Question#01: Identify the following statement as true or false.
\[\begin{align}
\left( \text{i} \right) \qquad&\sqrt{-3}\sqrt{-3}=3\quad\textbf{False} \\
\left( \text{ii} \right) \qquad&{{i}^{73}}=-i\quad \textbf{False} \\
\left( \text{iii} \right) \qquad&{{i}^{10}}=-1\quad \textbf{True} \\
\left( \text{iv} \right) \qquad&\text{Complex}\,\text{conjugate}\,\text{of}\,\left( -6i+{{i}^{2}} \right)\text{ is }\left( -1+6i \right) \quad \textbf{True} \\
\left( \text{v} \right) \qquad&\text{Difference of a complex number }z=a+bi\text{ and its conjugate}\\
& \text{is a real number.}\quad \textbf{False} \\
\left( \text{vi} \right) \qquad&\text{If }\left( a-1 \right)-\left( b+3 \right)i=5+8i,\text{ then }a=6\text{ and }b=-11. \quad \textbf{True} \\
\left( \text{vii} \right) \qquad&\text{Product of a complex number and its conjugate is always a}\\
&\text{non-negative real number.}\quad \textbf{True} \\
\end{align}\]
Question#02: Express each complex number in the standard form \(a+bi\), where \(a\) and \(b\) are real numbers.
\(i\)
\[\left( 2+3i \right)+\left( 7-2i \right)\]
Solution:
\[\begin{align}
& =\left( 2+3i \right)+\left( 7-2i \right) \\
& =2+3i+7-2i \\
& =9+i \\
\end{align}\]
\(ii\)
\[2(5+4i)-3(7+4i)\]
Solution:
\[\begin{align}
& =2(5+4i)-3(7+4i) \\
& =10+8i-21-12i \\
& =-11-4i \\
\end{align}\]
\(iii\)
\[-(-3+5i)-(4+9i)\]
Solution:
\[\begin{align}
& =-(-3+5i)-(4+9i) \\
& =3-5i-4-9i \\
& =-1-14i \\
\end{align}\]
\(iv\)
\[2{{i}^{2}}+6{{i}^{3}}+3{{i}^{16}}-6{{i}^{19}}+4{{i}^{25}}\]
Solution:
\[\begin{align}
& =2{{i}^{2}}+6{{i}^{3}}+3{{i}^{16}}-6{{i}^{19}}+4{{i}^{25}} \\
& =2\left( -1 \right)+6{{i}^{2}}.i+3{{\left( {{i}^{2}} \right)}^{8}}-6{{i}^{18}}.i+4{{i}^{24}}.i \\
& =-2+6\left( -1 \right).i+3{{\left( -1 \right)}^{8}}-6{{\left( -1 \right)}^{9}}.i+4{{\left( -1 \right)}^{12}}.i \\
& =-2-6i+3\left( 1 \right)-6\left( -1 \right).i+4\left( 1 \right).i \\
& =-2-\cancel{6i}+3+\cancel{6i}+4i \\
& =1+4i \\
\end{align}\]
Question#03: Simplify and write your answer in the form \(a+bi\).
\(i\)
\[\left( -7+3i \right)\left( -3+2i \right)\]
Solution:
\[\begin{align}
& =\left( -7+3i \right)\left( -3+2i \right) \\
& =-7\left( -3+2i \right)+3i\left( -3+2i \right) \\
& =21-14i-9i+6{{i}^{2}} \\
& =21-23i+6\left( -1 \right) \\
& =21-23i-6 \\
& =15-23i \\
\end{align}\]
\(ii\)
\[\left( 2-\sqrt{-4} \right)\left( 3-\sqrt{-4} \right)\]
Solution:
\[\begin{align}
& =\left( 2-\sqrt{-4} \right)\left( 3-\sqrt{-4} \right) \\
& =\left( 2-2i \right)\left( 3-2i \right) \\
& =2\left( 3-2i \right)-2i\left( 3-2i \right) \\
& =6-4i-6i+4{{i}^{2}} \\
& =6-10i+4\left( -1 \right) \\
& =6-10i-4 \\
& =2-10i \\
\end{align}\]
\(iii\)
\[{{\left( \sqrt{5}-3i \right)}^{2}}\]
Solution:
\[\begin{align}
& ={{\left( \sqrt{5}-3i \right)}^{2}} \\
& ={{\left( \sqrt{5} \right)}^{2}}+{{\left( 3i \right)}^{2}}-2\left( \sqrt{5} \right)\left( 3i \right) \\
& =5+9{{i}^{2}}-6\sqrt{5}i \\
& =5+9\left( -1 \right)-6\sqrt{5}i \\
& =5-9-6\sqrt{5}i \\
& =-4-6\sqrt{5}i \\
\end{align}\]
\(iv\)
\[\left( 2-3i \right)\left( \overline{3-2i} \right)\]
Solution:
\[\begin{align}
& =\left( 2-3i \right)\left( \overline{3-2i} \right) \\
& =\left( 2-3i \right)\left( 3+2i \right) \\
& =2\left( 3+2i \right)-3i\left( 3+2i \right) \\
& =6+4i-9i-6{{i}^{2}} \\
& =6-5i-6\left( -1 \right) \\
& =6-5i+6 \\
& =12-5i \\
\end{align}\]
Question#04: Simplify and write your answer in the form \(a+bi\).
\(i\)
\[\frac{-2}{1+i}\]
Solution:
\[\begin{align}
& =\frac{-2}{1+i} \\
& =\frac{-2}{1+i}\times \frac{1-i}{1-i} \\
& =\frac{-2\left( 1-i \right)}{{{1}^{2}}-{{i}^{2}}} \\
& =\frac{-2\left( 1-i \right)}{1-\left( -1 \right)} \\
& =\frac{-2\left( 1-i \right)}{1+1} \\
& =\frac{-\cancel{2}\left( 1-i \right)}{\cancel{2}} \\
& =-\left( 1-i \right) \\
& =1+i \\
\end{align}\]
\(ii\)
\[\frac{2+3i}{4-i}\]
Solution:
\[\begin{align}
& =\frac{2+3i}{4-i} \\
& =\frac{2+3i}{4-i}\times \frac{4+i}{4+i} \\
& =\frac{\left( 2+3i \right)\left( 4+i \right)}{{{4}^{2}}-{{i}^{2}}} \\
& =\frac{2\left( 4+i \right)+3i\left( 4+i \right)}{16-\left( -1 \right)} \\
& =\frac{8+2i+12i+3{{i}^{2}}}{16+1} \\
& =\frac{8+14i+3\left( -1 \right)}{17} \\
& =\frac{8+14i-3}{17} \\
& =\frac{5+14i}{17} \\
& =\frac{5}{17}+\frac{14}{17}i \\
\end{align}\]
\(iii\)
\[\frac{9-7i}{3+i}\]
Solution:
\[\begin{align}
& =\frac{9-7i}{3+i} \\
& =\frac{9-7i}{3+i}\times \frac{3-i}{3-i} \\
& =\frac{\left( 9-7i \right)\left( 3-i \right)}{{{3}^{2}}-{{i}^{2}}} \\
& =\frac{9\left( 3-i \right)-7i\left( 3-i \right)}{9-\left( -1 \right)} \\
& =\frac{27-9i-21i+7{{i}^{2}}}{9+1} \\
& =\frac{27-30i+7\left( -1 \right)}{10} \\
& =\frac{27-30i-7}{10} \\
& =\frac{20-30i}{10} \\
& =\frac{\cancel{10}\left( 2-3i \right)}{\cancel{10}} \\
& =2-3i \\
\end{align}\]
\(iv\)
\[\frac{2-6i}{3+i}-\frac{4+i}{3+i}\]
Solution:
\[\begin{align}
& =\frac{2-6i}{3+i}-\frac{4+i}{3+i} \\
& =\frac{2-6i-\left( 4+i \right)}{3+i} \\
& =\frac{2-6i-4-i}{3+i} \\
& =\frac{-2-7i}{3+i} \\
& =\frac{-2-7i}{3+i}\times \frac{3-i}{3-i} \\
& =\frac{\left( -2-7i \right)\left( 3-i \right)}{{{3}^{2}}-{{i}^{2}}} \\
& =\frac{-2\left( 3-i \right)-7i\left( 3-i \right)}{9-\left( -1 \right)} \\
& =\frac{-6+2i-21i+7{{i}^{2}}}{9+1} \\
& =\frac{-6-19i+7\left( -1 \right)}{10} \\
& =\frac{-6-19i-7}{10} \\
& =\frac{-13-19i}{10} \\
& =-\frac{13}{10}-\frac{19}{10}i \\
\end{align}\]
\(v\)
\[{{\left( \frac{1+i}{1-i} \right)}^{2}}\]
Solution:
\[\begin{align}
& ={{\left( \frac{1+i}{1-i} \right)}^{2}} \\
& =\frac{{{\left( 1+i \right)}^{2}}}{{{\left( 1-i \right)}^{2}}} \\
& =\frac{{{1}^{2}}+{{i}^{2}}+2i}{{{1}^{2}}+{{i}^{2}}-2i} \\
& =\frac{1+\left( -1 \right)+2i}{1+\left( -1 \right)-2i} \\
& =\frac{\cancel{1}-\cancel{1}+2i}{\cancel{1}-\cancel{1}-2i} \\
& =\frac{\cancel{2i}}{-\cancel{2i}} \\
& =-1 \\
& =-1+0i \\
\end{align}\]
\(vi\)
\[\frac{1}{\left( 2+3i \right)\left( 1-i \right)}\]
Solution:
\[\begin{align}
& =\frac{1}{\left( 2+3i \right)\left( 1-i \right)} \\
& =\frac{1}{2\left( 1-i \right)+3i\left( 1-i \right)} \\
& =\frac{1}{2-2i+3i-3{{i}^{2}}} \\
& =\frac{1}{2+i-3\left( -1 \right)} \\
& =\frac{1}{2+i+3} \\
& =\frac{1}{5+i} \\
& =\frac{1}{5+i}\times \frac{5-i}{5-i} \\
& =\frac{5-i}{{{5}^{2}}-{{i}^{2}}} \\
& =\frac{5-i}{25-\left( -1 \right)} \\
& =\frac{5-i}{25+1} \\
& =\frac{5-i}{26} \\
& =\frac{5}{26}-\frac{1}{26}i \\
\end{align}\]
Question#05: Calculate \(\overline{z},z+\overline{z},z-\overline{z},z\overline{z}\) for each of the following.
\(i\)
\[z=-i\]
Solution:
\[\begin{align}
&\left( a \right) &\overline{z}&=i \\
&\left( b \right) &z+\overline{z}&=-i+i \\
&&& =0 \\
& \left( c \right) &z-\overline{z}&=-i-i \\
&&& =-2i \\
&\left( d \right) &z\overline{z}&=\left( -i \right)\left( i \right) \\
&&& =-{{i}^{2}} \\
&&& =-\left( -1 \right) \\
&&& =1 \\
\end{align}\]
\(ii\)
\[z=2+i\]
Solution:
\[\begin{align}
&\left( a \right) &\overline{z}&=2-i \\
&\left( b \right) &z+\overline{z}&= \left( 2+i \right)+\left( 2-i \right) \\
&&& =2+\cancel{i}+2-\cancel{i} \\
&&&=4 \\
& \left( c \right) &z-\overline{z}&=\left( 2+i \right)-\left( 2-i \right) \\
&&& =2+i-2+i \\
&&& =2i \\
&\left( d \right) &z\overline{z}&=\left( 2+i \right)\left( 2-i \right) \\
&&& ={{2}^{2}}-{{i}^{2}} \\
&&& =4-\left( -1 \right) \\
&&& =4+1 \\
&&& =5 \\
\end{align}\]
\(iii\)
\[z=\frac{1+i}{1-i}\]
Solution:
\[\begin{align}
&& z&=\frac{1+i}{1-i}\times \frac{1+i}{1+i} \\
&& & =\frac{{{\left( 1+i \right)}^{2}}}{{{1}^{2}}-{{i}^{2}}} \\
&&& =\frac{{{1}^{2}}+{{i}^{2}}+2i}{1-\left( -1 \right)} \\
&& & =\frac{\cancel{1}-\cancel{1}+2i}{1+1} \\
&& & =\frac{\cancel{2}i}{\cancel{2}} \\
&&& =i \\
&\left( a \right) &\overline{z}&=-i \\
&\left( b \right) &z+\overline{z}&= i-i \\
&&&=0 \\
& \left( c \right) &z-\overline{z}&= i-\left( -i \right) \\
&&& =i+i \\
&&& =2i \\
&\left( d \right) &z\overline{z}&=\left( i \right)\left(-i \right) \\
&&& =-{{i}^{2}} \\
&&& =-\left( -1 \right) \\
&&& =1 \\
\end{align}\]
\(iv\)
\[z=\frac{4-3i}{2+4i}\]
Solution:
\[\begin{align}
&& z&=\frac{4-3i}{2+4i}\times \frac{2-4i}{2-4i} \\
&&& =\frac{\left( 4-3i \right)\left( 2-4i \right)}{{{2}^{2}}-{{\left( 4i \right)}^{2}}} \\
& &&=\frac{4\left( 2-4i \right)-3i\left( 2-4i \right)}{4-16{{i}^{2}}} \\
&&& =\frac{8-16i-6i+12{{i}^{2}}}{4-16\left( -1 \right)} \\
& &&=\frac{8-22i+12\left( -1 \right)}{4+16} \\
& &&=\frac{8-22i-12}{20} \\
&&& =\frac{-4-22i}{20} \\
& &&=-\frac{4}{20}-\frac{22}{20}i \\
&& & =-\frac{1}{5}-\frac{11}{10}i \\
&\left( a \right) &\overline{z}&=-\frac{1}{5}+\frac{11}{10}i \\
&\left( b \right) &z+\overline{z}& =\left( -\frac{1}{5}-\frac{11}{10}i \right)+\left( -\frac{1}{5}+\frac{11}{10}i \right) \\
&&& =-\frac{1}{5}-\cancel{\frac{11}{10}i}-\frac{1}{5}+\cancel{\frac{11}{10}i} \\
&&& =-\frac{1}{5}-\frac{1}{5} \\
&&& =-\frac{2}{5} \\
& \left( c \right) &z-\overline{z}&=\left( -\frac{1}{5}-\frac{11}{10}i \right)-\left( -\frac{1}{5}+\frac{11}{10}i \right) \\
& &&=-\cancel{\frac{1}{5}}-\frac{11}{10}i+\cancel{\frac{1}{5}}-\frac{11}{10}i \\
& &&=-\frac{11}{10}i-\frac{11}{10}i \\
&&& =\frac{-11i-11i}{10} \\
&& & =\frac{-22i}{10} \\
&& & =-\frac{11}{5}i \\
&\left( d \right) &z\overline{z}&=\left( -\frac{1}{5}-\frac{11}{10}i \right)\left( -\frac{1}{5}+\frac{11}{10}i \right) \\
&&& ={{\left( -\frac{1}{5} \right)}^{2}}-{{\left( \frac{11}{10}i \right)}^{2}} \\
&& & =\frac{1}{25}-\frac{121}{100}{{i}^{2}} \\
&&& =\frac{1}{25}-\frac{121}{100}\left( -1 \right) \\
&&& =\frac{1}{25}+\frac{121}{100} \\
&&& =\frac{4+121}{100} \\
&&& =\frac{\cancelto{5}{125}}{\cancelto{4}{100}} \\
& &&=\frac{5}{4} \\
\end{align}\]
Question#06: If \(z=2+3i\) and \(w=5-4i\) show that.
\(i\)
\[\overline{z+w}=\overline{z}+\overline{w}\]
Solution:
\[\begin{align}
\text{L}\text{.H}\text{.S}\text{.}&=\overline{z+w} \\
& =\overline{2+3i+5-4i} \\
& =\overline{7-i} \\
& =7+i \\
\text{R}\text{.H}\text{.S}\text{.}&=\overline{z}+\overline{w} \\
& =\overline{2+3i}+\overline{5-4i} \\
& =2-3i+5+4i \\
& =7+i \\
\text{Hence, L}\text{.H}\text{.S.}&=\text{R}\text{.H}\text{.S}\text{.} \\
\end{align}\]
\(ii\)
\[\overline{z-w}=\overline{z}-\overline{w}\]
Solution:
\[\begin{align}
\text{L}\text{.H}\text{.S}\text{.}&=\overline{z-w} \\
& =\overline{2+3i-\left( 5-4i \right)} \\
& =\overline{2+3i-5+4i} \\
& =\overline{-3+7i} \\
& =-3-7i \\
\text{R}\text{.H}\text{.S}\text{.}&=\overline{z}-\overline{w} \\
& =\overline{2+3i}-\left( \overline{5-4i} \right) \\
& =2-3i-\left( 5+4i \right) \\
& =2-3i-5-4i \\
& =-3-7i \\
\text{Hence, L}\text{.H}\text{.S.}&=\text{R}\text{.H}\text{.S}\text{.} \\
\end{align}\]
\(iii\)
\[\overline{zw}=\overline{z}\overline{w}\]
Solution:
\[\begin{align}
\text{L}\text{.H}\text{.S}\text{.} &=\overline{zw} \\
& =\overline{\left( 2+3i \right)\left( 5-4i \right)} \\
& =\overline{2\left( 5-4i \right)+3i\left( 5-4i \right)} \\
& =\overline{10-8i+15i-12{{i}^{2}}} \\
& =\overline{10+7i-12\left( -1 \right)} \\
& =\overline{10+7i+12} \\
& =\overline{22+7i} \\
& =22-7i \\
\text{R}\text{.H}\text{.S}\text{.}&=\overline{z}\overline{w} \\
& =\left( \overline{2+3i} \right)\left( \overline{5-4i} \right) \\
& =\left( 2-3i \right)\left( 5+4i \right) \\
& =2\left( 5+4i \right)-3i\left( 5+4i \right) \\
& =10+8i-15i-12{{i}^{2}} \\
& =10-7i-12\left( -1 \right) \\
& =10-7i+12 \\
& =22-7i \\
\text{Hence, L}\text{.H}\text{.S}\text{.} &=\text{R}\text{.H}\text{.S}\text{.} \\
\end{align}\]
\(iv\)
\[\overline{\left( \frac{z}{w} \right)}=\frac{{\bar{z}}}{{\bar{w}}}\]
Solution:
\[\begin{align}
\text{L}\text{.H}\text{.S}\text{.}&=\overline{\left( \frac{z}{w} \right)} \\
& =\overline{\left( \frac{2+3i}{5-4i} \right)} \\
& =\overline{\left( \frac{2+3i}{5-4i}\times \frac{5+4i}{5+4i} \right)} \\
& =\overline{\left( \frac{\left( 2+3i \right)\left( 5+4i \right)}{{{5}^{2}}-{{\left( 4i \right)}^{2}}} \right)} \\
& =\overline{\left( \frac{2\left( 5+4i \right)+3i\left( 5+4i \right)}{25-16{{i}^{2}}} \right)} \\
& =\overline{\left( \frac{10+8i+15i+12{{i}^{2}}}{25-16\left( -1 \right)} \right)} \\
& =\overline{\left( \frac{10+23i+12\left( -1 \right)}{25+16} \right)} \\
& =\overline{\left( \frac{10+23i-12}{41} \right)} \\
& =\overline{\left( \frac{-2+23i}{41} \right)} \\
& =\overline{-\frac{2}{41}+\frac{23}{41}i} \\
& =-\frac{2}{41}-\frac{23}{41}i \\
\text{R}\text{.H}\text{.S}\text{.}&=\frac{{\bar{z}}}{{\bar{w}}} \\
& =\frac{\overline{2+3i}}{\overline{5-4i}} \\
& =\frac{2-3i}{5+4i}\times \frac{5-4i}{5-4i} \\
& =\frac{\left( 2-3i \right)\left( 5-4i \right)}{{{5}^{2}}-{{\left( 4i \right)}^{2}}} \\
& =\frac{2\left( 5-4i \right)-3i\left( 5-4i \right)}{25-16{{i}^{2}}} \\
& =\frac{10-8i-15i+12{{i}^{2}}}{25-16\left( -1 \right)} \\
& =\frac{10-23i+12\left( -1 \right)}{25+16} \\
& =\frac{10-23i-12}{41} \\
& =\frac{-2-23i}{41} \\
& =-\frac{2}{41}-\frac{23}{41}i \\
\text{Hence, L}\text{.H}\text{.S}\text{.}&=\text{R}\text{.H}\text{.S}\text{.} \\
\end{align}\]
\(v\)
\[\frac{1}{2}\left( z+\overline{z} \right) \text{ is the real part of } z.\]
Solution:
\[\begin{align}
& =\frac{1}{2}\left( z+\overline{z} \right) \\
& =\frac{1}{2}\left( 2+3i+\overline{2+3i} \right) \\
& =\frac{1}{2}\left( 2+3i+2-3i \right) \\
& =\frac{1}{\cancel{2}}\left( \cancelto{2}{4} \right) \\
& =2 \\
& =\operatorname{Re}\left( z \right) \\
\text{Hence, }&\frac{1}{2}\left( z+\overline{z} \right)\text{is the real part of }z\text{.} \\
\end{align}\]
\(vi\)
\[\frac{1}{2i}\left( z-\overline{z} \right)\text{ is the imaginary part of }z.\]
Solution:
\[\begin{align}
& =\frac{1}{2i}\left( z-\overline{z} \right) \\
& =\frac{1}{2i}\left( 2+3i-\left( \overline{2+3i} \right) \right) \\
& =\frac{1}{2i}\left( 2+3i-\left( 2-3i \right) \right) \\
& =\frac{1}{2i}\left( \cancel{2}+3i-\cancel{2}+3i \right) \\
& =\frac{1}{\cancel{2i}}\left( \cancelto{3}{6i} \right) \\
& =3 \\
& =\operatorname{Im}\left( z \right) \\
\text{Hence, }&\frac{1}{2i}\left( z-\overline{z} \right)\text{ is the imaginary part of }z. \\
\end{align}\]
Question#07: Solve the following equation for real \(x\) and \(y\).
\(i\)
\[\left( 2-3i \right)\left( x+yi \right)=4+i\]
Solution:
\[\begin{align}
\left( 2-3i \right)\left( x+yi \right)&=4+i \\
x+yi&=\frac{4+i}{2-3i} \\
& =\frac{4+i}{2-3i}\times \frac{2+3i}{2+3i} \\
& =\frac{\left( 4+i \right)\left( 2+3i \right)}{{{2}^{2}}-{{\left( 3i \right)}^{2}}} \\
& =\frac{4\left( 2+3i \right)+i\left( 2+3i \right)}{4-9{{i}^{2}}} \\
& =\frac{8+12i+2i+3{{i}^{2}}}{4-9\left( -1 \right)} \\
& =\frac{8+14i+3\left( -1 \right)}{4+9} \\
& =\frac{8+14i-3}{13} \\
& =\frac{5+14i}{13} \\
x+yi&=\frac{5}{13}+\frac{14}{13}i \\
\text{So, }x&=\frac{5}{13},y=\frac{14}{13} \\
\end{align}\]
\(ii\)
\[\left( 3-2i \right)\left( x+yi \right)=2\left( x-2yi \right)+2i-1\]
Solution:
\[\begin{align}
&\left( 3-2i \right)\left( x+yi \right) =2\left( x-2yi \right)+2i-1 \\
&3\left( x+yi \right)-2i\left( x+yi \right) =2x-4yi+2i-1 \\
&3x+3yi-2xi-2y{{i}^{2}}=2x-1+i\left( 2-4y \right) \\
&3x+3yi-2xi-2y\left( -1 \right) =2x-1+i\left( 2-4y \right) \\
&3x+3yi-2xi+2y=2x-1+i\left( 2-4y \right) \\
&3x+2y+i\left( 3y-2x \right) =2x-1+i\left( 2-4y \right)\\
&\text{Comparing real and imaginary parts } \\
&3x+2y=2x-1\quad|\quad3y-2x=2-4y \\
&3x+2y-2x=-1\quad|\quad3y-2x+4y=2 \\
&x+2y=-1…(i)\quad|\quad-2x+7y=2…(ii) \\
&\text{Multipy equation (i) by 2 and add in equation (ii)} \\
&\cancel{2x}+4y=-\cancel{2} \\
&\underline{-\cancel{2x}+7y=\cancel{2}}\\
&\qquad 11y=0 \\
&\qquad y=0 \\
&\text{Put }y=0\text{ in equation (i), we have} \\
&x+2\left( 0 \right)=-1 \\
&x=-1 \\
\end{align}\]
\(iii\)
\[{{\left( 3+4i \right)}^{2}}-2\left( x-yi \right)=x+yi\]
Solution:
\[\begin{align}
& {{\left( 3+4i \right)}^{2}}-2\left( x-yi \right)=x+yi \\
& {{3}^{2}}+{{\left( 4i \right)}^{2}}+2\left( 3 \right)\left( 4i \right)-2x+2yi=x+yi \\
& 9+16{{i}^{2}}+24i-2x+2yi-x-yi=0 \\
& 9+16\left( -1 \right)+24i-3x+yi=0 \\
& 9-16+24i-3x+yi=0 \\
& -7+24i-3x+yi=0 \\
& -3x+yi=7-24i \\
& \text{Comparing real and imaginary parts} \\
& -3x=7\qquad |\qquad y=-24 \\
& x=-\frac{7}{3} \qquad |\\
\end{align}\]
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