### <Success>Complex numbers <br> lecture - 7</Success> <div style='float: left; width:100%;'> </div>

### <Success>Problem</Success> <div style='float: left;text-align:left; width:100%;font-size:30px;'> <p>Find the value of $\sin10^0 \sin 20^0 \cdots \sin 80^0 .$</p> <p><strong>Solution:</strong></p> <p>We have proved the identity,</p> <p>$\sin \frac {\pi}{n} \sin\frac {2\pi}{n} \cdots \sin \frac{(n-1)\pi} {n}=\frac{n}{2^{n-1}}$</p> <p>$n=18,$</p> <p>$\sin \frac{\pi}{18} \sin \frac{2\pi}{18} \cdots \sin \frac{17\pi}{18} =\frac{18}{2^{17}} $</p> </div>

### <Success>Solution</Success> <div style='float: left; text-align:left; width:100%;font-size:30px;'> <p><strong>Observation:</strong></p> <p><strong>(1)</strong> In the expression, $9^{th}$ term=$1$ ,$\quad$ put k=9, $\sin \frac {k\pi}{18} = \sin \frac {\pi}{2} =1$</p> <p><strong>(2)</strong> $\sin \frac {(18-k)\pi}{18}\ = \sin \frac {k\pi}{18}$</p> <p>$ [\sin (n - k) \pi = \sin\frac{k \pi}{n} ;$ $ k = 1, \cdots ,n]$</p> <p>$\therefore \sin \frac{\pi}{18} \sin \frac{2\pi}{18}\ \cdots \sin \frac{17\pi}{18}\ =\frac{18}{2^{17}} $</p> <p>$\sin ^2\frac{\pi}{18}\ \sin ^2\frac{2\pi}{18}\ \cdots \sin ^2\frac{8\pi}{18}\ =\frac{9}{2^{16}} $</p> <p>$\sin 10^0\ \sin 20^0 \cdots \sin 80^0\ =\frac{3}{2^{8}} $</p> </div>

### <Success>Exercise</Success> <div style='float: left; text-align:left; width:100%;font-size:30px;'> <p>Show that</p> <p><strong>(1)</strong> For $n=2m, \cos \frac {\pi}{n}\ \cos \frac{2\pi}{n}…\cos \frac {(m-1)\pi}{n} =\frac {\sqrt m}{2^{m-1}}$</p> <p><strong>(2)</strong> For $n=2m+1, \cos \frac {\pi}{n}\cos\frac{2\pi}{n}…\cos\frac {m\pi}{n} =\frac {1}{2^{m-1}}$</p> <p><strong>(3)</strong> For $n=2m, \tan\frac {\pi}{n}\tan\frac{2\pi}{n}…\tan\frac {(m-1)\pi}{n} =1$</p> <p><strong>(4)</strong> For $n=2m+1, \tan\frac {\pi}{n} \tan\frac{2\pi}{n}…\tan\frac {m\pi}{n} =\sqrt {2m+1}$</p> </div>

### <Success>Solution</Success> <div style='float: left; text-align:left; width:60%;font-size:30px;'> <p>W.L.G. $ P_k =$ cis$(\frac{2k\pi}{n})=\alpha^k,$ <br> $k=0,\cdots ,n-1$</p> <p>$ z^{n-1} +z^{n-2}+ \cdots +z+1=\Pi_{k=1}^{n-1}(z-\alpha^k)$</p> <p>$ z = 1,$ $n=\Pi_{k=1}^{n-1}(1-\alpha^k)$</p> <p>$ z = -1,$</p> </div> <div style='float: right; width:40%;'> <p><img alt="alt text" src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/ebcaa324-ba28-4f1c-ebac-c4cf4a7ba100/public"></p> <!----> </div>

### <Success>Solution</Success> <div style='float: left; text-align:left; width:100%;font-size:30px;'> <p>$|1-\alpha^k|^2 =4\biggl(\sin\frac{k\pi}{n}\biggl)^2, \quad k=0, \cdots n-1$</p> <p>$\Rightarrow|1-\alpha^k| =2\biggl|\sin\frac{k\pi}{n}\biggl|, \quad 0\le\frac{\pi}{n}\le \frac{k\pi}{n}\le\frac{(n-1)\pi}{n}\le\pi$</p> <p>$P_0P_k\ = 2 \sin\frac{k\pi}{n}$</p> <p>$\Rightarrow 2 \sin\frac{\pi}{n}\times2\ \sin\frac{2\pi}{n} \cdots 2\ \sin\frac{(n-1)\pi}{n}=n$</p> <p>$\ \sin\frac{\pi}{n}\times\ \sin\frac{2\pi}{n} \cdots \sin\frac{(n-1)\pi}{n}=\frac {n}{2^{n-1}}$</p> <p>Using this hint, we can show the result.</p> </div>

### <Success>Cube roots of unity</Success> <div style='float: left; text-align:left; width:100%;font-size:30px;'> <p>$\ n=3, \quad z^3=1 $</p> <p>$\rightarrow\ \biggl\{ 1,\ \text{cis}\frac {2\pi}{3},\quad \text{cis}\frac {4\pi}{3}\biggl\}$</p> <p>$\rightarrow\quad \omega=\ \text{cis} \frac{2\pi}{3} =\frac {-1}{2}+i\frac{\sqrt3}{2}$</p> <p>$\rightarrow\quad \omega^2=\ \text{cis} \frac{4\pi}{3} =\biggl(\text{cis}\frac{2\pi}{3}\biggl)^2=\omega^2=\bar{\omega}$</p> <p>$1, \omega,\omega^2$ are cube roots of unity.</p> </div>

### <Success>Remark</Success> <div style='float: left; text-align:left; width:60%;font-size:30px;'> <p>$(1)\ 1+\omega+\omega^2 =0$</p> <p>$(2)\ \omega^3 =1=\biggl(\text{cis}\frac{2\pi}{3}\biggl)^3=\text{cis}2\pi =1$</p> <p>$ \ \omega^{3n} =(\omega^3)^n =1 \quad n\in \Z$</p> </div> <div style='float: right; width:40%;'> <p><img alt="alt text" src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/d760022b-931f-4840-6620-9c2142fbfd00/public"></p> <!----> </div>

### <Success>Problem</Success> <div style='float: left; text-align:left; width:100%;font-size:30px;'> <p>Find the value of</p> <p>$4+5\biggl(\frac {-1}{2}+i\ \frac {{\sqrt3}}{2}\biggl)^{334} + 3\biggl(\frac {-1}{2}+i\ \frac{{\sqrt3}}{2}\biggl)^{365}$</p> </div> ====== ### <Success>Solution</Success> <div style='float: left; text-align:left; width:100%;font-size:30px;'> <p>$4+5\ \omega^{334}\ +3\ \omega^{365}$</p> <p>$=4+5\cdot \ \omega^{333}\ \cdot \omega+3\ \cdot \omega^{363} \cdot \omega^2$</p> <p>$=4+5 \omega+3\omega^2 \quad \quad (\therefore \omega^2=-\omega-1)$</p> <p>$=4+5 \omega-3\omega-3 $</p> <p>$=1+2 \biggl(\frac {-1}{2}+i\ \frac {{\sqrt3}}{2}\biggl)$</p> <p>$=i{\sqrt3}$</p> </div>

### <Success>Problem</Success> <div style='float: left; text-align:left; width:100%;font-size:30px;'> <p>Prove that $\biggl(\frac {\sqrt3+i}{{\sqrt3-i}}\biggl)^{3(2n+1)} =-1,\quad$ for all $n \in \Z.$</p> </div> ====== ### <Success>Solution</Success> <div style='float: left; text-align:left; width:100%;font-size:30px;'> <p>$\biggl(\frac {\sqrt3+i}{{\sqrt3-i}}\biggl)\cdot\biggl(\frac {\sqrt3+i}{{\sqrt3+i}}\biggl)$ $=\frac {(\sqrt3+i)^2}{4}$ $=i^2\ \biggl(\frac {1}{2}- \frac {{\sqrt3\ }i}{2}\biggl)^2=-\omega^2$</p> <p>$[\because \omega=\frac{-1}{2}+i\ \frac{{\sqrt3}}{2}]$</p> <p>$\biggl(\frac {{\sqrt3+i}}{{\sqrt3-i}}\biggl )^{3(2n+1)}=(-\omega^2)^{3(2n+1)}$ $=(-1)^{6n+3}(\omega^6)^{(2n+1)} =-1$</p> </div>

### <Success>Problem</Success> <div style='float: left;text-align:left; width:100%;font-size:30px;'> <p>Find the least positive integer ’n’ such that</p> <p>$(1+\omega^2)^n = (1+\omega^4)^n$</p> </div> ====== ### <Success>Solution</Success> <div style='float: left;text-align:left; width:100%;font-size:30px;'> <p>$1+\omega^2=-\omega$</p> <p>$\omega^4=\omega^3\cdot \omega=\omega$</p> <p>$(1+\omega^2)^n=(1+\omega^4)^n$</p> <p>$\Leftrightarrow (-\omega)^n=(-\omega^2)^n$</p> <p>$\Leftrightarrow \omega^n=\omega^{2n}$</p> <p>$\Leftrightarrow \omega^n=1$</p> <p>$n=3$</p> </div>

### <Success>Problem</Success> <div style='float: left; text-align:left; width:100%;font-size:30px;'> <p>Solve for $z \in \mathbb{C}: $ $ z^2 +z|z|+|z|^2=0 $</p> </div> ====== ### <Success>Solution</Success> <div style='float: left; text-align:left; width:100%;font-size:30px;'> <p>$ z^2 +z|z|+|z|^2=0 \rightarrow (1)$</p> <p>$ z=0 \quad (1)$ is satisfied</p> <p>Let $z \ne\ 0 $</p> <p>$\frac {z^2}{|z|^2}+ \frac {z}{|z|}+1=0$</p> <p>$\biggl (\frac {z}{|z|}\biggl)^2+ \biggl(\frac {z}{|z|}\biggl)+1=0$</p> <p>$\omega=\frac {z}{|z|}, \quad \omega^2+\omega+1=0\quad$ It has two solution $\text{cis}\frac{2\pi}{3},\ \text{cis}\frac{4\pi}{3}$</p> <p>$z=\lambda\omega, \quad \lambda \omega^2,\quad \lambda\ge 0$ are solution to (1)</p> </div>

### <Success>Solution</Success> <div style='float: left; text-align:left; width:40%;font-size:30px;'> <p>$|1-\omega| =|\omega-\omega^2|$ <br> $|1-\omega| =|\omega(1-\omega)|$ <br> $|1-\omega| =|1-\omega|$ <br> $|1-\omega| =|1-\omega^2|$</p> <p><img alt="alt text" src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/7b54ace8-018d-4650-2ed7-97c9e6409700/public"></p> </div> <div style='float: right; width:40%;'> <p><strong>Equilateral Triangle</strong></p> <p>$a=b=c$</p> <p>$\alpha=\beta=x$</p> <p><img alt="alt text" src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/fb02db88-9737-47bc-51da-3761ba8e1000/public"></p> <!----> </div>

### <Success>Theorem</Success> <div style='float: left; text-align:left; width:100%;font-size:30px;'> <p><strong>Proposition</strong></p> <p>Let T be a triangle with 15 vertices a,b,c in $\mathbb{C}$ (in anticlockwise direction).</p> <p>Then the following are equivalent:</p> <p><strong>(i)</strong> T is equilateral triangle</p> <p><strong>(ii)</strong> $a+\omega b+\omega^2c=0, \quad \omega=\text{cis}\frac {2\pi}{3}$</p> <p><strong>(iii)</strong> $a^2+b^2+c^2 = ab+bc+ca$</p> </div>

### <Success>Proof</Success> <div style='float: left; text-align:left; width:60%;font-size:30px;'> <p>$(i)\Leftrightarrow(ii)$</p> <p>$(i) \Rightarrow(ii) \quad a+\omega b+\omega^2c\ =0\rightarrow(1)$</p> <p>suppose (1) is satisfied</p> <p>$(a-z),(b-z),(c-z)$</p> <p>$a-z+\omega(b-z)+\omega^2(c-z)$</p> <p>$=a+\omega b+\omega^2c-z\ (1+\omega+\omega^2)$</p> <p>$=0\rightarrow(2)$</p> <p>$(1) \Leftrightarrow(2)$</p> </div> <div style='float: right; width:38%;'> <p><img alt="alt text" src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/7c950ccf-7c8d-409b-b284-696948b27300/public"> <img alt="alt text" src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/8032b402-3cbd-4ddb-aa3c-624c51502f00/public"></p> <!----> </div>

### <Success>Proof</Success> <div style='float: left; text-align:left; width:55%;font-size:30px;'> <p>T is equilateral $\Leftrightarrow c-b=\omega(b-a)$</p> <p>$\Leftrightarrow c+a\omega-b(1+\omega)=0$</p> <p>$\Leftrightarrow a\omega+b\omega^2+c=0$</p> <p>$\Leftrightarrow a+b\omega+c\omega^2=0$</p> </div> <div style='float: right; width:45%;'> <p><img alt="alt text" src="https://imagedelivery.net/YfdZ0yYuJi8R0IouXWrMsA/a5676cd8-e096-434e-7a83-8ab5301d0600/public"></p> <!----> </div>

### <Success>Proof</Success> <div style='float: left;text-align:left; width:100%;font-size:30px;'> <p><strong>(ii)</strong> $\Leftrightarrow$ <strong>(iii)</strong> (or) (i)$\Leftrightarrow$ <strong>(iii)</strong></p> <p>$a^2+b^2+c^2= ab+bc+ca$</p> <p>$\Leftrightarrow$ $(a-z)^2+(b-z)^2+(c-z)^2=(a-z)(b-z)+(b-z)(c-z)+(c-z)(a-z)$</p> <p>$z=a \quad$ shift the triangle T such that one of vertex is ‘0’.</p> <p>W.l.g. O,B,C are vertices of T.</p> <p>$b^2+c^2=bc \Leftrightarrow$ $\omega b+\omega^2c=0$</p> <p>$b^2+c^2=bc \Leftrightarrow$ $\omega=\frac {-b}{c}$</p> <p>$\frac{b}{c}+\frac{c}{b}+1$ $\Rightarrow \frac{b}{c}=1-\frac{c}{b}$</p> </div> ====== ### <Success>Proof</Success> <div style='float: left;text-align:left; width:100%;font-size:30px;'> <p>claim $\quad (\frac{-b}{c})^3=1$</p> <p>$\biggl(\frac{-b}{c}\biggl)^2=\frac {b}{c}\cdot\biggl(1-\frac {c}{b}\biggl) = \frac {b}{c}-1=-\frac {c}{b}$</p> <p>$\biggl(\frac{-b}{c}\biggl)^3=\biggl(\frac {-b}{c}\biggl)^2\ \biggl(\frac {-b}{c}\biggl)= \frac {-c}{b}\ \cdot\ \frac {-b}{c}=1$</p> </div>

<div> <h2><Success>Thank you</Success></h2> </div>