COMPLEX NUMBER - 4 (Practice Problems)
Examples
1. If $\alpha, \beta$ are two different complex numbers such that $|\alpha|=1$ or $|\beta|=1$ then the expression $\left|\frac{\beta-\alpha}{1-\bar{\alpha} \beta}\right|$ equals
(a). $\frac{1}{2}$
(b). $1$
(c). $2$
(d). None of these
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Solution:
$ \begin{aligned} \left|\frac{\beta-\alpha}{1-\bar{\alpha} \beta}\right| & =\left|\frac{\beta-\alpha}{\alpha \bar{\alpha}-\bar{\alpha} \beta}\right| \quad \quad(\alpha \bar{\alpha}=\mid \alpha \\ & =\left|\frac{\beta-\alpha}{\bar{\alpha}(\alpha-\beta)}\right|=\frac{|\beta-\alpha|}{|\bar{\alpha}(\alpha-\beta)|} \\ & =\frac{|(-1)(\alpha-\beta)|}{|\bar{\alpha}||\alpha-\beta|}=\frac{|-1||\alpha-\beta|}{|\alpha||\alpha-\beta|}=\frac{1}{|\alpha|}=1 \end{aligned} $
2. If $\mathrm{z}=\mathrm{x}+\mathrm{iy}$ such that $|\mathrm{z}+1|=|\mathrm{z}-1|$ and $\operatorname{amp}\left(\frac{\mathrm{z}-1}{\mathrm{z}+1}\right)=\pi / 4$, then
(a). $x=\sqrt{2}+1, y=0$
(b). $x=0, y=\sqrt{2}+1$
(c). $x=0, y=\sqrt{2}-1$
(d). $x=\sqrt{2}-1, y=0$
Show Answer
Solution:
let $z=x+i y$
$ \begin{aligned} & \because|\mathrm{z}+1|=|\mathrm{z}-1| \Rightarrow|(\mathrm{x}+1)+\mathrm{iy}|=|(\mathrm{x}-1)+\mathrm{iy}| \\ & \quad \Rightarrow \sqrt{(\mathrm{x}+1)^{2}+\mathrm{y}^{2}}=\sqrt{(\mathrm{x}-1)^{2}+\mathrm{y}^{2}} \\ & \quad \Rightarrow(\mathrm{x}+1)^{2}+\mathrm{y}^{2}=(\mathrm{x}-1)^{2}+\mathrm{y}^{2} \Rightarrow \mathrm{x}^{2}+2 \mathrm{x}+1=\mathrm{x}^{2}-2 \mathrm{x}+1 \\ & \quad \Rightarrow 4 \mathrm{x}=0 \Rightarrow \mathrm{x}=0 \end{aligned} $
$\begin{aligned} & \operatorname{amp}\left(\frac{\mathrm{z}-1}{\mathrm{z}+1}\right)=\pi / 4, \Rightarrow \operatorname{amp}\left(\frac{\mathrm{iy}-1}{\mathrm{iy}+1}\right)=\pi / 4………….(1) \\ & \because \frac{i y-1}{i y+1} \times \frac{i y-1}{i y-1}=\frac{1-y^2-2 i y}{-\left(y^2+1\right)}=\frac{-\left(1-y^2\right)}{\left(y^2+1\right)}+\frac{2 i y}{1+y^2} \\ & \frac{2 y}{y^2+1} \times \frac{-\left(y^2+1\right)}{\left(1-y^2\right)}=\tan \pi / 4 \Rightarrow \frac{-2 y}{1-y^2}=1 \\ & \Rightarrow 1-\mathrm{y}^2=-2 \mathrm{y} \\ & \Rightarrow \mathrm{y}^2-2 \mathrm{y}-1=0 \\ & y=\frac{2 \pm \sqrt{4+4}}{2}=\frac{2 \pm 2 \sqrt{2}}{2}=1 \pm \sqrt{2} \\ & \end{aligned}$
$1-\sqrt{2}$ can be neglected as it is negative but $\tan \theta$ lies in Ist quadrant.
3. For any complex number $z$, the maximum value of $|z|-|z-1|$ is
(a). $\frac{1}{2}$
(b). $1$
(c). $\frac{3}{2}$
(d). $2$
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Solution:
$\because\left|z _{1}-z _{2}\right| \geq|| z _{1}|-| z _{2}|| \geq\left|z _{1}\right|-\left|z _{2}\right|$
$|z|-|z-1| \leq|z-(z-1)|=1$
$\therefore \quad$ Maximum value of $|z|-|z-1|=1$
4. If $z _{1} \& z _{2}$ are two complex numbers such that $\frac{z _{1}-2 z _{2}}{2-z _{1} \cdot \bar{z} _{2}}$ is uni moduler, while $z _{2}$ is not uni moduler, then the value of $\left|z _{1}\right|$ is
(a). $1$
(b). $2$
(c). $3$
(d). $4$
Show Answer
Solution:
$\left|\frac{z _{1}-2 z _{2}}{2-z _{1} \bar{z} _{2}}\right|=1 \Rightarrow \frac{\left|z _{1}-2 z _{2}\right|}{\left|2-z _{1} \bar{z} _{2}\right|}=1$
$\Rightarrow\left|z _{1}-2 z _{2}\right|^{2}=\left|2-z _{1} \bar{z} _{2}\right|^{2}$
$\Rightarrow\left(\mathrm{z} _{1}-2 \mathrm{z} _{2}\right) \overline{\left(\mathrm{z} _{1}-2 \mathrm{z} _{2}\right)}=\left(2-\mathrm{z} _{1} \overline{\mathrm{z}} _{2}\right)\left(\overline{\left(2-\mathrm{z} _{1} \overline{\mathrm{z}} _{2}\right)}\right.$
$\Rightarrow\left(\mathrm{z} _{1}-2 \mathrm{z} _{2}\right)\left(\overline{\mathrm{z}} _{1}-2 \overline{\mathrm{z}} _{2}\right)=\left(2-\mathrm{z} _{1} \overline{\mathrm{z}} _{2}\right)\left(2-\overline{\mathrm{z}} _{1} \mathrm{z} _{2}\right)$
$\Rightarrow \mathrm{z} _{1} \overline{\mathrm{z}} _{1}-2 \mathrm{z} _{1} \overline{\mathrm{z}} _{2}-2 \mathrm{z} _{2} \overline{\mathrm{z}} _{1}+4 \mathrm{z} _{2} \overline{\mathrm{z}} _{2}$
$=4-2 \mathrm{z} _{1} \overline{\mathrm{z}} _{2}-\overline{\mathrm{z}} _{1} \mathrm{z} _{2}+\mathrm{z} _{1} \overline{\mathrm{z}} _{1} \overline{\mathrm{z}} _{2} \overline{\mathrm{z}} _{2}$
$\Rightarrow\left|z _{1}\right|^{2}+4\left|z _{2}\right|^{2}=4+\left|z _{1}\right|^{2}\left|z _{2}\right|^{2}$
$\Rightarrow\left|z _{1}\right|^{2}\left(1-\left|z _{2}\right|^{2}\right)-4\left(1-\left|z _{2}\right|^{2}\right)=0$
$\Rightarrow\left(\left|z _{1}\right|^{2}-4\right)\left(1-\left|z _{2}\right|^{2}\right)=0$
$ \left|z _{2}\right|=1,\left|z _{1}\right|=2 \quad \text { since }\left|z _{2}\right| \neq 1 _{1}\left(\text { not unimodular) so, }\left|z _{1}\right|=2\right. $
5. If $|z-3+2 i| \leq 4$, then the sum of least and greatest values of $|z|$ is
(a) $2 \sqrt{11}$
(b) $3 \sqrt{11}$
(c) $2 \sqrt{13}$
(d) $3 \sqrt{13}$
Show Answer
Solution:
$|\mathrm{z}-3+2 \mathrm{i}| \leq 4$
$\Rightarrow|z-3+2 \mathrm{i}| \geq|| \mathrm{z}|-| 3-2 \mathrm{i}||$
$\Rightarrow|\mathrm{z}-3+2 \mathrm{i}| \geq|| \mathrm{z}|-\sqrt{13}|$
From (1) & (2)
$\Rightarrow 4 \geq|\mathrm{z}-3+2 \mathrm{i}| \geq|| \mathrm{z}|-\sqrt{13}|$
$\Rightarrow|| \mathrm{z}|-\sqrt{13}| \leq 4$
$\Rightarrow-4 \leq|z|-\sqrt{13} \leq 4$
$\Rightarrow \sqrt{13}-4 \leq|z| \leq 4+\sqrt{13}$
$\therefore$ greatest value of $|z|$ is $\sqrt{13}+4$
& least value of $|z|$ is $\sqrt{13}-4$
$\therefore$ their sum $=\sqrt{13}+4+\sqrt{13}-4$
$=2 \sqrt{13}$
6. If $\mathrm{z} _{1}, \mathrm{z} _{2}, \mathrm{z} _{3}$ are three distinct complex numbers and $\mathrm{a}, \mathrm{b}, \mathrm{c}$ are three positive real numbers such that
$\frac{a}{\left|z _{2}-z _{3}\right|}=\frac{b}{\left|z _{3}-z _{1}\right|}=\frac{c}{\left|z _{1}-z _{2}\right|}$, then $\frac{a^{2}}{z _{2}-z _{3}}+\frac{b^{2}}{z _{3}-z _{1}}+\frac{c^{2}}{z _{1}-z _{2}}$ is
(a). $\frac{\mathrm{a}^{2}\left(\mathrm{z} _{2}{ }^{2}+\mathrm{z} _{3}{ }^{2}\right)+\mathrm{b}^{2}\left(\mathrm{z} _{1}{ }^{2}+\mathrm{z} _{3}{ }^{2}\right)+\mathrm{c}\left(\mathrm{z} _{1}{ }^{2}+\mathrm{z} _{2}{ }^{2}\right)}{\mathrm{z} _{1}{ }^{2}+\mathrm{z} _{2}{ }^{2}+\mathrm{z} _{3}{ }^{2}}$
(b). $0$
(c) $\frac{\mathrm{a}^{2}\left(\mathrm{~b}^{2}-\mathrm{c}^{2}\right) \mathrm{z} _{1}{ }^{2}+\mathrm{b}^{2}\left(\mathrm{c}^{2}-\mathrm{a}^{2}\right) \mathrm{z} _{2}{ }^{2}+\mathrm{c}^{2}\left(\mathrm{a}^{2}-\mathrm{b}^{2}\right) \mathrm{z} _{3}{ }^{2}}{\mathrm{z} _{1} \mathrm{z} _{2}+\mathrm{z} _{2} \mathrm{z} _{3}+\mathrm{z} _{3} \mathrm{z} _{1}}$
(d). None of these
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Solution:
Let $\frac{\mathrm{a}}{\left|\mathrm{z} _{2}-\mathrm{z} _{3}\right|}=\frac{\mathrm{b}}{\left|\mathrm{z} _{3}-\mathrm{z} _{1}\right|}=\frac{\mathrm{c}}{\left|\mathrm{z} _{1}-\mathrm{z} _{2}\right|}=\mathrm{k}$ (ray)
$\Rightarrow \quad \mathrm{a}^{2}=\mathrm{k}^{2}\left|\mathrm{z} _{2}-\mathrm{z} _{3}\right|^{2}$ etc
$\Rightarrow \frac{\mathrm{a}^{2}}{\left(\mathrm{z} _{2}-\mathrm{z} _{3}\right)}=\mathrm{k}^{2}\left(\overline{\mathrm{z}} _{2}-\overline{\mathrm{z}} _{3}\right)$
$ \frac{\mathrm{b}^{2}}{\mathrm{z} _{3}-\mathrm{z} _{1}}=\mathrm{k}^{2}\left(\overline{\mathrm{z}} _{3}-\overline{\mathrm{z}} _{1}\right) $
$ \frac{\mathrm{c}^{2}}{\mathrm{z} _{1}-\mathrm{z} _{2}}=\mathrm{k}^{2}\left(\overline{\mathrm{z}} _{1}-\overline{\mathrm{z}} _{2}\right) $
$\left.\left.\therefore \frac{\mathrm{a}^{2}}{\mathrm{z} _{2}-\mathrm{z} _{3}}+\frac{\mathrm{b}^{2}}{\mathrm{z} _{3}-\mathrm{z} _{1}}+\frac{\mathrm{c}^{2}}{\mathrm{z} _{1}-\mathrm{z} _{2}}=\mathrm{k}^{2} \right\rvert, \overline{\mathrm{z}} _{2}-\overline{\mathrm{z}} _{3}+\overline{\mathrm{z}} _{3}-\overline{\mathrm{z}} _{1}+\overline{\mathrm{z}} _{1}-\overline{\mathrm{z}} _{2}\right)=0$
Cube root of unity
$ \begin{aligned} & \mathrm{z}^{3}=1 \\ & \mathrm{z}^{3}-1=0 \\ & (\mathrm{z}-1)\left(\mathrm{z}^{2}+\mathrm{z}+1\right)=0 \\ & \mathrm{z}=1, \mathrm{z}^{2}+\mathrm{z}+1=0 \\ & \mathrm{z}=\frac{-1 \pm \sqrt{3} \mathrm{i}}{2} \\ & \mathrm{z}=1, \mathrm{z}=\frac{-1+\sqrt{3} \mathrm{i}}{\frac{2}{\downarrow}}, \mathrm{z}=\frac{-1-\sqrt{3} \mathrm{i}}{\frac{2}{\downarrow}} \end{aligned} $
$1+\omega+\omega^{2}=0$
$\omega^{3}=1$
$\mathrm{w}^{19}=\mathrm{w}^{3 \times 6+1}=(\mathrm{w} 3)^{2} \cdot \omega=\omega$
De Mover’s Theorem
$(\cos \theta+i \sin \theta)^{\mathrm{n}}=\operatorname{cosn} \theta+\mathrm{isinn} \theta$.
$n^{\text {th }}$ roots of unity
$ \begin{aligned} & 1=\cos 0+i \sin 0 \\ & 1^{1 / n}=(\cos 0+i \sin 0)^{1 / n} \\ & =\left[\cos \frac{2 r \pi+0}{n}+i \sin \frac{2 r \pi+0}{n}\right] \\ & =\cos \frac{2 r \pi}{n}+i \sin \frac{2 r \pi}{n} \end{aligned} $
$\mathrm{r}=0,1 / 2$ (n-1)
$=e^{\frac{i 2 r \pi}{n}}$ where $\mathrm{r}=0,1 / 2$ ..(n-1)
$=1, e^{\frac{i 2 r \pi}{n}}, e^{\frac{i 4 \pi}{n}}, e^{\frac{i 6 \pi}{n} \cdots \cdots \cdots \cdots} e^{\frac{i 2(n-1) \pi}{n}}$
$=1, \alpha, \alpha^{2} \ldots \ldots \ldots \ldots \ldots . . . \alpha^{\mathrm{n}-1}$ where $\alpha=\mathrm{e}^{\frac{\mathrm{i} 2 \mathrm{r} \pi}{\mathrm{n}}}$
(1) $1+\alpha+\alpha^{2}+\ldots$ $+\alpha^{\mathrm{n}-1}=0$
(2) $1 . \alpha \cdot \alpha^{2}$ $\alpha^{\mathrm{n}-1}=(-1)^{\mathrm{n}-1}$
Important Relations
$\begin{array}{ll}1. & x^2+x+1=(x-\omega)\left(x-\omega^2\right) \\ 2. & x^2-x+1=(x+\omega)\left(x-\omega^2\right) \\ 3. & x^2+2 y+y^2=(x-y \omega)\left(x-y \omega^2\right) \\ 4. & x^2-x y+y^2=(x+y \omega)\left(x+y \omega^2\right) \\ 5. & x^2+y^2=(x+i y)(x-i y) \\ 6. & x^3+y^3=(x+y)(x+y \omega)\left(x+y \omega^2\right) \\ 7. & x^3-y^3=(x-y)(x-y \omega)\left(x-y \omega^2\right) \\ 8. & x^2+y^2+z^2-x y-y z-z x=\left(x+y \omega+z \omega^2\right)\left(x+y \omega^2+z \omega\right) \\ 9. & x^3+y^3+z^3-3 x y z=(x+y+z)\left(x+y \omega+z \omega^2\right)\left(x+y w^2+z \omega\right)\end{array}$
Exercises
1. If $w$ is a cube root of unity, then $\omega+\omega^{\left(\frac{1}{2}+\frac{3}{8}+\frac{9}{32}+\frac{27}{128}+\ldots \ldots \ldots a\right)}$ is
(a). $+1$
(b). $\mathrm{i}^{2}$
(c). $0$
(d). None of these
Show Answer
Solution:
$\frac{1}{2}+\frac{3}{8}+\frac{9}{32}+\frac{27}{128}+\ldots \ldots \ldots . . . \alpha=\frac{1}{2}\left[1+\frac{3}{4}+\frac{9}{16}+\frac{27}{64}+\ldots \ldots \ldots . \alpha\right]=\frac{1}{2} \quad \frac{1}{1-\frac{3}{4}}=\frac{1}{2} \times \frac{4}{1}=2$
$\therefore \mathrm{w}+\mathrm{w}^{2}=-1=\mathrm{i}^{2}$
2. $\alpha, \beta$ real $\gamma$ are the roots of $x^{3}-3 x^{2}+3 x+7=0$
( $w$ is the cube root of unity), then $\left(\frac{\alpha-1}{\beta-1}+\frac{\beta-1}{\gamma-1}+\frac{\gamma-1}{\alpha-1}\right)$ is
(a). $3 / \omega$
(b). $\omega^{2}$
(c). $2 \omega^{2}$
(d). $3 \omega^{2}$
Show Answer
Solution:
we have
$ \begin{aligned} & x^{3}-3 x^{2}+3 x+7=0 \\ & \Rightarrow \quad(\mathrm{x}-1)^{3}+8=0 \quad \Rightarrow(\mathrm{x}-1)^{3}=(-2)^{3} \\ & \Rightarrow\left[\frac{x-1}{-2}\right]^{3}=1 \quad \Rightarrow \frac{x-1}{-2}=1, \omega, \omega^{2} \\ & \Rightarrow \quad \mathrm{x}=-2+1, \mathrm{x}-1=-2 \omega, \mathrm{x}-1,=-2 \omega^{2} \\ & x=-1, x=1-2 \omega, x=1-2 \omega^{2} \\ & \alpha=-1, \beta=1-2 \omega, \gamma=1-2 \omega^{2} \\ & \alpha-1=-2, \beta-1=-2 \omega, \gamma-1=-2 \omega^{2} \\ & \therefore \mathrm{G} \cdot \mathrm{E}=\frac{+2}{+2 \omega}+\frac{-2 \omega}{-2 \omega^{2}}+\frac{+2 \omega^{2}}{+2} \\ & =\omega^{2}+\omega^{2}+\omega^{2} \\ & =3 \omega^{2} \end{aligned} $
3. The common roots of equation
$z^{3}+2 z^{2}+2 z+1=0 \& z^{1985}+z^{100}+1=0$ are
(a). $1, \omega$
(b). $1, \omega^{2}$
(c). $\omega, \omega^{2}$
(d). None of these
Show Answer
Solution:
$ \begin{array}{ll} & z^{3}+2 z^{2}+2 z+1=0 \ldots \ldots \ldots \ldots \ldots . . . .(1) \\ \Rightarrow \quad & (z+1)\left(z^{2}+z+1\right)=0 \\ \Rightarrow \quad & z=-1, \omega, \omega^{2} \text { are the roots of }(1) \\ & z=-1, z^{1985}+z^{100}+1=1 \neq 0 \\ & z=\omega,(\omega)^{1985}+\omega^{100}+1=\omega^{2}+\omega+1=0 \\ & z=\omega^{2}, \omega^{3970}+\omega^{200}+1=0 \end{array} $
$\therefore$ The common roots are $\mathrm{w}, \mathrm{w}^{2}$
$\therefore$ (c) is correct.
4. If $\alpha, \beta, \gamma$ are the cube roots of $p, p<0$ then for arang $x, y \& z \frac{x \alpha+y \beta+z \gamma}{x \beta+y \gamma+z \alpha}$ is
(a). $\mathrm{w}^{2}$
(b). $-1$
(c) $\frac{x^{3}+y^{3}+z^{3}}{\alpha^{3}+\beta^{3}+\gamma^{3}}$
(d). None of these
Show Answer
Solution:
Let $\mathrm{x}^{3}=\mathrm{p}$
$\Rightarrow \mathrm{x}=(1 . \mathrm{p})^{1 / 3}$
$\Rightarrow \quad \mathrm{x}=\mathrm{p}^{1 / 3} \cdot 1^{1 / 3}$
$x=p^{1 / 3}, p^{1 / 3} \omega, p^{1 / 3} \omega^{2}$
$\therefore \quad$ Given expression $=\frac{x \cdot p^{1 / 3}+y p^{1 / 3} \omega+z p^{1 / 3} \omega^{2}}{x \cdot p^{1 / 3} \omega+y \cdot p^{1 / 3} \omega^{2}+z \cdot p^{1 / 3}}$
$=\frac{x+y \omega+z \omega^{2}}{x \omega+y \omega^{2}+z}=\frac{\omega^{2}\left(x+\omega y+\omega^{2} z\right)}{\left(x+\omega y+z \omega^{2}\right)} \quad$ (multiply and divide by $\left.\omega^{2}\right)$
$=\omega^{2}$
5. The value of the expression
$\left(1+\frac{1}{\omega}\right)\left(1+\frac{1}{\omega^{2}}\right)+\left(2+\frac{1}{\omega}\right)\left(2+\frac{1}{\omega^{2}}\right)+\left(3+\frac{1}{\omega}\right)\left(3+\frac{1}{\omega^{2}}\right)+\ldots \ldots \ldots+\left(n+\frac{1}{\omega}\right)\left(n+\frac{1}{\omega^{2}}\right)$
Show Answer
Solution:
$\mathrm{T} _{\mathrm{k}}=\left(\mathrm{k}+\frac{1}{\omega}\right)\left(\mathrm{k}+\frac{1}{\omega^{2}}\right)=\left(\mathrm{k}+\omega^{2}\right)(\mathrm{k}+\omega)=\mathrm{k}^{2}-\mathrm{k}+1$
$\therefore \mathrm{S} _{\mathrm{n}}=\sum _{\mathrm{k}=1}^{\mathrm{n}} \mathrm{T} _{\mathrm{k}}=\sum _{\mathrm{k}=1}^{\mathrm{n}}\left(\mathrm{k}^{2}-\mathrm{k}+1\right)$
$=\sum \mathrm{k}^{2}-\sum \mathrm{k}+\sum 1$
$=\frac{\mathrm{n}(\mathrm{n}+1)(2 \mathrm{n}+1)}{6}-\frac{\mathrm{n}(\mathrm{n}+1)}{2}+\mathrm{n}$
$=n\left[\frac{2 n^{2}+3 n+1}{6}-\frac{n+1}{2}+1\right]$
$=n\left[\frac{2 n^{2}+3 n+1-3 n-3+6}{6}\right]$
$=\mathrm{n}\left(\frac{2 \mathrm{n}^{2}+4}{6}\right)$
$=\frac{\mathrm{n}\left(\mathrm{n}^{2}+2\right)}{3}$
6. If $w$ and $w^{2}$ are the cube roots of unity and
$\frac{1}{a+\omega}+\frac{1}{b+\omega}+\frac{1}{c+\omega}=2 \omega^{2} \& \frac{1}{a+\omega^{2}}+\frac{1}{b+\omega^{2}}+\frac{1}{c+\omega^{2}}=2 \omega$ then the value of $\frac{1}{a+1}+$ $\frac{1}{\mathrm{~b}+1}+\frac{1}{\mathrm{c}+1}$ is
(a). $1$
(b). $-1$
(c). $\omega$
(d). $2$
Show Answer
Solution:
Note that $\omega \& \omega^{2}$ are roots of
$\frac{1}{\mathrm{a}+\mathrm{x}}+\frac{1}{\mathrm{~b}+\mathrm{x}}+\frac{1}{\mathrm{c}+\mathrm{x}}=\frac{2}{\mathrm{x}}$
$\frac{3 \mathrm{x}^{2}+2(\mathrm{a}+\mathrm{b}+\mathrm{c}) \mathrm{x}+\mathrm{bc}+\mathrm{ca}+\mathrm{ab}}{(\mathrm{a}+\mathrm{x})(\mathrm{b}+\mathrm{x})(\mathrm{c}+\mathrm{x})}=\frac{2}{\mathrm{x}}$
$\Rightarrow \mathrm{x}^{3}-(\mathrm{bc}+\mathrm{ca}+\mathrm{ab}) \mathrm{x}-2 \mathrm{abc}=0$
The roots of the equation are $\omega \& \omega^{2}$
If $\alpha$ in the third root of this equation then $\alpha+\omega^{+} \omega^{2}=0 \Rightarrow \alpha=1$
Put $\alpha=1$ on equation (1)
$\frac{1}{\mathrm{a}+1}+\frac{1}{\mathrm{~b}+1}+\frac{1}{\mathrm{c}+1}=2$.
Practice questions
1. The value of $\left(1-\omega+\omega^{2}\right)^{5}+\left(1+\omega-\omega^{2}\right)^{5}$, where $\omega$ and $\omega^{2}$ are the complex cube roots of unity, is
(a). $0$
(b). $32 \omega$
(c). $-32$
(d). $32$
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Answer: (d)2. For any two complex numbers $z _{1}$ and $z _{2}$ and any real numbers $a$ and $b$, $\left|\mathrm{az} _{1}-\mathrm{bz} _{2}\right|^{2}+\left|\mathrm{bz} \mathrm{z} _{1}+\mathrm{az} _{2}\right|^{2}$ is equal to
(a). $\left(a^{2}+b^{2}\right)\left(\left|z _{1}\right|+\left|z _{2}\right|\right)$
(b). $\left(a^{2}+b^{2}\right)\left(\left|z _{1}\right|^{2}+\left|z _{2}\right|^{2}\right)$
(c). $\left(a^{2}+b^{2}\right)\left(\left|z _{1}\right|^{2}-\left|z _{2}\right|^{2}\right)$
(d). none of these
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Answer: (b)3. For any two complex numbersz $z _{1}$ and $z _{2}$, the relation $\left|z _{1}+z _{2}\right|=\left|z _{1}\right|+\left|z _{2}\right|$ holds, if
(a). $ \arg \left(z _{1}\right)=\arg \left(z _{2}\right)$
(b). $ \arg \left(\mathrm{z} _{1}\right)+\arg \left(\mathrm{z} _{2}\right)=\frac{\pi}{2}$
(c). $\mathrm{z} _{1} \mathrm{Z} _{2}$
(d). $\left|\mathrm{z} _{1}\right|=\left|\mathrm{z} _{2}\right|$
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Answer: (a)4. If $\omega$ and $\omega^{2}$ are the two imaginary cube roots of unity, then the equation whose roots are $a \omega^{317}$ and $a \omega^{382}$ is
(a). $\mathrm{x}^{2}+\mathrm{ax}-\mathrm{a}^{2}=0$
(b). $x^{2}+a^{2} x+a^{2}=0$
(c). $x^{2}+a x+a^{2}=0$
(d). $ x^{2}-a^{2} x+a=0$
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Answer: (c)5. If $1, \mathrm{a} _{1}, \mathrm{a} _{2}$ $\ldots \mathrm{a} _{\mathrm{n}-1}$ are roots of unity, then the value of $\left(1-\mathrm{a} _{1}\right)\left(1-\mathrm{a} _{2}\right)$
(a). $0$
(b). $1$
(c). $\mathrm{n}$
(d). $\mathrm{n}^{2}$
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Answer: (c)6. If $\omega \neq 1$ is a cube root of unity, then the value of
$ \left|\begin{array}{ccc} 1+\omega^{2}+2 \omega^{100} & \omega^{2} & 1 \\ 1& 1+\mathrm{w}^{100}+2 \omega^{200} & \omega \\ \omega & \omega^{2} & 2+\omega^{100}+\omega^{200} \end{array}\right| $
(a). $0$
(b). $1$
(c). $\omega$
(d). $\omega^{2}$
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Answer: (a)7. If the area of the triangle formed by the points $\mathrm{z}, \mathrm{z}+\mathrm{iz}$ and iz is 50 sq. unit, then $|\mathrm{z}|$ is equal to
(a). $5$
(b). $8$
(c). $10$
(d). none of these
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Answer: (c)8. Let $\omega$ be an imaginary roots of $x^{n}=1$. Then, $(5-\omega)\left(5-\omega^{2}\right)$ $\left(5-\omega^{\mathrm{n}-1}\right)$ is
(a). $1$
(b). $\frac{5^{\mathrm{n}}+1}{4}$
(c). $4^{\mathrm{n}}-1$
(d). $\frac{5^{\mathrm{n}}-1}{4}$
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Answer: (d)9. If $\omega(\neq 1)$ is a cube root of unity and $(1+\omega)^{7}=\mathrm{A}+\mathrm{B} \omega$
(a). $0, 1$
(b). $1, 1$
(c). $1, 0$
(d). $-1,1$
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Answer: (b)10. If $x=\omega-\omega^{2}-2$ then the value of $\mathrm{x}^{4}+3 \mathrm{x}^{3}+2 \mathrm{x}^{2}-11 \mathrm{x}-6$ is
(a). $1$
(b). $-1$
(c). $2$
(d). none of these
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Answer: (a)11. If $z=\cos \theta+i \sin \theta$, then $\frac{z^{2 n}-1}{z^{2 n}+1}$ is equal to ( $n$ is an integer)
(a). $i \cot n \theta$
(b). $ i \tan n \theta$
(c). $\tan n$
(d). $\cot n \theta$
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Answer: (b)12. The cube root of unity
(a). lie on the circle $|z|=1$
(b). are collinear
(c). form an equilateral triangle
(d). none of these
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Answer: (c)13. The complex number $z$ satisfying the equation $|z|-4=|z-i|-|z+5 i|=0$, is
(a). $\sqrt{3}-\mathrm{i}$
(b). $2 \sqrt{3}$
(c). $-2 \sqrt{3}-2 \mathrm{i}$
(d). $0$
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Answer: (c)14. If $z _{1}=8+4 i, z _{2}=6+4 i$ and $\arg \left(\frac{z-z _{1}}{z-z _{2}}\right)=\frac{\pi}{4}$, then $z$ satisfies
(a). $|z-7-4 i|=1$
(b). $|\mathrm{z}-7-5 \mathrm{i}|=\sqrt{2}$
(c). $|z-4 i|=8$
(d). $|\mathrm{z}-7 \mathrm{i}|=\sqrt{18}$
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Answer: (b)15. If $z=\cos \theta+i \sin \theta$, then
(a). $\mathrm{z}^{\mathrm{n}}+\frac{1}{\mathrm{z}^{\mathrm{n}}}=2 \cos n \theta$
(b). $\mathrm{z}^{\mathrm{n}}+\frac{1}{\mathrm{z}^{\mathrm{n}}}=2^{\mathrm{n}} \cos n \theta$
(c). $ z^{n}-\frac{1}{z^{n}}=2^{n} i\sin n \theta$
(d). $z^{n}-\frac{1}{z^{n}}=(2 i)^{n} \sin n \theta$
BETA
