Find the principal argument of
Question: Find the principal argument of $(-2 i)$ Solution: Let, z = -2i Let $0=r \cos \theta$ and $-2=r \sin \theta$ By squaring and adding, we get $(0)^{2}+(-2)^{2}=(r \cos \theta)^{2}+(r \sin \theta)^{2}$ $\Rightarrow 0+4=r^{2}\left(\cos ^{2} \theta+\sin ^{2} \theta\right)$ $\Rightarrow 4=r^{2}$ $\Rightarrow r=2$ $\therefore \cos \theta=0$ and $\sin \theta=-1$ Since, lies in fourth quadrant, we have $\theta=-\frac{\pi}{2}$ Since, $\theta \in(-\pi, \pi]$ it is principal argument....
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Question: If $|z|=6$ and $\arg (z)=\frac{3 \pi}{4}$, find $z$ Solution: We have, $|z|=6$ and $\arg (z)=\frac{3 \pi}{4}$ Let $z=r(\cos \theta+i \sin \theta)$ We know that, $|z|=r=6$ And $\arg (z)=\theta=\frac{3 \pi}{4}$ Thus, $z=r(\cos \theta+i \sin \theta)=6\left(\cos \frac{3 \pi}{4}+i \sin \frac{3 \pi}{4}\right)$...
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Question: Prove that $\arg (z)+\arg ^{(\bar{z})}=0$ Solution: Let $z=r(\cos \theta+i \sin \theta)$ $\Rightarrow \arg (z)=\theta$ Now, $\bar{z}=r(\cos \theta-i \sin \theta)=r(\cos (-\theta)+i \sin (-\theta))$ $\Rightarrow \arg (\bar{z})=-\theta$ Thus, $\arg (z)+\arg (\bar{z})=\theta-\theta=0$ Hence proved....
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Question: If $z=(\sqrt{5}+3 i)$, find $z^{-1}$. Solution: We have, $z=(\sqrt{5}+3 i)$ $\Rightarrow \overline{\mathrm{z}}=(\sqrt{5}-3 \mathrm{i})$ $\Rightarrow|\mathrm{z}|^{2}=(\sqrt{5})^{2}+(3)^{2}$ $=5+9=14$ $\therefore$ The multiplicative inverse of $(\sqrt{5}+3 \mathrm{i})$, $\mathrm{z}^{-1}=\frac{\overline{\mathrm{z}}}{|\mathrm{z}|^{2}}=\frac{\sqrt{5}-3 \mathrm{i}}{14}$ $\mathrm{z}^{-1}=\frac{\sqrt{5}}{14}+\frac{3}{14} \mathrm{i}$...
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Question: If $z=(1-i)$, find $z^{-1}$ Solution: We have, $z=(1-i)$ $\Rightarrow \overline{\mathrm{z}}=1+\mathrm{i}$ $\Rightarrow|\mathrm{z}|^{2}=(1)^{2}+(-1)^{2}=2$ $\therefore$ The multiplicative inverse of $(1-\mathrm{i})$, $z^{-1}=\frac{\bar{z}}{|z|^{2}}=\frac{1+i}{2}$ $z^{-1}=\frac{1}{2}+\frac{1}{2} i$...
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Question: Find the conjugate of $\frac{1}{(3+4 \mathrm{i})}$. Solution: Let $\mathrm{z}=\frac{1}{3+4 \mathrm{i}}$ $=\frac{1}{3+4 i} \times \frac{3-4 i}{3-4 i}=\frac{3-4 i}{9+16}$ $=\frac{3}{25}-\frac{4}{25} i$ $\Rightarrow \bar{z}=\frac{3}{25}+\frac{4}{25} i$...
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Question: Solve for $x: x^{2}-5 i x-6=0$ Solution: We have, $x^{2}-5 i x-6=0$ Here, $b^{2}-4 a c=(-5 i)^{2}-4 \times 1 \times-6$ $=25 i^{2}+24=-25+24=-1$ Therefore, the solutions are given by $\mathrm{x}=\frac{-(-5 \mathrm{i}) \pm \sqrt{-1}}{2 \times 1}$ $x=\frac{5 i \pm i}{2 \times 1}$ $x=\frac{5 i \pm i}{2}$ Hence, x= 3i and x = 2i...
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Question: Solve for $x:(1-i) x+(1+i) y=1-3 i$ Solution: We have, $(1-\mathrm{i}) \mathrm{x}+(1+\mathrm{i}) \mathrm{y}=1-3 \mathrm{i}$ $\Rightarrow x-i x+y+i y=1-3 i$ $\Rightarrow(x+y)+i(-x+y)=1-3 i$ On equating the real and imaginary coefficients we get, $\Rightarrow x+y=1$ (i) and $-x+y=-3$ (ii) From (i) we get $x=1-y$ Substituting the value of x in (ii), we get $-(1-y)+y=-3$ $\Rightarrow-1+y+y=-3$ $\Rightarrow 2 y=-3+1$ $\Rightarrow y=-1$ $\Rightarrow x=1-y=1-(-1)=2$ Hence, $x=2$ and $y=-1$...
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Question: Express $\frac{3-\sqrt{-16}}{1-\sqrt{-9}}$ in the form $(a+i b)$. Solution: We have, $\frac{3-\sqrt{-16}}{1-\sqrt{-9}}$ We know that $\sqrt{-1}=\mathrm{i}$ Therefore, $\frac{3-\sqrt{-16}}{1-\sqrt{-9}}=\frac{3-4 i}{1-3 i}$ $\frac{3-\sqrt{-16}}{1-\sqrt{-9}}=\frac{3-4 i}{1-3 i} \times \frac{1+3 i}{1+3 i}$ $\frac{3-\sqrt{-16}}{1-\sqrt{-9}}=\frac{3+9 i-4 i-12 i^{2}}{(1)^{2}-(3 i)^{2}}$ $\frac{3-\sqrt{-16}}{1-\sqrt{-9}}=\frac{15+5 i}{1+9}=\frac{15}{10}+\frac{5 i}{10}=\frac{3}{2}+\frac{1}{2} ...
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Question: Express $\frac{(3+i \sqrt{5})(3-\sqrt{5})}{(\sqrt{3}+\sqrt{2} i)-(\sqrt{3}-\sqrt{2} i)}$ in the form $(a+i b)$ Solution: We have, $\frac{(3+i \sqrt{5})(3-i \sqrt{5})}{(\sqrt{3}+\sqrt{2 i})-(\sqrt{3}-\sqrt{2 i})}$ $=\frac{(3)^{2}-(i \sqrt{5})^{2}}{\sqrt{3}+\sqrt{2 i}-\sqrt{3}+\sqrt{2 i}}\left[\because(a+b)(a-b)=a^{2}-b^{2}\right]$ $=\frac{9+5}{2 \sqrt{2 i}} \times \frac{\sqrt{2 i}}{\sqrt{2 i}}$ $=\frac{14 \sqrt{2 i}}{2(\sqrt{2 i})^{2}}$ $=\frac{7 \sqrt{2 i}}{-2}$ $=\frac{-7 \sqrt{2 i}}{...
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Question: Express $(2-3 i)^{3}$ in the form $(a+i b)$ Solution: We have, $(2-3 i)^{3}$ $=2^{3}-3 \times 2^{2} \times 3 i-3 \times 2 \times(3 i)^{2}-(3 i)^{3}$ $=8-36 i+54+27 i$ $=46-9 i .$...
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Question: Find the least positive integer n for which $\left(\frac{1+i}{1-i}\right)^{n}=1$ Solution: We have, $\left(\frac{1+i}{1-i}\right)^{n}=1$ Now, $\frac{1+\mathrm{i}}{1-\mathrm{i}}=\frac{1+\mathrm{i}}{1-\mathrm{i}} \times \frac{1+\mathrm{i}}{1+\mathrm{i}}$ $=\frac{(1+i)^{2}}{1^{2}-i^{2}}$ $=\frac{1^{2}+2 i+i^{2}}{1-(-1)}$ $=\frac{1+2 i-1}{2}$ $=i$ $\therefore\left(\frac{1+\mathrm{i}}{1-\mathrm{i}}\right)^{\mathrm{n}}=(\mathrm{i})^{\mathrm{n}}=1 \Rightarrow \mathrm{n}$ is multiple of 4 $\th...
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Question: Evaluate: $\left(i^{41}+\frac{1}{i^{71}}\right)$ Solution: We have $\left(i^{41}+\frac{1}{i^{71}}\right)$ $\mathrm{i}^{41}=\mathrm{i}^{40} \cdot \mathrm{i}=\mathrm{i}$ $\mathrm{i}^{71}=\mathrm{i}^{68} \cdot \mathrm{i}^{3}=-\mathrm{i}$ Therefore $\left(i^{41}+\frac{1}{i^{71}}\right)=i-\frac{1}{i}=\frac{i^{2}-1}{i}$ $\left(i^{41}+\frac{1}{i^{71}}\right)=-\frac{2}{i} \times \frac{i}{i}$ $\left(i^{41}+\frac{1}{i^{71}}\right)=-\frac{2 i}{i^{2}}=2 i$ Hence, $\left(i^{41}+\frac{1}{i^{71}}\rig...
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Question: Evaluate $\left(1+i^{10}+i^{20}+i^{30}\right)$ Solution: We have, $1+\mathrm{i}^{10}+\mathrm{i}^{20}+\mathrm{i}^{30}$ $=1+\left(i^{4}\right)^{2} \cdot i^{2}+\left(i^{4}\right)^{5}+\left(i^{4}\right)^{7} \cdot i^{2}$ We know that, $i^{4}=1$ $\Rightarrow 1+(1)^{2} \cdot i^{2}+(1)^{5}+(1)^{7} \cdot i^{2}$ $=1+i^{2}+1+i^{2}$ $=1-1+1-1$ $=0$...
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Question: Find the sum ( $\mathrm{i}+\mathrm{i}^{2}+\mathrm{i}^{3}+\mathrm{i}^{4}+\ldots .$ up to 400 terms)., where $\mathrm{n} \mathrm{N}$. Solution: We have, $\mathrm{i}+\mathrm{i}^{2}+\mathrm{i}^{3}+\mathrm{i}^{4}+\ldots$ up to 400 terms We know that given series is GP where a=i , r = i and n = 400 Thus, $S=\frac{a\left(1-r^{n}\right)}{1-r}$ $=\frac{\mathrm{i}\left(1-(\mathrm{i})^{400}\right)}{1-\mathrm{i}}$ $=\frac{\mathrm{i}\left(1-\left(\mathrm{i}^{4}\right)^{100}\right)}{1-\mathrm{i}}$ $...
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Question: Find the sum $\left(\mathrm{i}^{\mathrm{n}}+\mathrm{i}^{\mathrm{n}+1}+\mathrm{i}^{\mathrm{n}+2}+\mathrm{i}^{\mathrm{n}+3}\right)$, where $\mathrm{n} \mathrm{N}$. Solution: We have $\mathrm{i}^{\mathrm{n}}+\mathrm{i}^{\mathrm{n}+1}+\mathrm{i}^{\mathrm{n}+2}+\mathrm{i}^{\mathrm{n}+3}$ $=i^{n}+i^{n} \cdot i+i^{n} \cdot i^{2}+i^{n} \cdot i^{3}$ $=i^{n}\left(1+i+i^{2}+i^{3}\right)$ $=i^{n}(1+i-1-i)$ $=i^{n}(0)=0$...
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Question: Evaluate $(\sqrt{36} \times \sqrt{-25})$ Solution: We have, $(\sqrt{36} \times \sqrt{-25})$ $=6 \times \sqrt{-1 \times 25}$ $=6 \times(\sqrt{-1} \times \sqrt{25})$ $=6 \times(\sqrt{-1} \times 5)$ $=6 \times 5 i=30 i$...
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Question: Evaluate $\left(i^{4 n+1}-i^{4 n-1}\right)$ Solution: We have, $i^{4 n+1}-i^{4 n-1}$ $=i^{4 n} \cdot i-i^{4 n} \cdot i^{-1}$ $=\left(i^{4}\right)^{n} \cdot i-\left(i^{4}\right)^{n} \cdot i^{-1}$ $=(1)^{n} \cdot i-(1)^{n} \cdot i^{-1}$ $=i-i^{-1}$ $=\mathrm{i}-\frac{1}{\mathrm{i}}$ $=\frac{\mathrm{i}^{2}-1}{\mathrm{i}}$ $=\frac{-1-1}{\mathrm{i}}$ $=\frac{-2}{\mathrm{i}} \times \frac{\mathrm{i}}{\mathrm{i}}$ $=\frac{-2 \mathrm{i}}{\mathrm{i}^{2}}=\frac{-2 \mathrm{i}}{-1}$ $=2 \mathrm{i}$...
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Question: Evaluate $\left(\frac{\mathrm{i}^{180}+\mathrm{i}^{178}+\mathrm{i}^{176}+\mathrm{i}^{174}+\mathrm{i}^{172}}{\mathrm{i}^{170}+\mathrm{i}^{168}+\mathrm{i}^{166}+\mathrm{i}^{164}+\mathrm{i}^{162}}\right)$ Solution: We have, $\left(\frac{\mathrm{i}^{180}+\mathrm{i}^{178}+\mathrm{i}^{176}+\mathrm{i}^{174}+\mathrm{i}^{172}}{\mathrm{i}^{170}+\mathrm{i}^{168}+\mathrm{i}^{166}+\mathrm{i}^{164}+\mathrm{i}^{162}}\right)$ $=\left(\frac{\mathrm{i}^{180}+\mathrm{i}^{178}+\mathrm{i}^{176}+\mathrm{i}^...
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Question: Evaluate $\left(i^{57}+i^{70}+i^{91}+i^{101}+i^{104}\right)$ Solution: We have, $\mathrm{i}^{57}+\mathrm{i}^{70}+\mathrm{i}^{91}+\mathrm{i}^{101}+\mathrm{i}^{104}$ $=\left(i^{4}\right)^{14} \cdot i+\left(i^{4}\right)^{17} \cdot i^{2}+\left(i^{4}\right)^{22} \cdot i^{3}+\left(i^{4}\right)^{25} \cdot i+\left(i^{4}\right)^{26}$ We know that, $i^{4}=1$ $\Rightarrow(1)^{14} \cdot i+(1)^{17} \cdot i^{2}+(1)^{22} \cdot i^{3}+(1)^{25} \cdot i+(1)^{26}$ $=i+i^{2}+i^{3}+i+1$ $=i-1-i+i+1$ $=i$...
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Question: Evaluate$\frac{1}{\mathrm{i}^{78}}$ Solution: we have $\frac{1}{\mathrm{i}^{78}}$ $=\frac{1}{\left(i^{4}\right)^{19} \cdot i^{2}}$ We know that, $i^{4}=1$ $\Rightarrow \frac{1}{1^{19} \cdot \mathrm{i}^{2}}$ $\Rightarrow \frac{1}{\mathrm{i}^{2}}=\frac{1}{-1}$ $\Rightarrow \frac{1}{\mathrm{i}^{78}}=-1$...
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Question: $\sqrt{1-i}$ Solution: Let, $(a+i b)^{2}=1-i$ Now using, $(a+b)^{2}=a^{2}+b^{2}+2 a b$ $\Rightarrow a^{2}+(b i)^{2}+2 a b i=1-i$ Since $i^{2}=-1$ $\Rightarrow a^{2}-b^{2}+2 a b i=1-i$ Now, separating real and complex parts, we get $\Rightarrow a^{2}-b^{2}=1 \ldots \ldots \ldots \ldots \ldots$ eq. 1 $\Rightarrow 2 \mathrm{ab}=-1 \ldots \ldots . .$ eq. 2 $\Rightarrow \mathrm{a}=-\frac{1}{2 b}$ Now, using the value of a in eq.1, we get $\Rightarrow\left(-\frac{1}{2 b}\right)^{2}-b^{2}=1$ ...
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Question: $\sqrt{-8i}$ Solution: Let, $(a+i b)^{2}=0-8 i$ Now using, $(a+b)^{2}=a^{2}+b^{2}+2 a b$ $\Rightarrow a^{2}+(b i)^{2}+2 a b i=0-8 i$ Since $i^{2}=-1$ $\Rightarrow a^{2}-b^{2}+2 a b i=0-8 i$ Now, separating real and complex parts, we get $\Rightarrow a^{2}-b^{2}=0$ ..eq.1 $\Rightarrow 2 \mathrm{ab}=-8 \ldots \ldots \ldots$ eq. 2 $\Rightarrow \mathrm{a}=-\frac{4}{b}$ Now, using the value of a in eq.1, we get $\Rightarrow\left(-\frac{4}{b}\right)^{2}-b^{2}=0$ $\Rightarrow 16-b^{4}=0$ $\Ri...
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Question: $\sqrt{7-30 \sqrt{-2}}$ Solution: Let, $(a+i b)^{2}=7-30^{\sqrt{2}}_{i}$ Now using, $(a+b)^{2}=a^{2}+b^{2}+2 a b$ $\Rightarrow a^{2}+(b i)^{2}+2 a b i=7-30^{\sqrt{2}}_{i}$ Since $i^{2}=-1$ $\Rightarrow a^{2}-b^{2}+2 a b i=7-30^{\sqrt{2}} i$ Now, separating real and complex parts, we get $\Rightarrow a^{2}-b^{2}=7$ ..eq.1 $\Rightarrow 2 \mathrm{ab}=30^{\sqrt{2}} \ldots \ldots . . \mathrm{eq} .2$ $\Rightarrow a=\frac{15 \sqrt{2}}{b}$ Now, using the value of a in eq.1, we get $\Rightarrow...
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Question: $\sqrt[3]{8+27}=\sqrt[3]{8}+\sqrt[3]{27}$ Solution: False We have, $\sqrt[3]{8+27} \neq \sqrt[3]{8}+\sqrt[3]{27}$ $\because$ $\sqrt[3]{8}+\sqrt[3]{27}=\sqrt[3]{2 \times 2 \times 2}+\sqrt[3]{3 \times 3 \times 3}=2+3=5$ But $\sqrt[3]{8+27}=\sqrt[3]{35} \neq \sqrt[3]{125}=5$...
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