$=\frac{1+2\text{i}\cos\theta+\text{i}\cos\theta(1+2\text{i}\cos\theta)}{1^2+(2\cos\theta)^2}$
$=\frac{1+2\text{i}\cos\theta+\text{i}\cos\theta-2\cos^2\theta)}{1+4\cos^2\theta}$
$=\frac{1-2\cos^2\theta+3\text{i}\cos\theta}{1+4\cos^2\theta}$
$=\frac{1-2\cos^2\theta}{1+4\cos^2\theta}+\frac{3\cos\theta}{1+4\cos^2\theta}\text{i}$
We know that z is purely real if and on;y if Im z=0
$\therefore \ \frac{3\cos\theta}{1+4\cos^2\theta}=0$ $\big(\because$ z is given to be purely real $\big)$
$\Rightarrow3\cos\theta=0$
$\Rightarrow\cos\theta=0$
$\Rightarrow\cos\theta=\cos\frac{\pi}{2}$
$\therefore$ The general solution is given by
$\theta=2\text{n}\pi\pm\frac{\pi}{2},\text{n}\in\text{z}$
View full question & answer→Question 343 Marks
$\text{x}^6+\text{x}^4+\text{x}^2+1,$ when $\text{x}=\frac{1+\text{i}}{\sqrt{2}}.$
AnswerWe have
$\text{x}=\frac{1+\text{i}}{\sqrt{2}}$
$\Rightarrow\sqrt{2}\text{x}=1+\text{i}$
$\Rightarrow\sqrt({2}\text{x})^2=(1+\text{i})^2$ (squaring both sides)
$\Rightarrow2\text{x}^2=1^2+(\text{i})^2+2\times1\times\text{i}$
$=1-1+2\text{i}$
$\Rightarrow2\text{x}^2=2\text{i}$
$\Rightarrow\text{x}^2=\text{i}$
$\Rightarrow(\text{x}^2)^2=(\text{i})^2$ (squaring both sides)
$\Rightarrow\text{x}^4=-1$
$\Rightarrow\text{x}^4+1=0 \ ...(1)$
Now
$\text{x}^6+\text{x}^4+\text{x}^2+1$
$=\text{x}^6+\text{x}^4+\text{x}^2+1$
$=\text{x}^2(\text{x}^4+1)+1(\text{x}^4+1)$
$=\text{x}^2\times0+1\times0$ (using(i))
$=0$
View full question & answer→