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.

Find real and imaginary parts of $(2+4 i)^{-1}$.

Solution.

Suppose $z=2+4i$.

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

$$\begin{align} Im(2+4i)^{-1} & = Im(z^{-1}) = -\dfrac{Im(z)}{|z|^2} \\ & =-\dfrac{4}{2^2+4^2} = -\dfrac{4}{20}\\ &= \dfrac{1}{5}. \end{align}$$

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 $\dfrac{12}{169}$.

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 $\dfrac{-13}{53}$.

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 $\dfrac{36}{25}$.

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 $\dfrac{-720}{1681}$.

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 $-1$.

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