ضع كلاً مما يأتي بالصيغة العادية للعدد المركب:

$(2+3i{)}^2+(12+2i)$

$\begin{array}{rl}(2+3i{)}^2+(12+2i)& =4+12i+9i^2+(12+2i)=(-5+12i)+(12+2i)\\ & =7+14i\end{array}$

تمارين (1-1)

(1)- ضع كلاً مما يأتي بالصيغة العادية للعدد المركب:

$i^5$

$i^5=i^4\cdot i=\left(1\right).i=i=0+i$

$i^6$

$i^6=i^4\cdot i^2=\left(1\right)\cdot (-1)=-1=-1+0i$

$i^{124}$

$i^{124}={\left(i^4\right)}^{31}=(1{)}^{31}=1=1+0i$

$i^{999}$

$i^{999}={\left(i^4\right)}^{249}{i}^3=\left(1\right)\cdot i^2\cdot i=-i=0-i$

$i^{4n+1}$

$i^{4n+1}=i^{4n}\cdot i={\left(i^4\right)}^n\cdot i=\left(1\right)\cdot i=i=0+i$

$(2+3i{)}^2+(12+2i)$

$\begin{array}{rl}(2+3i{)}^2+(12+2i)& =4+12i+9i^2+(12+2i)=(-5+12i)+(12+2i)\\ & =7+14i\end{array}$

$(10+3i)(0+6i)$

$(10+3i)(0+6i)=0+60i+0+18i^2=-18+60i$

$(1+i{)}^4-(1-i{)}^4$

$$\begin{gathered} (1+i{)}^4-(1-i{)}^4={\left[\right(1+i{)}^2]}^2-{\left[\right(1-i{)}^2]}^2 \\ ={[1+2i+i^2]}^2-{[1-2i+i^2]}^2=(2i{)}^2-(-2i{)}^2=4i^2-4i^2=0=0+0i \end{gathered}$$

$\frac{12+i}{i}\times \frac{-i}{-i}$

$\frac{12+i}{i}\times \frac{-i}{-i}=\frac{-12i-i^2}{-i^2}=\frac{1-12i}{1}=1-12i$

$\frac{3+4i}{3-4i}$

$\frac{3+4i}{3-4i}=\frac{3+4i}{3-4i}\times \frac{3+4i}{3+4i}=\frac{9+12i+12i+16i^2}{3^2+4^2}=\frac{-7+24i}{9+16}=\frac{-7+24i}{25}=\frac{-7}{25}+\frac{24}{25}i$

$\frac{i}{2+3i}$

$\frac{i}{2+3i}=\frac{i}{2+3i}\times \frac{2-3i}{2-3i}=\frac{2i-3i^2}{2^2+3^2}=\frac{3+2i}{4+9}=\frac{3+2i}{13}=\frac{3}{13}+\frac{2}{13}i$

${\left(\frac{3+i}{1+i}\right)}^3$

${\left(\frac{3+i}{1+i}\right)}^3={(\frac{3+i}{1+i}\times \frac{1-i}{1-i})}^3$

ملاحظة: عدد مركب بصورة كسرية مرفوع لقوة يضرب داخل القوس في المرافق أولاً ثم يبسط وبعدها نتخلص من القوة المرفوع لها.

$\begin{array}{rl}& ={\left(\frac{3-3i+i-i^2}{1^2+1^2}\right)}^3={\left(\frac{4-2i}{2}\right)}^3=(2-i{)}^3=(2-i{)}^2(2-i)\\ & =(4-4i+i^2)(2-i)=(3-4i)(2-i)=6-3i-8i+4i^2\\ & =2-11i\end{array}$

$\frac{2+3i}{1-i}\times \frac{1+4i}{4+i}$

$$\begin{gathered} \frac{2+3i}{1-i}\times \frac{1+4i}{4+i}=\frac{2+8i+3i+12i^2}{4+i-4i-i^2}=\frac{-10+11i}{5-3i}\times \frac{5+3i}{5+3i}=\frac{-50-30i+55i+33i^2}{5^2+3^2} \\ =\frac{-83+25i}{25+9}=\frac{-83+25i}{34}=\frac{-83}{34}+\frac{25}{34}i \end{gathered}$$

$(1+i{)}^3+(1-i{)}^3$

$\begin{array}{rl}& (1+i{)}^3+(1-i{)}^3\\ & =(1+i{)}^2(1+i)+(1-i{)}^2(1-i)\\ & =(1+2i+i^2)(1+i)+(1-2i+i^2)(1-i)=\left(2i\right)(1+i)+(-2i)(1-i)\\ & =2i+2i^2-2i+2i^2=-4=-4+0i\end{array}$

(2)- جد قيمة كل من $x,y$ الحقيقيتين اللتين تحققان المعادلة الآتية:

$y+5i=(2x+i)(x+2i)$

$$\begin{gathered} \begin{array}{rl}& y+5i=2x^2+4xi+xi+2i^2\Rightarrow y+5i=(2x^2-2)+(4x+x)i\\ & y+5i=(2x^2-2)+5xi\end{array} \\ \begin{array}{rl}& y=(2x^2-2)\dots \dots \left(1\right)\\ & 5=5x\dots \dots .\left(2\right)\Rightarrow \frac{5}{5}=\frac{5x}{5}\Rightarrow x=1\end{array} \end{gathered}$$

نعوض في المعادلة (1)

$y=2(1{)}^2-2=2-2=0\Rightarrow y=0$

$8i=(x+2i)(y+2i)+1$

$\begin{array}{rl}& 8i=xy+2xi+2yi+4i^2+1\\ & 8i=(xy-3)+(2x+2y)i\\ & xy-3=0\Rightarrow xy=3\Rightarrow y=\frac{3}{x}\dots \dots .\left(1\right)\\ & 2x+2y=8]÷2\Rightarrow x+y=4\Rightarrow y=4-x\dots \dots .(2)\end{array}$

نعوض (1) في (2)

$$\begin{gathered} 4-x=\frac{3}{x}\stackrel{x\times }{⟹}x(4-x)=3 \\ \begin{array}{rl}& 4x-x^2=3\Rightarrow x^2-4x+3=0\\ & (x-3)(x-1)=0\end{array} \end{gathered}$$

إما: $x-3=0\Rightarrow x=3$

نعوض في المعادلة (2)

$y=1$

أو: $x-1=0\Rightarrow x=1$

نعوض في المعادلة (2)

$y=3$

$\left(\frac{1-i}{1+i}\right)+(x+yi)=(1+2i{)}^2$

$\begin{array}{rl}& (x+yi)=(1+2i{)}^2-\left(\frac{1-i}{1+i}\right)\Rightarrow (x+yi)=(1+4i+4i^2)-(\frac{1-i}{1+i}\times \frac{1-i}{1-i})\\ & (x+yi)=-3+4i-\left(\frac{1-i-i+i^2}{1^2+1^2}\right)\\ & (x+yi)=-3+4i-\left(\frac{-2i}{2}\right)\Rightarrow (x+yi)=-3+4i+i\\ & x+yi=-3+5i\Rightarrow x=-3 , y=5\end{array}$

$\frac{2-i}{1+i}x+\frac{3-i}{2+i}y=\frac{1}{i}$

$\begin{array}{rl}& [\frac{2-i}{1+i}\times \frac{1-i}{1-i}]x+[\frac{3-i}{2+i}\times \frac{2-i}{2-i}]y=\frac{i^4}{i}\\ & \left[\frac{2-2i-i+i^2}{1^2+1^2}\right]x+\left[\frac{6-3i-2i+i^2}{2^2+1^2}\right]y=i^3\\ & \left[\frac{1-3i}{2}\right]x+\left[\frac{5-5i}{5}\right]y=-i]\times 10\Rightarrow 5(1-3i)x+2(5-5i)y=-10i\\ & 5x-15xi+10y-10yi=0-10i\\ & 5x+10y=0...\left(1\right)\\ & -15x-10y=-10...\left(2\right)\\ & -10x=-10\Rightarrow x=1\end{array}$

نعوض في المعادلة (1)

$5\left(1\right)+10y=0\Rightarrow 10y=-5\Rightarrow y=\frac{-5}{10}\Rightarrow y=\frac{-1}{2}$

(3)- أثبت أن:

$\frac{1}{(2-i{)}^2}-\frac{1}{(2+i{)}^2}=\frac{8}{25}i$

$\begin{array}{rl}& \mathbf{\text{LHS }}\frac{1}{(2-i{)}^2}-\frac{1}{(2+i{)}^2}=\frac{1}{4-4i+i^2}-\frac{1}{4+4i+i^2}\\ & =\frac{1}{4-4i-1}-\frac{1}{4+4i-1}=\frac{1}{3-4i}\times \frac{3+4i}{3+4i}-\frac{1}{3+4i}\times \frac{3-4i}{3-4i}\\ & =\frac{3+4i}{9+16}-\frac{3-4i}{9+16}=\frac{3+4i}{25}-\frac{3-4i}{25}=\frac{3+4i-3+4i}{25}=\frac{8}{25}i \mathbf{\text{RHS}}\end{array}$

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

$$\begin{gathered} \mathbf{\text{LHS }}\frac{(1-i{)}^2}{1+i}+\frac{(1+i{)}^2}{1-i}=\frac{1-2i+i^2}{1+i}+\frac{1+2i+i^2}{1-i}=\frac{-2i}{1+i}+\frac{2i}{1-i} \\ \begin{array}{rl}& =\frac{-2i}{1+i}\times \frac{1-i}{1-i}+\frac{2i}{1-i}\times \frac{1+i}{1+i}=\frac{-2i+2i^2}{1^2+1^2}+\frac{2i+2i^2}{1^2+1^2}\\ & =\frac{-2i-2}{2}+\frac{2i-2}{2}=\frac{-2i}{2}-\frac{2}{2}+\frac{2i}{2}-\frac{2}{2}=-i-1+i-1=-2 \mathbf{\text{RHS}}\end{array} \end{gathered}$$

(4)- حلل كلاً من الأعداد $29 ، 125 \text{، }41 \text{، }85$ إلى حاصل ضرب عاملين من الصورة $a+bi$ حيث $a,b$ عددان نسبيان.

$\begin{array}{rl}& 85=81+4=81-4i^2=(9-2i)(9+2i)\\ & 41=25+16=25-16i^2=(5-4i)(5+4i)\\ & 125=121+4=121-4i^2=(11-2i)(11+2i)\\ & 29=25+4=25-4i^2=(5-2i)(5+2i)\end{array}$

(5)- جد قيمة $x,y$ الحقيقيتين إذا علمت أن $\frac{3+i}{2-i} , \frac{6}{x+yi}$ مترافقان.

$\begin{array}{rl}& \frac{6}{x+yi}=\frac{3-i}{2+i}\Rightarrow (x+yi)(3-i)=12+6i\Rightarrow x+yi=\frac{12+6i}{3-i}\\ & x+yi=\frac{12+6i}{3-i}\times \frac{3+i}{3+i}=\frac{36+12i+18i+6i^2}{3^2+1^2}=\frac{36-6+30i}{9+1}\\ & x+yi=\frac{30+30i}{10}\Rightarrow x+yi=3+3i\\ & x=3 , y=3\end{array}$