mathematics-class-11-unit-05-chapter-01-complex-numbers-l-1

<p><Success style="float:center;text-align:center;font-size:28px;" ><B>Complex number lecture 1</B></Success></p>
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<p>Let $z_k=\cos \left(\frac{2 k \pi}{10}\right)+i \sin \left(\frac{2 k \pi}{10}\right)$ where $k \in$ $\{1,2, \ldots, 9\}$. Then which of the following statement(s) is (are) true?</p>
<p>$(A)$ For each $z_k$ there exists $z_j$ such that $z_k z_j=1$;</p>
<p>$(B)$ $\frac{\left|1-z_1\right|\left|1-z_2\right|\ldots\left|1-z_9\right|}{10} =1$;</p>
<p>$(C) \sum_{k=1}^9 \cos \left(\frac{2 k \pi}{10}\right)=1;$</p>
<p>$(D) \sum_{k=1}^9 \sin \left(\frac{2 k \pi}{10}\right)=0;$
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<p><strong>(A)</strong> we have</p>
<p>$ z_k=\cos \left(\frac{2 k \pi}{10}\right)+i \sin \left(\frac{2 k \pi}{10}\right)$</p>
<p>$\text { where } k \in \{1,2, \ldots, 9 \}$</p>
<p>$= e^{i 2 k \pi/ 10}  $</p>
<p>$= 10^{\text {th }} \text { roots of unity } $</p>
<p>$\left(\text { i.e } z_k^{10}=1\right. \text { for all } k \in \{1,2, \ldots, 9\})$</p>
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<p>In particular, if $j=10-k$, then</p>
<p>$z_k z_j=z_k z_{10-k}=z_{10}=1$</p>
<p>Also, since $k \in \{1,2, \ldots, 9\}$,</p>
<p>$j=10-k \in \{1,2, \ldots, 9\}$.</p>
<p>Thus, for every $z_k \ ∃ \ z_j$ such that $z_k z_j=1$.</p>
<p>Option (A) is correct.</p>
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<p><strong>(B)</strong> Recall that for a complex number $z = x + i y,$ the modulus or the absolute value of $z,$ denoted as $|z|,$</p>
<h4 id="z--sqrt-x2--y2">$|z| = \sqrt{ x^2 + y^2}$</h4>
<p><strong>Fact:</strong>   If $z_1$ and $z_2$ are two complex numbers, then</p>
<h4 id="z_1-z_2--z_1-z_2">$|z_1, z_2| = |z_1| |z_2|$.</h4>
<p>Inductively,</p>
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<p>$\left|z_1 z_2 \cdots z_m\right| = | z_1|| z_2|\cdots| z_m \mid$</p>
<p>In our problem
$\left|1-z_1\right| |1-z_2 \mid \cdots\left|1-z_9\right|.$</p>
<p>$=\left|\left(1-z_1\right)\left(1-z_2\right)\cdots\left(1-z_9\right)\right|$</p>
<p>Notice that</p>
<p>$x^{10}-1=(x-1)\left(x-z_1\right)\left(x-z_2\right) \cdots\left(x-z_9\right) . . . (1)$</p>
<p>Also,</p>
<p>$x^{10}-1=(x-1)\left(x^9+x^8+x^7+\cdots+x+1\right) . . .(2)$</p>
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<p>Comparing equations (1) & (2), we get</p>
<p>$x^9+x^8+\cdots+x+1 =\left(x-z_1\right)\left(x-z_2\right) \cdots\left(x-z_9\right) . . . (*)$</p>
<p>Taking $\ x=1$, we get</p>
<p>$10=\left(1-z_1\right)\left(1-z_2\right) \cdots\left(1-z_9\right)$</p>
<p>$\Rightarrow  \frac{\left|1-z_1\right|\left|1-z_2\right| \cdots\left|1-z_9\right|}{10} =  \frac{\left|\left(1-z_1\right)\left(1-z_2\right) \cdots\left(1-z_9\right)\right|}{10}$</p>
<p>$=\frac{10}{10} = 1$</p>
<p>Option (B) is correct.</p>
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<p><strong>(C) and (D):</strong> Comparing the coefficient of $x^8$ in the equation $(*)$, we get
$ 1=-\left(z_1+z_2+\cdots+z_9\right)$</p>
<p>$\Rightarrow  \sum_{k=1}^9\left[\cos \left(\frac{2 k \pi}{10}\right)+i \sin \left(\frac{2 k \pi}{10}\right)\right]=-1 $</p>
<p>$\Rightarrow  \sum_{k=1}^9 \cos \left(\frac{2 k \pi}{10}\right)+i \sum_{k=1}^9 \sin \left(\frac{2 k \pi}{10}\right)=-1$</p>
<p>$\Rightarrow \quad \sum_{k=1}^9 \cos \left(\frac{2 k \pi}{10}\right)=-1 $</p>
<p>$\text{and} \quad \sum_{k=1}^9 \sin \left(\frac{2 k \pi}{10}\right)=0$</p>
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<p>$\text{ we’ve}$</p>
<p>$x^n-1=(x-1)\left(x-z_1\right)\left(x-z_2\right) \cdots\left(x-z_{n-1}\right) … (1)$</p>
<p>Also,</p>
<p>$x^n-1=(x-1)\left(x^{n-1}+x^{n-2}+\cdots \cdot+x+1\right) … (2)$</p>
<p>$\text{By comparing}$ (1) & (2), $\text{are get}$</p>
<p>$\left(x-z_1\right) \left(x-z_2\right) \cdots\left(x-z_{n-1}\right)$</p>
<p>$= x^{n-1}+x^{n-2}+\cdots+x+1$</p>
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<p>$\therefore \text { Taking } x=1, $</p>
<p>$\text{we get}$</p>
<p>$\left(1-z_1\right)\left(1-z_2\right) \cdots\left(1-z_{n-1}\right)=n $</p>
<p>$ \therefore \frac{\left.\left|1-z_1\right||1-z_2\right| \cdots\left|1-z_{n-1}\right|}{n} $</p>
<p>$=\frac{\left.\mid\left(1-z_1\right)\left(1-z_2\right) \cdots\left(1-z_{n-1}\right)\right|}{n}$</p>
<p>$ =\frac{n}{n}={1}$</p>
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<p>$x^n-1=(x-1)\left(x-z_1\right)\left(x-z_2\right)\cdots\left(x-z_{n-1}\right) … (1)$</p>
<p>From equation (1), we get</p>
<p>$1+z_1+z_2+\cdots+z_{n-1}=0 $</p>
<p>$\Rightarrow z_1+z_2+\cdots+z_{n-1}=-1$</p>
<p>$\text { we’ve } \sum_{k=1}^{n-1} z_k=-1 $</p>
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<p>$\Rightarrow \sum_{k=1}^{n-1} \cos \left(\frac{2 k \pi}{n}\right)+i \sin \left(\frac{2 k \pi}{n}\right)=-1 $</p>
<p>$\Rightarrow \sum_{k=1}^{n-1} \cos \left(\frac{2 k \pi}{n}\right)+i \sum_{k=1}^{n-1} \sin \left(\frac{2 k \pi}{n}\right)=-1 $</p>
<p>$\Rightarrow \sum_{k=1}^{n-1} \cos \left(\frac{2 k \pi}{n}\right)=-1 $</p>
<p>$\text{and}\quad\sum_{k=1}^{n-1} \sin \left(\frac{2 k \pi}{n}\right)=0$</p>
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<p>$\omega=e^{2 \pi i / 3} \text { or } e^{4 \pi i / 3}$</p>
<p><strong>(1)</strong> Since $\quad\omega^2=1$,</p>
<p>$\Rightarrow \omega^3-1=0 $
$\Rightarrow(\omega-1)\left(\omega^2+\omega+1\right)=0$</p>
<p>$\text{Since} ~\omega \neq 1 \Rightarrow \omega^2+\omega+1=0$</p>
<p><strong>(2)</strong> $\text{Since} ~\omega^2=1 \Rightarrow \omega^{-1}=\omega^2$</p>
<p><strong>(3)</strong> $ ~\omega^{3 k}=1 \text{for every integer} ~k$</p>
<p>$\left(Why? \quad \omega^{3 k}=\left(\omega^3\right)^k=1^k=1\right)$</p>
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<p>$\text { Given equation is} $</p>
<p>$ E_1^{4 n+3}+E_2^{4 n+3}+E_3^{4 n+3}=0 $</p>
<p>$\Rightarrow  \left(\omega E_3\right)^{4 n+3}+\left(\omega^2 E_3\right)^{4 n+3}+E_3^{4 n+3}=0$</p>
<p>$\Rightarrow {\left[\omega^{4 n+3}+\left(\omega^2\right)^{4 n+3}+1\right] E_3^{4 n+3}=0 }$</p>
<p>Note that $E_3 \neq 0$.</p>
<p>$\biggl( \bold{\text{why ?}} \ \ E_3  =-3+2 \omega+3 \omega^2 \ = ~ -3+2 \omega+3(-\omega-1) =-6-\omega \neq 0 \biggl )$</p>
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<p>$\omega^{4 n+3}+\left(\omega^2\right)^{4 n+3}+1=0$</p>
<p>$\Rightarrow\omega^{n+3(n+1)}+\left(\omega^2\right)^{n+3(n+1)}+1=0 $</p>
<p>$\Rightarrow \omega^n+\left(\omega^2\right)^n+1=0$</p>
<p>$\text{i. e.} \quad\omega^n+\omega^{2 n}+1=0$</p>
<p>$\text { If } ~n=3 k \text { for } k \in  z, \text { then }$</p>
<p>$\omega^n+\omega^{2 n}+1 =\omega^{3 k}+\left(\omega^2\right)^{3 k}+1 $</p>
<p>$=1+1+1=3 \neq 0$</p>
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<li>Let $z_ 1$ and $z_ 2$ be two distinct complex numbers and let $z=(1-t) z_ 1+t z_ 2$ for some real number $t$ with $0< t< 1$. For a complex number $z$, let $\operatorname{Arg}(z)$ denote the principal argument with $0 \leq \operatorname{Arg}(z)< 2 \pi$. Then which of the statement(s) is (are) true?</li>
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<p>$(A) \left|z-z_ 1\right|+\left|z-z_ 2\right|=\left|z_ 1-z_2\right|;$</p>
<p>$(B) \operatorname{Arg}\left(z-z_1\right)=\operatorname{Arg}\left(z-z_ 2\right);$</p>
<p>$(C) \left|\begin{array}{cc}z-z_1 & \bar{z}-\overline{z_1} \\ z_2-z_1 & \overline{z_2}-\overline{z_1}\end{array}\right|=0$</p>
<p>$(D) \operatorname{Arg}\left(z-z_1\right)=\operatorname{Arg}\left(z_ 2-z_ 1\right).$</p>
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<p>By the parallelogram law,</p>
<p>$z_1+w=z \Rightarrow w=z-z_1 $</p>
<p>$\therefore\left|z-z_1\right|+\left|z-z_2\right|=\left|z_1-z_2\right| $</p>
<p>Option (A) is correct.</p>
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<p><strong>(B) and (D)</strong></p>
<p>$z  = x + i y $</p>
<p>$ =r(\cos \theta + i \sin \theta) $</p>
<p>for some $r \geqslant 0 \ $ & $ \ \theta \in  {\R} $</p>
<p>$\theta \rightarrow \text { argument of } z$</p>
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<p>Let</p>
<p>$Arg (z) =  \text{ the principle argument of } z$</p>
<p>$Arg (z) = \text{ the unique } \theta \text {~such that }$</p>
<p>$z = r(\cos \theta + i \sin \theta) \text { and } $</p>
<p>$ 0\leqslant \theta < 2 \pi$.</p>
<p>$Arg (z-z_1) = Arg (z_2 -z_1)$</p>
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<p><strong>(C)</strong> $\text{ Let A } = \left[\begin{array}{ll}z-z_1 & \bar{z}-\overline{z_1} \\ z_2-z_1 & \overline{z_2}-\overline{z_1}\end{array}\right]$</p>
<p>$z-z_1=\lambda\left(z_2-z_1\right) \ \ \text{ for some } {\lambda \in \R.} $</p>
<p>Recall for $z = x + i y,$ the complex conjugate if $z$ is</p>
<p>$\bar {z} = x - i y $</p>
<p><ins><strong>Verify</strong></ins> if $z_1 \text { and } z_2 $ are two  complex  numbers, then</p>
<p>(i) $ \overline{z_1 + z_2} = \bar {z_1} + \bar {z_2} $</p>
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<p><strong>(ii)</strong> $ \overline{z_1 z_2} = \bar{z_1}. \bar {z_2}$</p>
<p>Since $\quad z - z_1 = \lambda (z_2 - z_1),$</p>
<p>$\rArr \overline {z-z_1} = \overline{\lambda (z_2 - z_1)}$
$\rArr \bar {z} - \bar{z_1} = \bar{\lambda} \overline{(z_2 - z_1)}$</p>
<p>$= \lambda (\bar{z_2}- \bar{z_1}) \text{ as } \lambda \in \R$</p>
<p>$\therefore|A|=\left|\begin{array}{cc}\lambda\left(z_2-z_1\right) & \lambda\left(\bar{z}_2-\bar{z}_1\right) \\z_2-z_1 & \overline{z_2}-\overline{z_1}\end{array}\right|$
$ = 0 $</p>
<p>Options (c) and (D) are correct.</p>
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