الجذور التكعيبية للواحد صحيح

$$\begin{gathered} Z^3=1\Rightarrow Z^3-1=0\Rightarrow (Z-1)(Z^2+z+1)=0 \\ \text{either }Z=1 \mathrm{الأول} \mathrm{الجذر} \\ \text{or}{z}^2+z+1=0 \mathrm{بالدستور} a=1,b=1,c=1 \\ \begin{array}{rl}& Z=\frac{-b\pm \sqrt{b^2-4ac}}{2a}=\frac{-\left(1\right)\pm \sqrt{(1{)}^2-(4\left)\right(1\left)\right(1)}}{2\left(1\right)}\\ & Z=\frac{-1\pm \sqrt{1-4}}{2}=\frac{-1\pm \sqrt{-3}}{2}\\ & Z=\frac{-1\pm \sqrt{3}i}{2}=\frac{-1}{2}\pm \frac{\sqrt{3}}{2}i\\ & \text{either }z=\frac{-1}{2}+\frac{\sqrt{3}}{2}i=\omega \mathrm{الثاني} \mathrm{الجذر}\\ & \text{or }z=\frac{-1}{2}-\frac{\sqrt{3}}{2}i={\omega }^2 \mathrm{الثالث} \mathrm{الجذر}\end{array} \end{gathered}$$

هناك ثلاثة جذور للواحد الصحيح وهي $(1,\omega ,{\omega }^2)$ حيث أن الرمز (w) يقرأ أوميكا "Omega"

خواص الجذور التكعيبية للواحد الصحيح:

1. الجذران $\omega ,{\omega }^2$ جذران تخيليان مترافقان.
2. مجموع الجذور الثلاثة يساوي صفر أي $1+\omega +{\omega }^2=0$
3. حاصل ضرب الجذور الثلاثة يساوي واحد أي $\omega {\omega }^2=1$

استنتاجات لخواص الجذور:

1. مجموع أي جذرين = سالب الجذر الآخر مثلاً $1+{\omega }^2=-\omega \mid ,\omega +1=-{\omega }^2,\omega +{\omega }^2=-1$
2. أي جذر = سالب (مجموع الجذرين الآخرين) مثلاً $1=-\omega -{\omega }^2,{\omega }^2=-1-\omega ,\omega =-1-{\omega }^2$
3. ${\omega }^3=1\Rightarrow \omega =\frac{1}{{\omega }^2}\Rightarrow {\omega }^2=\frac{1}{\omega }$
4. كل $\omega$ هي مرافق ${\omega }^2$ وبالعكس يمكن استبدال أحدهما بالآخر كما في المثالين التالين: $2\omega +5{\omega }^2=3{\omega }^2+5\omega$ و$4\omega +2{\omega }^2=4{\omega }^2+2\omega$
5. $\omega -{\omega }^2={\omega }^2-\omega =\pm \sqrt{3}i$ لاحظ: $(\frac{-1}{2}+\frac{\sqrt{3}}{2}i)-(\frac{-1}{2}-\frac{\sqrt{3}}{2}i)=\frac{2\sqrt{3}i}{2}=+\sqrt{3}i$ و$(\frac{-1}{2}-\frac{\sqrt{3}}{2}i)-(\frac{-1}{2}+\frac{\sqrt{3}}{2}i)=\frac{-2\sqrt{3}i}{2}=-\sqrt{3}i$
6. $\omega .{\omega }^2={\omega }^3=1$ لاحظ: $(\frac{-1}{2}+\frac{\sqrt{3}}{2}i)\cdot (\frac{-1}{2}-\frac{\sqrt{3}}{2}i)={\left(\frac{-1}{2}\right)}^2+{\left(\frac{\sqrt{3}}{2}\right)}^2=\frac{1}{4}+\frac{3}{4}=\frac{4}{4}=1$
7. نستخدم $(r=0,1,2,3,\dots )،{\omega }^{3n+r}={\omega }^r$ حيث $\left(r\right)$ عدد صحيح.
8. نستخدم ${\omega }^3$ في عمليات التبسيط.

ومن هذه الاستنتاجات نتوصل إلى أن ناتج $\omega$ مرفوعة إلى قوة معينة هو أحد جذور الوا$(1,\omega ,{\omega }^2)$ لاحظ الأمثلة التالية:

$$\begin{gathered} \begin{array}{rl}& 1) {\omega }^4={\omega }^3\cdot \omega =(1)\cdot \omega =\omega \\ & 2) {\omega }^5={\omega }^3\cdot {\omega }^2=(1)\cdot {\omega }^2={\omega }^2\\ & 3) {\omega }^6={\left({\omega }^3\right)}^2={\omega }^3\cdot {\omega }^3=(1)\cdot (1)=1\\ & 4) {\omega }^{81}={\left({\omega }^3\right)}^{27}=(1{)}^{27}=1\\ & 5) {\omega }^{-58}={\omega }^{-58}{\omega }^{60}={\omega }^{-58+60}={\omega }^2\\ & 6) {\omega }^{-4}=\frac{1}{{\omega }^4}=\frac{1}{{\omega }^3\cdot \omega }=\frac{1}{\omega }=\frac{{\omega }^3}{\omega }={\omega }^2\\ & 7) {\omega }^{-22}=\frac{1}{{\omega }^{22}}=\frac{1}{{\left({\omega }^3\right)}^7\cdot \omega }=\frac{1}{(1{)}^7\cdot \omega }=\frac{{\omega }^3}{\omega }={\omega }^2\\ & 8) {\omega }^{6n+5}={\omega }^{6n}\cdot {\omega }^5={\left({\omega }^3\right)}^{2n}\cdot {\omega }^5=(1{)}^{2n}{\omega }^5={\omega }^5={\omega }^3\cdot {\omega }^2={\omega }^2\\ & 9) {\omega }^7+{\omega }^5+1=0\\ & \mathbf{\text{LHS:}}{\omega }^7+{\omega }^5+1={\left({\omega }^3\right)}^2\cdot \omega +{\omega }^3\cdot {\omega }^2+1=\omega +{\omega }^2+1=0\mathbf{:}\mathbf{\text{RHS}}\\ & 10) {\omega }^{-6n-5}={\omega }^{(-1)(6n+5)}={\omega }^{-1}\cdot {\omega }^{6n+5}=\frac{{\omega }^{6n+5}}{\omega }=\frac{{\left({\omega }^3\right)}^{2n}\cdot {\omega }^5}{\omega }\\ & \frac{1\cdot {\omega }^3\cdot {\omega }^2}{\omega }=\frac{{\omega }^2}{\omega }=\omega \end{array} \\ \begin{array}{rl}& 11) 12+2\omega +2{\omega }^2=2(1+\omega +{\omega }^2)=2(0)=0\\ & {12) (2+5\omega +5{\omega }^2)}^3={[2+5(\omega +{\omega }^2)]}^3=[2-5{]}^3=(-3{)}^3=-27\\ & 13) -4{(2+\omega +2{\omega }^2)}^9=-4{[\omega +2(1+{\omega }^2)]}^9=-4[\omega -2\omega {]}^9\\ & -4\left[\right(-\omega {)}^9]=-4(-\omega {\omega }^9)=4\end{array} \\ \begin{array}{rl}& 14) (3-2\omega {)}^2+{(3-2{\omega }^2)}^2=9-12\omega +4{\omega }^2+9-12{\omega }^2+4{\omega }^4\\ & =9-12\omega +4{\omega }^2+9-12{\omega }^2+4\omega =18-8\omega -8{\omega }^2=18-8(\omega +{\omega }^2)\\ & =18+8=26\end{array} \end{gathered}$$

(1)- أثبت أن:

${(\frac{1}{1+{\omega }^2}-\frac{1}{1+\omega })}^2=-3$

$\begin{array}{rl}\mathbf{\text{LHS:}}& {(\frac{1}{1+{\omega }^2}-\frac{1}{1+\omega })}^2={(\frac{1}{-\omega }-\frac{1}{-{\omega }^2})}^2={(\frac{{\omega }^3}{-\omega }-\frac{{\omega }^3}{-{\omega }^2})}^2\\ & ={(-{\omega }^2+\omega )}^2={\omega }^4-2{\omega }^3+{\omega }^2=\omega -2+{\omega }^2=-1-2=-3 \mathbf{:}\mathbf{RHS}\end{array}$

${(\frac{1}{2-\omega }-\frac{1}{2-{\omega }^2})}^2=\frac{-3}{49}$

$\begin{array}{rl}& \mathbf{\text{LHS:}}{(\frac{1}{2-\omega }-\frac{1}{2-{\omega }^2})}^2={\left(\frac{(2-{\omega }^2)-(2-\omega )}{(2-{\omega }^2)(2-\omega )}\right)}^2={\left(\frac{2-{\omega }^2-2+\omega }{4-2\omega -2{\omega }^2+{\omega }^3}\right)}^2\\ & ={\left(\frac{-{\omega }^2+\omega }{4-2(\omega +{\omega }^2)+1}\right)}^2={\left(\frac{-{\omega }^2+\omega }{5+2}\right)}^2=\frac{{(-{\omega }^2+\omega )}^2}{(7{)}^2}=\frac{{\omega }^4-2{\omega }^3+{\omega }^2}{49}\\ & =\frac{\omega -2+{\omega }^2}{49}=\frac{-1-2}{49}=\frac{-3}{49}\mathbf{:}\mathbf{\text{RHS}}\end{array}$

${(\frac{1}{\omega }-\frac{1}{{\omega }^2})}^2(2+\frac{2}{\omega })\left(\frac{-1}{1+{\omega }^2}\right)=6$

$\begin{array}{rl}\mathbf{\text{LHS:}}& {(\frac{1}{\omega }-\frac{1}{{\omega }^2})}^2(2+\frac{2}{\omega })\left(\frac{-1}{1+{\omega }^2}\right)={(\frac{{\omega }^3}{\omega }-\frac{{\omega }^3}{{\omega }^2})}^2(2+\frac{2{\omega }^3}{\omega })\left(\frac{-{\omega }^3}{-\omega }\right)\\ & ={({\omega }^2-\omega )}^2(2+2{\omega }^2)\left({\omega }^2\right)=({\omega }^4-2{\omega }^3+{\omega }^2)\left(2\right)(1+{\omega }^2)\left({\omega }^2\right)\\ & =(\omega -2+{\omega }^2)\left(2\right)(-\omega )\left({\omega }^2\right)=(-1-2)(-2{\omega }^3)=(-3)(-2)=6\mathbf{:}\mathbf{\text{RHS}}\end{array}$

$\frac{{\omega }^{14}+{\omega }^7-1}{{\omega }^{10}+{\omega }^5-2}=\frac{2}{3}$

$\mathbf{LHS}\mathbf{:}\frac{{\omega }^{14}+{\omega }^7-1}{{\omega }^{10}+{\omega }^5-2}=\frac{{\left({\omega }^3\right)}^4{\omega }^2+{\left({\omega }^3\right)}^2\cdot \omega -1}{{\left({\omega }^3\right)}^3\cdot \omega +{\omega }^3{\omega }^2-2}=\frac{{\omega }^2+\omega -1}{\omega +{\omega }^2-2}=\frac{-1-1}{-1-2}=\frac{-2}{-3}=\frac{2}{3}\mathbf{:}\mathbf{\text{RHS}}$

$(1-\frac{2}{{\omega }^2}+{\omega }^2)(1+\omega -\frac{5}{\omega })=18$

$\begin{array}{rl}\mathbf{\text{LHS:}}& (1-\frac{2}{{\omega }^2}+{\omega }^2)(1+\omega -\frac{5}{\omega })=(1-\frac{2{\omega }^3}{{\omega }^2}+{\omega }^2)(1+\omega -\frac{5{\omega }^3}{\omega })\\ & =(1-2\omega +{\omega }^2)(1+\omega -5{\omega }^2)=(1+{\omega }^2-2\omega )(1+\omega -5{\omega }^2)\\ & =(-\omega -2\omega )(-{\omega }^2-5{\omega }^2)=(-3\omega )(-6{\omega }^2)=18{\omega }^3=18\mathbf{:}\mathbf{\text{RHS}}\end{array}$

${(1+{\omega }^2)}^3+(1+\omega {)}^3=-2$

$\begin{array}{rl}& \mathbf{\text{LHS:}}{(1+{\omega }^2)}^3+(1+\omega {)}^3=(-\omega {)}^3+{(-{\omega }^2)}^3=(-\omega {)}^3+{(-{\omega }^3)}^2=-{\omega }^3-{\left({\omega }^3\right)}^2\\ & =-1-1=-2\mathbf{:}\mathbf{\text{RHS}}\end{array}$

$\frac{\omega }{{(2+5\omega +2{\omega }^2)}^2}+\frac{{\omega }^2}{{(2+2\omega +5{\omega }^2)}^2}=-\frac{1}{9}$

$$\begin{gathered} \mathbf{\text{LHS:}}\frac{\omega }{{[2(1+{\omega }^2)+5\omega ]}^2}+\frac{{\omega }^2}{{\left[2\right(1+\omega )+5{\omega }^2]}^2}=\frac{\omega }{[-2\omega +5\omega {]}^2}+\frac{{\omega }^2}{{[-2{\omega }^2+5{\omega }^2]}^2} \\ =\frac{\omega }{[3\omega {]}^2}+\frac{{\omega }^2}{{\left[3{\omega }^2\right]}^2}=\frac{\omega }{9{\omega }^2}+\frac{{\omega }^2}{9{\omega }^4}=\frac{\omega }{9{\omega }^2}+\frac{1}{9{\omega }^2}=\frac{\omega +1}{9{\omega }^2}=\frac{-{\omega }^2}{9{\omega }^2}=-\frac{1}{9}\mathbf{:}\mathbf{RHS} \end{gathered}$$

(2)- إذا كان $(x+yi)={(1+2\omega +\frac{1}{\omega })}^2$ أثبت أن $x^2+y^2=1$

$\begin{array}{c}{(1+2\omega +\frac{1}{\omega })}^2={(1+2\omega +\frac{{\omega }^3}{\omega })}^2={(1+2\omega +{\omega }^2)}^2=(-\omega +2\omega {)}^2={\omega }^2=\frac{-1}{2}-\frac{\sqrt{3}}{2}i\\ x^2+y^2={\left(\frac{-1}{2}\right)}^2+{(-\frac{\sqrt{3}}{2})}^2=\frac{1}{4}+\frac{3}{4}=\frac{4}{4}=1\end{array}$

(3)- جد بأبسط صورة:

$\omega (1+i{)}^4-{(5+3\omega +5{\omega }^2)}^2$

$\begin{array}{rl}& \omega (1+i{)}^4-{(5+3\omega +5{\omega }^2)}^2=\omega {\left[\right(1+i{)}^2]}^2-{[3\omega +5(1+{\omega }^2)]}^2\\ & =\omega {(1+2i+i^2)}^2-(3\omega -5\omega {)}^2=\omega (2i{)}^2-(-2\omega {)}^2\\ & =-4\omega -4{\omega }^2=-4(\omega +{\omega }^2)=-4(-1)=4\end{array}$

$\frac{5\omega +3}{3{\omega }^2+5}$

$\frac{5\omega +3}{3{\omega }^2+5}={\left(\frac{5\omega +3{\omega }^3}{3{\omega }^2+5}\right)}^3={\left(\frac{\omega (5+3{\omega }^2)}{3{\omega }^2+5}\right)}^3={\omega }^3=1$

(4)- كون المعادلة التربيعية التي جذراها: ${(4+5\omega +5{\omega }^2)}^2,{(3+5\omega +4{\omega }^2)}^6$

$\begin{array}{rl}& h={(4+5\omega +5{\omega }^2)}^2={[4+5(\omega +{\omega }^2)]}^2=(4-5{)}^2=(-1{)}^2=1\\ & k={(3+5\omega +4{\omega }^2)}^6=[3+5\omega +4(-1-\omega ){]}^6=[3+5\omega -4-4\omega {]}^6=[-1+\omega {]}^6\\ & k={\left[\right(-1+\omega {)}^2]}^3={(1-2\omega +{\omega }^2)}^3={(1+{\omega }^2-2\omega )}^3=(-\omega -2\omega {)}^3=(-3\omega {)}^3=-27\\ & (h+k)=\left(1\right)+(-27)=-26\\ & \left(\text{h.k}\right)=\left(1\right)(-27)=-27\\ & x^2-(-26)x-27=0\Rightarrow x^2+26x-27=0 \mathrm{التربيعية} \mathrm{المعادلة}\end{array}$

(5)- كون المعادلة التربيعية التي جذراها: $(\frac{2}{\omega }+3\omega i),(\frac{2}{{\omega }^2}+3{\omega }^{2}i)$

$\begin{array}{rl}& \mathrm{h}=(\frac{2}{\omega }+3\omega \mathrm{i})=(\frac{2{\omega }^3}{\omega }+3\omega i)=(2{\omega }^2+3\omega i)\\ & \mathrm{k}=(\frac{2}{{\omega }^2}+3{\omega }^{2}i)=(\frac{2{\omega }^3}{{\omega }^2}+3{\omega }^{2}i)=(2\omega +3{\omega }^2\mathrm{i})\\ & (h+k)=(2{\omega }^2+3\omega \mathrm{i})+(2\omega +3{\omega }^2\mathrm{i})=2({\omega }^2+\omega )+3(\omega +{\omega }^2)i=-2-3i\\ & h\cdot k=(2{\omega }^2+3\omega \mathrm{i})(2\omega +3{\omega }^2\mathrm{i})=(4{\omega }^3-9{\omega }^3)+(6{\omega }^4+6{\omega }^2)i\\ & h\cdot k=(4-9)+6(\omega +{\omega }^2)i=-5-6i\\ & x^2-(-2-3i)x+(-5-6i)=0 \mathrm{التربيعية} \mathrm{المعادلة}\end{array}$

(6)- كون المعادلة التربيعية التي جذراها: $h=1-{\omega }^{2}i,k=1-\omega i$

$\begin{array}{rl}& h+k=(1-{\omega }^{2}i)+(1-\omega i)=(1+1)+(-{\omega }^2-\omega )i=2+i\\ & h\cdot k=(1-{\omega }^{2}i)(1-\omega i)=(1-{\omega }^3)+(-{\omega }^2-\omega )i=i\\ & x^2-(2+i)x+i=0 \mathrm{التربيعية} \mathrm{المعادلة}\end{array}$

(7)- كون المعادلة التربيعية التي جذراها: $(\frac{3i}{\omega }-\frac{2\omega }{i}),(\frac{3i}{{\omega }^2}-\frac{2{\omega }^2}{i})$

$\begin{array}{rl}& h=(\frac{3i}{\omega }-\frac{2\omega }{i})=(\frac{3{\omega }^{3}i}{\omega }-\frac{2\omega }{i}\cdot \frac{-i}{-i})=(3{\omega }^{2}i+2\omega i)=(3{\omega }^2+2\omega )i\\ & k=(\frac{3i}{{\omega }^2}-\frac{2{\omega }^2}{i})=(\frac{3{\omega }^{3}i}{{\omega }^2}-\frac{2{\omega }^2}{i}\cdot \frac{-i}{-i})=(3\omega i+2{\omega }^{2}i)=(3\omega +2{\omega }^2)i\\ & h+k=(3{\omega }^2+2\omega )i+(3\omega +2{\omega }^2)i=(5{\omega }^2+5\omega )i=5({\omega }^2+\omega )i=-5i\\ & h\cdot k=(3{\omega }^2+2\omega )i\cdot (3\omega +2{\omega }^2)i=-(9{\omega }^3+6{\omega }^4+6{\omega }^2+4{\omega }^3)\\ & h\cdot k=-\left[\right(13)+6(\omega +{\omega }^2)]=-(13-6)=-7\\ & x^2-(-5i)x+(-7)=0 \mathrm{التربيعية} \mathrm{المعادلة}\end{array}$

(8)- كون المعادلة التربيعية التي جذراها: $\frac{2}{1-{\omega }^2},\frac{2}{1-\omega }$

$$\begin{gathered} h=\frac{2}{1-{\omega }^2},k=\frac{2}{1-\omega } \\ h+k=\frac{2}{1-{\omega }^2}+\frac{2}{1-\omega }=\frac{2(1-\omega )+2(1-{\omega }^2)}{(1-{\omega }^2)(1-\omega )} \\ \begin{array}{rl}& h+k=\frac{2-2\omega +2-2{\omega }^2}{1-\omega -{\omega }^2+{\omega }^3}=\frac{4-2(\omega +{\omega }^2)}{2-\omega -{\omega }^2}=\frac{4+2}{2+1}=\frac{6}{3}=2\\ & h\cdot k=\left(\frac{2}{1-{\omega }^2}\right)\left(\frac{2}{1-\omega }\right)=\frac{4}{1-\omega -{\omega }^2+{\omega }^3}=\frac{4}{3}\\ & x^2-2x+\frac{4}{3}=0 \mathrm{التربيعة} \mathrm{المعادلة}\end{array} \end{gathered}$$

(9)- أوجد قيمة كل مما يأتي:

${(\frac{1}{3+4\omega +5{\omega }^2}-\frac{1}{3+5\omega +4{\omega }^2})}^2$

$$\begin{gathered} {(\frac{1}{3+4\omega +5{\omega }^2}-\frac{1}{3+5\omega +4{\omega }^2})}^2 \\ \begin{array}{rl}& ={(\frac{1}{3+4(-1-{\omega }^2)+5{\omega }^2}-\frac{1}{3+5(-1-{\omega }^2)+4{\omega }^2})}^2\\ & ={(\frac{1}{3-4-4{\omega }^2+5{\omega }^2}-\frac{1}{3-5-5{\omega }^2+4{\omega }^2})}^2\\ & ={(\frac{1}{-1+{\omega }^2}-\frac{1}{-2-{\omega }^2})}^2\\ & ={\left(\frac{-2-{\omega }^2-(-1+{\omega }^2)}{(-1+{\omega }^2)(-2-{\omega }^2)}\right)}^2={\left(\frac{-2+1-{\omega }^2-{\omega }^2}{2+{\omega }^2-2{\omega }^2-{\omega }^4}\right)}^2={\left(\frac{-1-2{\omega }^2}{2+(-{\omega }^2-\omega )}\right)}^2\\ & ={\left(\frac{-1-2{\omega }^2}{2+1}\right)}^2={\left(\frac{-1-2{\omega }^2}{3}\right)}^2=\frac{1+4{\omega }^2+4{\omega }^4}{9}=\frac{1+4{\omega }^2(1+{\omega }^2)}{9}\\ & =\frac{1+4{\omega }^2(-\omega )}{9}=\frac{1-4{\omega }^3}{9}=\frac{1-4}{9}=\frac{-3}{9}=\frac{-1}{3}\end{array} \end{gathered}$$

${(\sqrt{1+{\omega }^{13}}+\sqrt{1+{\omega }^{14}})}^2$

$\begin{array}{rl}& {(\sqrt{1+{\omega }^{13}}+\sqrt{1+{\omega }^{14}})}^2\\ & ={(\sqrt{1+\omega }+\sqrt{1+{\omega }^2})}^2={(\sqrt{-{\omega }^2}+\sqrt{-\omega })}^2\\ & ={(\sqrt{{\omega }^2}i+\sqrt{\omega }i)}^2={(\omega i+{\omega }^{2}i)}^2={\left(i(\omega +{\omega }^2)\right)}^2=(-i{)}^2\\ & =i^2=-1\end{array}$

$\frac{1-{\omega }^2+\omega }{1+i+\omega +\omega i}$

$$\begin{gathered} \frac{1-{\omega }^2+\omega }{1+i+\omega +\omega i} \\ =\frac{1+\omega -{\omega }^2}{1+i+\omega (1+i)}=\frac{-{\omega }^2-{\omega }^2}{(1+i)(-{\omega }^2)}=\frac{-2{\omega }^2}{(1+i)(-{\omega }^2)} \\ =\frac{2}{1+i}\cdot \frac{1-i}{1-i}=\frac{2(1-i)}{2}=1-i \end{gathered}$$

${(\frac{\sqrt{2}}{\omega }+3\sqrt{2}\omega +\sqrt{2})}^2(\frac{1}{\omega }+4\omega +1)$

$$\begin{gathered} {(\frac{\sqrt{2}}{\omega }+3\sqrt{2}\omega +\sqrt{2})}^2(\frac{1}{\omega }+4\omega +1) \\ \begin{array}{rl}& ={(\frac{\sqrt{2}{\omega }^3}{\omega }+3\sqrt{2}\omega +\sqrt{2})}^2(\frac{{\omega }^3}{\omega }+4\omega +1)\\ & ={(\sqrt{2}{\omega }^2+3\sqrt{2}\omega +\sqrt{2})}^2({\omega }^2+4\omega +1)\\ & ={\left[\sqrt{2}({\omega }^2+3\omega +1)\right]}^2({\omega }^2+4\omega +1)\\ & ={\left(\sqrt{2}[({\omega }^2+1)+3\omega ]\right)}^2({\omega }^2+1+4\omega )=\left(\sqrt{2}\right[(-\omega )+3\omega \left]{)}^2\right(-\omega +4\omega )\\ & =\left(\sqrt{2}\right(2\omega \left){)}^2\right(3\omega )=\left(8{\omega }^2\right)(3\omega )=24{\omega }^3=24\end{array} \end{gathered}$$

(10)- إذا كان $a=(\frac{-1}{2}+\frac{\sqrt{-3}}{2})$ فأثبت $(a^{12}+a^{22}+a^{23}=0)$ وكذلك $(a^9\cdot a^{16}\cdot a^{32}=1)$

$\begin{array}{rl}& \because (\frac{-1}{2}+\frac{\sqrt{-3}}{2})=\frac{-1}{2}+\frac{\sqrt{3}i}{2}=\omega \\ & a^{12}+a^{22}+a^{23}={\omega }^{12}+{\omega }^{22}+{\omega }^{23}={\left({\omega }^3\right)}^4+{\left({\omega }^3\right)}^7\omega +{\left({\omega }^3\right)}^7{\omega }^2=1+\omega +{\omega }^2=0\\ & a^9\cdot a^{16}\cdot a^{32}={\omega }^9\cdot {\omega }^{16}\cdot {\omega }^{32}=1\cdot \omega \cdot {\omega }^2={\omega }^3=1\end{array}$

(11)- برهن أن $\sqrt{\frac{1}{1+{\omega }^2}+\frac{2+{\omega }^2}{{\omega }^4}}=i$

$\sqrt{\frac{1}{1+{\omega }^2}+\frac{2+{\omega }^2}{{\omega }^4}}=\sqrt{\frac{1}{-\omega }+\frac{2+{\omega }^2}{\omega }}=\sqrt{\frac{1-(2+{\omega }^2)}{-\omega }}=\sqrt{\frac{-1-{\omega }^2}{-\omega }}=\sqrt{\frac{\omega }{-\omega }}=\sqrt{-1}=i$

(12)- برهن أن ${(\omega -{\omega }^2)}^8=81$

$\begin{array}{rl}{(\omega -{\omega }^2)}^8& ={\left[{(\omega -{\omega }^2)}^2\right]}^4={[{\omega }^2-2\omega \cdot {\omega }^2+{\omega }^4]}^4={[{\omega }^2+{\omega }^4-2]}^4=[-1-2{]}^4\\ & =[-3{]}^4={\left[\right(-3{)}^2]}^2=[9{]}^2=81\end{array}$

(13)- برهن أن ${(1+{\omega }^4)}^3+{(1-{\omega }^7-{\omega }^8)}^3=7$

$\begin{array}{rl}{(1+{\omega }^4)}^3+{(1-{\omega }^7-{\omega }^8)}^3& =(1+\omega {)}^3+{(1-\omega -{\omega }^2)}^3={(-{\omega }^2)}^3+{(1-(\omega +{\omega }^2))}^3\\ & =(-{\omega }^6)+(1-(-1){)}^3=-1+(2{)}^3=-1+8=7\end{array}$

(14)- أوجد الناتج $(\frac{1}{{\omega }^4}-\frac{1}{{\omega }^2})(2{\omega }^6+\frac{2}{\omega })\left(\frac{-{\omega }^6}{1+{\omega }^5}\right)$

$\begin{array}{rl}& (\frac{1}{{\omega }^4}-\frac{1}{{\omega }^2})(2{\omega }^6+\frac{2}{\omega })\left(\frac{-{\omega }^6}{1+{\omega }^5}\right)=(\frac{1}{\omega }-\frac{1}{{\omega }^2})(2+\frac{2}{\omega })\left(\frac{-1}{1+{\omega }^2}\right)\\ & =(\frac{{\omega }^3}{\omega }-\frac{{\omega }^3}{{\omega }^2})(2+\frac{2{\omega }^3}{\omega })\left(\frac{-1}{-\omega }\right)=({\omega }^2-\omega )(2+2{\omega }^2)\left(\frac{{\omega }^3}{\omega }\right)\\ & =({\omega }^2-\omega )(2+2{\omega }^2)\left({\omega }^2\right)=({\omega }^2-\omega )[2\left({\omega }^2\right)+2{\omega }^2\left({\omega }^2\right)]=({\omega }^2-\omega )[2{\omega }^2+2\omega ]\\ & =({\omega }^2-\omega )\left[2({\omega }^2+\omega )\right]=({\omega }^2-\omega )\left[2\right(-1\left)\right]=-2(\pm \sqrt{3}i)=\mp 2\sqrt{3}i\end{array}$

(15)- جد ناتج $x,y$ والتي تحقق المعادلة التالية: $x+yi=\frac{-8}{{\omega }^2}$

يحل بطريقتين:

الطريقة الأولى:

$\begin{array}{rl}& x+yi=\frac{-8}{{\omega }^2}=-8\omega =-8(\frac{-1}{2}+\frac{\sqrt{3i}}{2})=4-4\sqrt{3}i\\ & x+yi=4-4\sqrt{3}i\Rightarrow x=4,y=-4\sqrt{3}\end{array}$

الطريقة الثانية:

$\begin{array}{rl}& x+yi=\frac{-8}{{\omega }^2}=\frac{-8}{\frac{-1}{2}-\frac{\sqrt{3}i}{2}}=\frac{-8}{\frac{-1}{2}-\frac{\sqrt{3}i}{2}}\times \frac{\frac{-1}{2}+\frac{\sqrt{3}i}{2}}{\frac{-1}{2}+\frac{\sqrt{3}i}{2}}=\frac{4-4\sqrt{3}i}{{\left(\frac{1}{2}\right)}^2+{\left(\frac{\sqrt{3}}{2}\right)}^2}\\ & x+yi=\frac{4-4\sqrt{3}i}{\frac{1}{4}+\frac{3}{4}}=\frac{4-4\sqrt{3}i}{1}=4-4\sqrt{3}i\Rightarrow x+yi=4-4\sqrt{3}i\\ & x=4,y=-4\sqrt{3}\end{array}$

طريقة حل المسائل التي تحتوي على w فهناك بعض الطرق الأساسية التي تستخدم في تبسيط حل المسائل وهي كالآتي:

الطريقة الأولى: إيجاد العامل المشترك

(16)- جد قيمة ما يأتي:

$\sqrt{\frac{2+11\omega +11{\omega }^2}{2-5\omega -5{\omega }^2}}$

$\sqrt{\frac{2+11\omega +11{\omega }^2}{2-5\omega -5{\omega }^2}}=\sqrt{\frac{2+11(\omega +{\omega }^2)}{2-5(\omega +{\omega }^2)}}=\sqrt{\frac{2-11}{2+5}}=\sqrt{\frac{-9}{7}}=\frac{3i}{\sqrt{7}}$

$\sqrt{\frac{1+10\omega +10{\omega }^2}{1-3\omega -3{\omega }^2}}$

$\sqrt{\frac{1+10\omega +10{\omega }^2}{1-3\omega -3{\omega }^2}}=\sqrt{\frac{1+10(\omega +{\omega }^2)}{1-3(\omega +{\omega }^2)}}=\sqrt{\frac{1-10}{1+3}}=\sqrt{\frac{-9}{4}}=\frac{3i}{2}$

(17)- جد قيم $x,y$ التي تحقق المعادلة: $x+yi={(\sqrt{1+{\omega }^{32}}+\sqrt{1+{\omega }^{61}})}^2-\frac{3+i}{1+i}$

$\begin{array}{rl}& x+yi={(\sqrt{1+{\omega }^{32}}+\sqrt{1+{\omega }^{61}})}^2-\frac{3+i}{1+i}={(\sqrt{1+{\omega }^2}+\sqrt{1+\omega })}^2-\frac{3+i}{1+i}\times \frac{1-i}{1-i}\\ & x+yi={(\sqrt{-\omega }+\sqrt{-{\omega }^2})}^2-\frac{3-3i+i-i^2}{1^2+1^2}={(\sqrt{-\omega {\omega }^3}+i\omega )}^2-\frac{4-2i}{2}\\ & x+yi={(\sqrt{-{\omega }^4}+i\omega )}^2-\frac{4}{2}-\frac{2i}{2}={(i{\omega }^2+i\omega )}^2-\frac{4-2i}{2}={\left[i({\omega }^2+\omega )\right]}^2-(2-i)\\ & x+yi=\left[i\right(-1){]}^2-(2-i)=-1-(2-i)=-3+i\\ & x=-3,y=1\end{array}$

(18)- جد قيم $x,y$ التي تحقق المعادلة: $x\omega +y\omega i={\left(\frac{i{\omega }^2+i}{{\omega }^2}\right)}^{\frac{1}{2}}$

$\begin{array}{rl}& x\omega +y\omega i={\left(\frac{i{\omega }^2+i}{{\omega }^2}\right)}^{\frac{1}{2}}\Rightarrow \omega (x+yi)={\left(\frac{i({\omega }^2+1)}{{\omega }^2}\right)}^{\frac{1}{2}}\\ & \Rightarrow \omega (x+yi)={\left(\frac{i(-\omega )}{{\omega }^2}\right)}^{\frac{1}{2}}={\left(\frac{-i}{\omega }\right)}^{\frac{1}{2}}\Rightarrow \omega (x+yi)={\left(\frac{-i\left({\omega }^3\right)}{\omega }\right)}^{\frac{1}{2}}\\ & \Rightarrow \omega (x+yi)={(-i{\omega }^2)}^{\frac{1}{2}}=\sqrt{-i{\omega }^2}\Rightarrow \omega (x+yi)=\omega \sqrt{-i}\\ & (x+yi)=\sqrt{-i}\text{بلتربيع}\\ & (x+yi{)}^2=-i\\ & x^2+2xyi-y^2=0-i\Rightarrow x^2-y^2+2xyi=0-i\\ & x^2-y^2=0\dots \dots \text{(1)}\\ & 2xy=-1\dots \dots \left(2\right)\\ & y=\frac{-1}{2x}\dots \dots .(\ast ) \left(\left(1\right) \mathrm{في} (*) \mathrm{معادلة} \mathrm{نعوض}\right)\\ & x^2-\frac{1}{4x^2}=0\stackrel{4x^2\times \mathrm{نضرب}}{⟹}4x^2-1=0\Rightarrow (2x^2-1)(2x^2+1)=0\\ & \text{either }2x^2-1=0\Rightarrow x^2=\frac{1}{2}\Rightarrow x=\pm \frac{1}{\sqrt{2}}\Rightarrow y=\frac{-1}{2(\pm \frac{1}{\sqrt{2}})}=\frac{-1}{\sqrt{2}\sqrt{2}(\pm \frac{1}{\sqrt{2}})}=\mp \frac{1}{\sqrt{2}}\\ & \text{or}2x^2+1=0\Rightarrow 2x^2=-1 \mathrm{يهمل}\\ & \frac{-1}{\sqrt{2}}+\frac{1}{\sqrt{2}}i,\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{2}}i \mathrm{هما} \mathrm{الجذران}\end{array}$

(19)- جد قيمة الآتي: $(1-i{)}^4-{(5+3\omega +5{\omega }^2)}^3$

$\begin{array}{rl}& (1-i{)}^4-{(5+3\omega +5{\omega }^2)}^3={\left[\right(1-i{)}^2]}^2-(5+3\omega +5(-1-\omega ){)}^3\\ & {[1-2i+i^2]}^2-(5+3\omega -5-5\omega {)}^3=[-2i{]}^2-(-2\omega {)}^3=4i^2+8{\omega }^3=-4+8=4\end{array}$

(20)- كون المعادلة التربيعية التي جدراها ${(2\omega +2{\omega }^2-1)}^2,{(2-2\omega -2{\omega }^2)}^2$

$$\begin{gathered} \begin{array}{rl}& {(2\omega +2{\omega }^2-1)}^2=(2\omega +2(-1-\omega )-1{)}^2=(2\omega -2-2\omega -1{)}^2=(-3{)}^2=9 \mathrm{الأول} \mathrm{الجذر}\\ & {(2-2\omega -2{\omega }^2)}^2=(2-2\omega -2(-1-\omega ){)}^2=(2-2\omega +2+2\omega {)}^2=(4{)}^2=16 \mathrm{الثاني} \mathrm{الجذر}\end{array} \\ 9+16=25 \mathrm{الجذرين} \mathrm{مجموع} \\ \left(9\right)\cdot \left(16\right)=144 \mathrm{الجذرين} \mathrm{ضرب} \\ \therefore x^2-(\mathrm{الجذرين} \mathrm{مجموع})x+\mathrm{الجرين} \mathrm{ضرب} \mathrm{حاصل}=0 \\ \therefore x^2-25x+144=0 \mathrm{التربيعة} \mathrm{المعادلة} \end{gathered}$$

الطريقة الثانية: طريقة الاستبدال

(21)- جد ناتج ${(3+2\omega +4{\omega }^2)}^2$

$\begin{array}{rl}{(3+2\omega +4{\omega }^2)}^2& =[3+2\omega +4(-1-\omega ){]}^2=(3+2\omega -4-4\omega {)}^2\\ & =(-1-2\omega {)}^2=1+4\omega +4{\omega }^2=1+4(\omega +{\omega }^2)=1-4=-3\end{array}$

الطريقة الثالثة: معاملات البسط والمقام متساوية

(22)- جد ناتج $\frac{10\omega +3}{3{\omega }^2+10}$

$\frac{10\omega +3}{3{\omega }^2+10}=\frac{10\omega +3{\omega }^3}{3{\omega }^2+10}=\frac{\omega (10+3{\omega }^2)}{3{\omega }^2+10}=\omega$

(23)- أثبت أن ${(\frac{a-b{\omega }^2}{a\omega -b}-\frac{d-c\omega }{d{\omega }^2-c})}^4=9$

$\begin{array}{rl}& {(\frac{a-b{\omega }^2}{a\omega -b}-\frac{d-c\omega }{d{\omega }^2-c})}^4={(\frac{a-b{\omega }^2}{a\omega -b{\omega }^3}-\frac{d{\omega }^3-c\omega }{d{\omega }^2-c})}^4\\ & ={(\frac{a-b{\omega }^2}{\omega (a-b{\omega }^2)}-\frac{\omega (d{\omega }^2-c)}{d{\omega }^2-c})}^4={(\frac{1}{\omega }-\omega )}^4={({\omega }^2-\omega )}^4=(\pm \sqrt{3}i{)}^4\\ & ={\left[\right(\pm \sqrt{3}i{)}^2]}^2={\left(3i^2\right)}^2=(-3{)}^2=9\end{array}$

الطريقة الرابعة: إيجاد المضاعف المشترك

(24)- أثبت أن ${(\frac{5}{3-\omega }-\frac{5}{3-{\omega }^2})}^2=\frac{-75}{169}$

$$\begin{gathered} \begin{array}{rl}& {(\frac{5}{3-\omega }-\frac{5}{3-{\omega }^2})}^2=25{(\frac{1}{3-\omega }-\frac{1}{3-{\omega }^2})}^2=\left(25\right){\left(\frac{(3-{\omega }^2)-(3-\omega )}{(3-\omega )(3-{\omega }^2)}\right)}^2\\ & =\left(25\right){\left(\frac{3-{\omega }^2-3+\omega }{9-3{\omega }^2-3\omega +{\omega }^3}\right)}^2=\left(25\right){\left(\frac{-{\omega }^2+\omega }{10-3{\omega }^2-3\omega }\right)}^2\end{array} \\ \begin{array}{rl}& =\left(25\right){\left(\frac{-{\omega }^2+\omega }{10+3(-{\omega }^2-\omega )}\right)}^2=\left(25\right){\left(\frac{-{\omega }^2+\omega }{10+3\left(1\right)}\right)}^2=\left(25\right){\left(\frac{-{\omega }^2+\omega }{13}\right)}^2\\ & =\left(25\right)\left(\frac{{\omega }^4-2{\omega }^3+{\omega }^2}{169}\right)=\left(25\right)\left(\frac{{\omega }^4+{\omega }^2-2}{169}\right)=\left(25\right)\left(\frac{({\omega }^4+{\omega }^2)-2}{169}\right)\\ & =\left(25\right)\left(\frac{(\omega +{\omega }^2)-2}{169}\right)=\left(25\right)\left(\frac{(-1)-2}{169}\right)=\left(25\right)\left(\frac{-3}{169}\right)=\frac{-75}{169}\end{array} \end{gathered}$$

(25)- جد قيمة x $4^x-2^{x-\frac{1}{2}+\frac{\sqrt{3}}{2}i}-2^{x-\frac{1}{2}-\frac{\sqrt{3}}{2}i}+\frac{1}{2}=0$

$\begin{array}{rl}& 4^x-2^{x+\omega }-2^{x+{\omega }^2}+2^{-1}=0\\ & {\left(\right[2{]}^2)}^x-2^{x+\omega }-2^{x+{\omega }^2}+2^{\omega +{\omega }^2}=0 \mathrm{الأسس} \mathrm{نجزأ}\\ & 2^{2x}-2^x{2}^{\omega }-2^x{2}^{{\omega }^2}+2^{\omega }{2}^{{\omega }^2}=0\\ & 2^x(2^x-2^{\omega })-2^{{\omega }^2}(2^x-2^{\omega })=0\\ & (2^x-2^{\omega })(2^x-2^{{\omega }^2})=0\\ & \text{either }(2^x-2^{\omega })=0\Rightarrow 2^x=2^{\omega }\Rightarrow x=\omega \\ & \text{or}(2^x-2^{{\omega }^2})=0\Rightarrow 2^x=2^{{\omega }^2}\Rightarrow x={\omega }^2\end{array}$