====== Question 9, Exercise 1.2 ======
Solutions of Question 9 of Exercise 1.2 of Unit 01: Complex Numbers. This is unit of Model Textbook of Mathematics for Class XI published by National Book Foundation (NBF) as Federal Textbook Board, Islamabad, Pakistan. 
====Question 9(i)====
Find real and imaginary parts of $(2+4 i)^{-1}$.

**Solution.** 

Suppose $z=2+4i$.

\begin{align}
Re(2+4i)^{-1} & = Re(z^{-1}) = \dfrac{Re(z)}{|z|^2} \\
& =\dfrac{2}{2^2+4^2} = \dfrac{2}{20}\\ 
&= \dfrac{1}{10}.
\end{align}

\begin{align}
Im(2+4i)^{-1} & = Im(z^{-1}) = -\dfrac{Im(z)}{|z|^2} \\
& =-\dfrac{4}{2^2+4^2} = -\dfrac{4}{20}\\ 
&= \dfrac{1}{5}.
\end{align}
GOOD
====Question 9(ii)====
Find real and imaginary parts of  $(3-\sqrt{-4})^{-2}$.

**Solution.** 

Suppose $z=3 - \sqrt{-4}=3-2i$. 

We will use the following formulas:	
\[\text{Re}(z^{-2}) = \frac{(\text{Re}(z))^2 - (\text{Im}(z))^2}{|z|^4},\]
\[\text{Im}(z^{-2}) = \frac{-2 \text{Re}(z) \text{Im}(z)}{|z|^4}. \]
First, note $Re(z)=3$, $Im(z)=-2$ and
\begin{align}
|z| &= \sqrt{3^2 + (-2)^2} \\&= \sqrt{13}.
\end{align}
Then 
\[|z|^4=169.\]
Using in above formulas
\begin{align}
Re((3 - 2i)^{-2}) &= \frac{(3)^2 - (-2)^2}{(\sqrt{13})^4} \\
&= \frac{9 - 4}{169} = \frac{5}{169}.
\end{align}
\begin{align}
Im((3 - 2i)^{-2}) &= \frac{-2 \cdot 3 \cdot (-2)}{(\sqrt{13})^4}\\ 
&= \frac{12}{169}.
\end{align}
Therefore, the real part is \(\dfrac{5}{169}\) and the imaginary part is \(\dfrac{12}{169}\).	GOOD
====Question 9(iii)====
Find real and imaginary parts of  $\left(\dfrac{7+2 i}{3-i}\right)^{-1}$.

**Solution.** 

We use the following formulas:
	
\[Re\left(\left(\frac{x_1 + i y_1}{x_2 + i y_2}\right)^{-1}\right) = \frac{x_1 x_2 + y_1 y_2}{x_1^2 + y_1^2},\]
\[Im\left(\left(\frac{x_1 + i y_1}{x_2 + i y_2}\right)^{-1}\right) = \frac{x_1 y_2 - x_2 y_1}{x_1^2 + y_1^2}.\]
	
For \(z_1 = 7 + 2i\) and \(z_2 = 3 - i\), we have:
	
\[x_1 = 7, \quad y_1 = 2, \quad x_2 = 3, \quad y_2 = -1.\]
	
Using in above formulas:
	
\begin{align}
Re\left(\left(\frac{7 + 2i}{3 - i}\right)^{-1}\right) &= \frac{7 \cdot 3 + 2 \cdot (-1)}{7^2 + 2^2}\\ 
&= \frac{21 - 2}{49 + 4} = \frac{19}{53}.
\end{align}

\begin{align}
Im\left(\left(\frac{7 + 2i}{3 - i}\right)^{-1}\right) &= \frac{7 \cdot (-1) - 3 \cdot 2}{7^2 + 2^2}\\ 
&= \frac{-7 - 6}{49 + 4} = \frac{-13}{53}.
\end{align}
	
Therefore, the real part is \(\dfrac{19}{53}\) and the imaginary part is \(\dfrac{-13}{53}\).

====Question 9(iv)====
Find real and imaginary parts of  $\left(\dfrac{4+2 i}{2+5 i}\right)^{-2}$.

**Solution.** 

We will use the following formulas:
	
\begin{align}
&Re\left(\left(\frac{x_{1}+i y_{1}}{x_{2}+i y_{2}}\right)^{-2}\right) \\
&= \frac{\left(x_{2}^{2}-y_{2}^{2}\right)\left(x_{1}^{2}-y_{1}^{2}\right) + 4 x_{2} x_{1} y_{2} y_{1}}{\left(x_{1}^{2}+y_{1}^{2}\right)^{2}}\end{align}
	
\begin{align}&Im\left(\left(\frac{x_{1}+i y_{1}}{x_{2}+i y_{2}}\right)^{-2}\right)\\& = \frac{-2 \left[x_{1} y_{1} \left(x_{2}^{2}-y_{2}^{2}\right) - x_{2} y_{2} \left(x_{1}^{2}-y_{1}^{2}\right)\right]}{\left(x_{1}^{2}+y_{1}^{2}\right)^{2}}.\end{align}
	
Given \(z_1 = 4 + 2i\) and \(z_2 = 2 + 5i\), we have:
	
\[x_1 = 4, \quad y_1 = 2, \quad x_2 = 2, \quad y_2 = 5.\]
	
First, we need to compute \(x_1^2 + y_1^2\):
	
\[x_1^2 + y_1^2 = 4^2 + 2^2 = 20.\]
	
Next, compute \(x_2^2 - y_2^2\):
	
\[x_2^2 - y_2^2 = 2^2 - 5^2  = -21.\]
	
Then, compute \(x_1^2 - y_1^2\):
	
\[x_1^2 - y_1^2 = 4^2 - 2^2  = 12.\]
	
Now, compute \(4 x_2 x_1 y_2 y_1\):
	
\[4 x_2 x_1 y_2 y_1 = 4 \cdot 2 \cdot 4 \cdot 5 \cdot 2 = 320.\]
	
Now, we can compute the real part:
	
\begin{align}
&Re\left(\left(\frac{4 + 2i}{2 + 5i}\right)^{-2}\right)\\ 
= &\frac{\left(x_2^2 - y_2^2\right)\left(x_1^2 - y_1^2\right) + 4 x_2 x_1 y_2 y_1}{\left(x_1^2 + y_1^2\right)^2} \\
= &\frac{(-21) \cdot 12 + 320}{20^2}\\
= &\frac{-252 + 320}{400}\\
= &\frac{17}{100}.
\end{align}
	
Next, we compute the imaginary part:
	
\begin{align}
&Im\left(\left(\frac{4 + 2i}{2 + 5i}\right)^{-2}\right)\\ &= \frac{-2 \left[x_1 y_1 \left(x_2^2 - y_2^2\right) - x_2 y_2 \left(x_1^2 - y_1^2\right)\right]}{\left(x_1^2 + y_1^2\right)^2}
\end{align}
First, compute \(x_1 y_1 \left(x_2^2 - y_2^2\right)\):
\[x_1 y_1 \left(x_2^2 - y_2^2\right) = 4 \cdot 2 \cdot (-21) = -168.\]
	
Next, compute \(x_2 y_2 \left(x_1^2 - y_1^2\right)\):
	
\[x_2 y_2 \left(x_1^2 - y_1^2\right) = 2 \cdot 5 \cdot 12 = 120.\]
	
Thus,
\begin{align}
&Im\left(\left(\frac{4 + 2i}{2 + 5i}\right)^{-2}\right) \\ &= \frac{-2 \left[-168 - 120\right]}{400} \\
&= \frac{-2 \cdot (-288)}{400}\\
&= \frac{36}{25}
\end{align}
So, the real part of \(\left(\dfrac{4+2i}{2+5i}\right)^{-2}\) is \(\dfrac{17}{100}\) and the imaginary part is \(\dfrac{36}{25}\).

====Question 9(v)====
Find real and imaginary parts of  $\left(\dfrac{5-4 i}{5+4 i}\right)^{2}$.

**Solution.** 

We will use the following formulas:
	
\[\text{Re}\left(\left(\frac{x_1 + i y_1}{x_2 + i y_2}\right)^2\right) = \frac{\left(x_1^2 - y_1^2\right)\left(x_2^2 - y_2^2\right) + 4 x_1 x_2 y_1 y_2}{\left(x_2^2 + y_2^2\right)^2}.\]
	
\[\text{Im}\left(\left(\frac{x_1 + i y_1}{x_2 + i y_2}\right)^2\right) = \frac{2 \left[x_1 y_1 \left(x_2^2 - y_2^2\right) - x_2 y_2 \left(x_1^2 - y_1^2\right)\right]}{\left(x_2^2 + y_2^2\right)^2}.\]
	
For \(z_1 = 5 - 4i\) and \(z_2 = 5 + 4i\), we have:
	
\[x_1 = 5, \quad y_1 = -4, \quad x_2 = 5, \quad y_2 = 4.\]

First, calculate:
\[x_2^2 + y_2^2 = 5^2 + 4^2 = 25 + 16 = 41\]
Now, calculate the real part:
	
\begin{align}
&\text{Re}\left(\left(\frac{5 - 4i}{5 + 4i}\right)^2\right) \\
= &\frac{\left(5^2 - (-4)^2\right)\left(5^2 - 4^2\right) + 4 \cdot 5 \cdot 5 \cdot (-4) \cdot 4}{\left(5^2 + 4^2\right)^2}\\
= &\frac{\left(25 - 16\right)\left(25 - 16\right) + 4 \cdot 5 \cdot 5 \cdot (-4) \cdot 4}{41^2} \\
= &\frac{9 \cdot 9 + 4 \cdot 5 \cdot 5 \cdot (-4) \cdot 4}{1681} \\
= &\frac{-1519}{1681}
\end{align}
Next, we calculate the imaginary part:
\begin{align}
&\text{Im}\left(\left(\frac{5 - 4i}{5 + 4i}\right)^2\right)\\
=& \frac{2 \left[5 \cdot (-4) \left(5^2 - 4^2\right) - 5 \cdot 4 \left(5^2 - (-4)^2\right)\right]}{\left(5^2 + 4^2\right)^2}\\
=&\frac{2 \left[5 \cdot (-4) \left(25 - 16\right) - 5 \cdot 4 \left(25 - 16\right)\right]}{41^2}\\
=& \frac{2 \left[5 \cdot (-4) \cdot 9 - 5 \cdot 4 \cdot 9\right]}{1681} \\
=& \frac{2 \left[-20 \cdot 9 - 20 \cdot 9\right]}{1681}\\
=& \frac{-720}{1681} 
\end{align}
Therefore, the real part of \(\left(\dfrac{5 - 4i}{5 + 4i}\right)^2\) is \(\dfrac{-1519}{1681}\) and the imaginary part is \(\dfrac{-720}{1681}\).

====Question 9(vi)====
Find real and imaginary parts of  $\dfrac{3-7 i}{2+5 i}$.

**Solution.** 

We will use the given formulas:
	
\[\text{Re}\left(\frac{x_{1}+i y_{1}}{x_{2}+i y_{2}}\right)=\frac{x_{1} x_{2}+y_{1} y_{2}}{x_{2}^{2}+y_{2}^{2}}\]
	
\[\text{Im}\left(\frac{x_{1}+i y_{1}}{x_{2}+i y_{2}}\right)=\frac{x_{2} y_{1}-x_{1} y_{2}}{x_{2}^{2}+y_{2}^{2}}.\]
	
Given \(z_1 = 3 - 7i\) and \(z_2 = 2 + 5i\), we have:
\[x_1 = 3, \quad y_1 = -7, \quad x_2 = 2, \quad y_2 = 5.\]
	
First, compute \(x_2^2 + y_2^2\):
	
\[x_2^2 + y_2^2 = 2^2 + 5^2 = 29\]
	
Next, compute \(x_1 x_2 + y_1 y_2\):
\[x_1 x_2 + y_1 y_2 = 3 \cdot 2 + (-7) \cdot 5 = -29\]
	
Then, compute \(x_2 y_1 - x_1 y_2\):
	
\[x_2 y_1 - x_1 y_2 = 2 \cdot (-7) - 3 \cdot 5 = -29\]
	
Now, we can compute the real parts:
	
\begin{align}
Re\left(\frac{3 - 7i}{2 + 5i}\right) &= \frac{x_1 x_2 + y_1 y_2}{x_2^2 + y_2^2}\\ 
&= \frac{-29}{29} = -1.\\
\end{align}

Now, we can compute the imaginary parts:	
\begin{align}
Im\left(\frac{3 - 7i}{2 + 5i}\right) &= \frac{x_2 y_1 - x_1 y_2}{x_2^2 + y_2^2} \\
&= \frac{-29}{29} = -1.
\end{align}
	
So, the real part of \(\dfrac{3 - 7i}{2 + 5i}\) is \(-1\) and the imaginary part is \(-1\).

====Go to ====
<text align="left"><btn type="primary">[[math-11-nbf:sol:unit01:ex1-2-p8|< Question 8]]</btn></text>
<text align="right"><btn type="success">[[math-11-nbf:sol:unit01:ex1-2-p10|Question 10 >]]</btn></text>