Samacheer Kalvi Class 12 Maths Solutions Chapter 2 Complex Numbers Exercise 2.7

Question 1. Write in polar form of the following complex numbers.
(i) \( 2 + i2 \sqrt{3} \)
(ii) \( 3 – i \sqrt{3} \)
(iii) \( -2 - i2 \)
(iv) \( \frac{i-1}{\cos{\frac{\pi}{3}}+i\sin{\frac{\pi}{3}}} \)
Answer:
(i) Given complex number is \( z = 2 + i2 \sqrt{3} \).
Let \( z = r (\cos \theta + i \sin \theta) \).
Equating real and imaginary parts:
\( r \cos \theta = 2 \) (positive)
\( r \sin \theta = 2 \sqrt{3} \) (positive)
To find \( r \):
\( r^2 = (2)^2 + (2\sqrt{3})^2 \)
\( r^2 = 4 + 12 \)
\( r^2 = 16 \)
\( r = 4 \)
Since \( \cos \theta \) and \( \sin \theta \) are both positive, \( \theta \) lies in the 1st quadrant.
\( \cos \theta = \frac{2}{4} = \frac{1}{2} \)
\( \sin \theta = \frac{2\sqrt{3}}{4} = \frac{\sqrt{3}}{2} \)
This means \( \theta = \frac{\pi}{3} \). This angle is called the principal argument.
The general argument is \( 2k\pi + \theta \), where \( k \) is an integer.
So, argument \( = 2k\pi + \frac{\pi}{3} \)
The polar form is \( z = r (\cos \theta + i \sin \theta) \).
Therefore, \( 2 + 2i\sqrt{3} = 4 \left( \cos \left(2k\pi + \frac{\pi}{3}\right) + i \sin \left(2k\pi + \frac{\pi}{3}\right) \right) \), where \( k \in Z \).
This representation clearly shows the magnitude and direction of the complex number.
In simple words: First, find the length (modulus) of the complex number. Then, find its angle (argument) by looking at the signs of its real and imaginary parts to know which quarter of the graph it's in. Finally, write it in the special polar format.

🎯 Exam Tip: Remember to express the general argument as \( 2k\pi + \alpha \) where \( \alpha \) is the principal argument, to account for all possible rotations.

(ii) Given complex number is \( z = 3 - i \sqrt{3} \).
Let \( z = r (\cos \theta + i \sin \theta) \).
Equating real and imaginary parts:
\( r \cos \theta = 3 \) (positive)
\( r \sin \theta = -\sqrt{3} \) (negative)
To find \( r \):
\( r^2 = (3)^2 + (-\sqrt{3})^2 \)
\( r^2 = 9 + 3 \)
\( r^2 = 12 \)
\( r = \sqrt{12} = 2\sqrt{3} \)
Since \( \cos \theta \) is positive and \( \sin \theta \) is negative, \( \theta \) lies in the IV quadrant.
\( \cos \theta = \frac{3}{2\sqrt{3}} = \frac{\sqrt{3}}{2} \)
\( \sin \theta = \frac{-\sqrt{3}}{2\sqrt{3}} = -\frac{1}{2} \)
This means the principal argument \( \theta = -\frac{\pi}{6} \).
The general argument is \( 2k\pi + \theta \).
So, argument \( = 2k\pi - \frac{\pi}{6} \)
The polar form is \( z = r (\cos \theta + i \sin \theta) \).
Therefore, \( 3 - i\sqrt{3} = 2\sqrt{3} \left( \cos \left(2k\pi - \frac{\pi}{6}\right) + i \sin \left(2k\pi - \frac{\pi}{6}\right) \right) \), where \( k \in Z \).
The angle for complex numbers can be clockwise (negative) or counter-clockwise (positive).
In simple words: Find the length of the complex number first. Then, check the signs of the real and imaginary parts to figure out which quarter the number is in. After that, find the angle and write the number in its polar (angle-and-length) form.

🎯 Exam Tip: Be careful with the signs of the real and imaginary parts to correctly identify the quadrant and determine the principal argument.

(iii) Given complex number is \( z = -2 - i2 \).
Let \( z = r (\cos \theta + i \sin \theta) \).
Equating real and imaginary parts:
\( r \cos \theta = -2 \)
\( r \sin \theta = -2 \)
To find \( r \):
\( r^2 = (-2)^2 + (-2)^2 \)
\( r^2 = 4 + 4 \)
\( r^2 = 8 \)
\( r = \sqrt{8} = 2\sqrt{2} \)
Since \( \cos \theta \) and \( \sin \theta \) are both negative, \( \theta \) lies in the III quadrant.
\( \cos \theta = \frac{-2}{2\sqrt{2}} = -\frac{1}{\sqrt{2}} \)
\( \sin \theta = \frac{-2}{2\sqrt{2}} = -\frac{1}{\sqrt{2}} \)
For \( |\cos \theta| = \frac{1}{\sqrt{2}} \) and \( |\sin \theta| = \frac{1}{\sqrt{2}} \), the reference angle is \( \frac{\pi}{4} \).
In the III quadrant, \( \theta = \pi + \frac{\pi}{4} = \frac{5\pi}{4} \) (or \( \theta = -\pi + \frac{\pi}{4} = -\frac{3\pi}{4} \)). Let's use the latter for principal argument.
The principal argument \( \theta = -\pi + \frac{\pi}{4} = -\frac{3\pi}{4} \).
The general argument is \( 2k\pi + \theta \).
So, argument \( = 2k\pi - \frac{3\pi}{4} \)
The polar form is \( z = r (\cos \theta + i \sin \theta) \).
Therefore, \( -2 - i2 = 2\sqrt{2} \left( \cos \left(2k\pi - \frac{3\pi}{4}\right) + i \sin \left(2k\pi - \frac{3\pi}{4}\right) \right) \), where \( k \in Z \).
This form gives a clear picture of the complex number's position and distance from the origin.
In simple words: Calculate the length of the complex number. Then, based on the negative real and imaginary parts, place the angle in the third quarter of the graph. Write the angle with \( 2k\pi \) to show all possible rotations, and then put it into the polar form.

🎯 Exam Tip: When both real and imaginary parts are negative, the argument is in the third quadrant. You can find it as \( -\pi + \text{reference angle} \) for the principal argument.

(iv) Given complex number is \( z = \frac{i-1}{\cos{\frac{\pi}{3}}+i\sin{\frac{\pi}{3}}} \).
First, simplify the denominator. We know \( \cos{\frac{\pi}{3}} = \frac{1}{2} \) and \( \sin{\frac{\pi}{3}} = \frac{\sqrt{3}}{2} \).
So, \( z = \frac{i-1}{\frac{1}{2}+i\frac{\sqrt{3}}{2}} \)
To simplify, multiply the numerator and denominator by the conjugate of the denominator:
\( z = \frac{i-1}{\frac{1}{2}+i\frac{\sqrt{3}}{2}} \times \frac{\frac{1}{2}-i\frac{\sqrt{3}}{2}}{\frac{1}{2}-i\frac{\sqrt{3}}{2}} \)
\( z = \frac{(i-1)(\frac{1}{2}-i\frac{\sqrt{3}}{2})}{(\frac{1}{2})^2 + (\frac{\sqrt{3}}{2})^2} \)
\( z = \frac{i\frac{1}{2} - i^2\frac{\sqrt{3}}{2} - \frac{1}{2} + i\frac{\sqrt{3}}{2}}{\frac{1}{4} + \frac{3}{4}} \)
\( z = \frac{\frac{1}{2}i + \frac{\sqrt{3}}{2} - \frac{1}{2} + \frac{\sqrt{3}}{2}i}{1} \)
\( z = \left(\frac{\sqrt{3}-1}{2}\right) + i \left(\frac{\sqrt{3}+1}{2}\right) \)
Now, let \( z = r (\cos \theta + i \sin \theta) \).
Equating real and imaginary parts:
\( r \cos \theta = \frac{\sqrt{3}-1}{2} \) (positive)
\( r \sin \theta = \frac{\sqrt{3}+1}{2} \) (positive)
To find \( r \):
\( r^2 = \left(\frac{\sqrt{3}-1}{2}\right)^2 + \left(\frac{\sqrt{3}+1}{2}\right)^2 \)
\( r^2 = \frac{3 - 2\sqrt{3} + 1}{4} + \frac{3 + 2\sqrt{3} + 1}{4} \)
\( r^2 = \frac{4 - 2\sqrt{3}}{4} + \frac{4 + 2\sqrt{3}}{4} \)
\( r^2 = \frac{4 - 2\sqrt{3} + 4 + 2\sqrt{3}}{4} \)
\( r^2 = \frac{8}{4} = 2 \)
\( r = \sqrt{2} \)
Since \( \cos \theta \) and \( \sin \theta \) are both positive, \( \theta \) lies in the 1st quadrant.
The argument \( \theta \) can be found using \( \tan \theta = \frac{r \sin \theta}{r \cos \theta} = \frac{\frac{\sqrt{3}+1}{2}}{\frac{\sqrt{3}-1}{2}} = \frac{\sqrt{3}+1}{\sqrt{3}-1} \).
We know that \( \tan(\frac{\pi}{4} + \frac{\pi}{6}) = \frac{\tan \frac{\pi}{4} + \tan \frac{\pi}{6}}{1 - \tan \frac{\pi}{4} \tan \frac{\pi}{6}} = \frac{1 + \frac{1}{\sqrt{3}}}{1 - 1 \cdot \frac{1}{\sqrt{3}}} = \frac{\frac{\sqrt{3}+1}{\sqrt{3}}}{\frac{\sqrt{3}-1}{\sqrt{3}}} = \frac{\sqrt{3}+1}{\sqrt{3}-1} \).
So, \( \theta = \frac{\pi}{4} + \frac{\pi}{6} = \frac{3\pi + 2\pi}{12} = \frac{5\pi}{12} \).
The general argument is \( 2k\pi + \frac{5\pi}{12} \).
The polar form is \( z = r (\cos \theta + i \sin \theta) \).
Therefore, \( \frac{i-1}{\cos{\frac{\pi}{3}}+i\sin{\frac{\pi}{3}}} = \sqrt{2} \left( \cos \left(2k\pi + \frac{5\pi}{12}\right) + i \sin \left(2k\pi + \frac{5\pi}{12}\right) \right) \), where \( k \in Z \).
This process simplifies a complex fraction into a more understandable polar form.
In simple words: First, make the complex fraction simpler by getting rid of the fraction in the denominator. Then, find its length and angle just like the other problems. Put these values into the polar form.

🎯 Exam Tip: When dealing with complex fractions, always rationalize the denominator first by multiplying by its conjugate before finding the modulus and argument.

Question 2. Find the rectangular form of the complex numbers
(i) \( (\cos \frac{\pi}{12} + i\sin \frac{\pi}{12}) \) (This is implicitly a part of a larger problem in the source, which combines it with \( \cos \frac{\pi}{6} + i\sin \frac{\pi}{6} \). We will solve the combined form shown in the source.)
(ii) \( \frac{\cos \frac{\pi}{6} - i\sin \frac{\pi}{6}}{2 (\cos \frac{\pi}{3} + i\sin \frac{\pi}{3})} \)
Answer:
(i) The problem in the source shows an expansion of a product of complex numbers, hinting at \( (\cos \frac{\pi}{6} + i\sin \frac{\pi}{6}) (\cos \frac{\pi}{12} + i\sin \frac{\pi}{12}) \). Let's find its rectangular form.
Using De Moivre's theorem for multiplication, \( (\cos A + i\sin A)(\cos B + i\sin B) = \cos(A+B) + i\sin(A+B) \).
So, \( \left(\cos \frac{\pi}{6} + i\sin \frac{\pi}{6}\right) \left(\cos \frac{\pi}{12} + i\sin \frac{\pi}{12}\right) \)
\( = \cos \left(\frac{\pi}{6} + \frac{\pi}{12}\right) + i \sin \left(\frac{\pi}{6} + \frac{\pi}{12}\right) \)
\( = \cos \left(\frac{2\pi + \pi}{12}\right) + i \sin \left(\frac{2\pi + \pi}{12}\right) \)
\( = \cos \left(\frac{3\pi}{12}\right) + i \sin \left(\frac{3\pi}{12}\right) \)
\( = \cos \left(\frac{\pi}{4}\right) + i \sin \left(\frac{\pi}{4}\right) \)
Now, convert to rectangular form:
\( = \frac{1}{\sqrt{2}} + i \frac{1}{\sqrt{2}} \)
This form is simple to understand as real and imaginary parts. An alternative calculation shown in the source directly uses angle sum formulas.
Alternatively (Aliter solution from source):
\( \cos \left(\frac{\pi}{6} + \frac{\pi}{12}\right) + i \sin \left(\frac{\pi}{6} + \frac{\pi}{12}\right) \)
\( = \cos \frac{3\pi}{12} + i \sin \frac{3\pi}{12} \)
\( = \cos \frac{\pi}{4} + i \sin \frac{\pi}{4} \)
\( = \frac{1}{\sqrt{2}} + i \frac{1}{\sqrt{2}} = \frac{1+i}{\sqrt{2}} \)
In simple words: When multiplying complex numbers in polar form, you add their angles and multiply their lengths. Here, we add the angles \( \frac{\pi}{6} \) and \( \frac{\pi}{12} \). Then, we find the cosine and sine of the new angle to get the rectangular form, which is like \( x + iy \).

🎯 Exam Tip: Remember De Moivre's theorem for products and quotients of complex numbers in polar form. It simplifies calculations significantly.

(ii) Given complex number is \( z = \frac{\cos \frac{\pi}{6} - i\sin \frac{\pi}{6}}{2 (\cos \frac{\pi}{3} + i\sin \frac{\pi}{3})} \).
We can write \( \cos \frac{\pi}{6} - i\sin \frac{\pi}{6} \) as \( \cos(-\frac{\pi}{6}) + i\sin(-\frac{\pi}{6}) \).
Using De Moivre's theorem for division, \( \frac{r_1(\cos A + i\sin A)}{r_2(\cos B + i\sin B)} = \frac{r_1}{r_2}(\cos(A-B) + i\sin(A-B)) \).
Here, \( r_1 = 1 \), \( A = -\frac{\pi}{6} \)
And \( r_2 = 2 \), \( B = \frac{\pi}{3} \)
So, \( z = \frac{1}{2} \left( \cos \left(-\frac{\pi}{6} - \frac{\pi}{3}\right) + i \sin \left(-\frac{\pi}{6} - \frac{\pi}{3}\right) \right) \)
\( z = \frac{1}{2} \left( \cos \left(-\frac{\pi}{6} - \frac{2\pi}{6}\right) + i \sin \left(-\frac{\pi}{6} - \frac{2\pi}{6}\right) \right) \)
\( z = \frac{1}{2} \left( \cos \left(-\frac{3\pi}{6}\right) + i \sin \left(-\frac{3\pi}{6}\right) \right) \)
\( z = \frac{1}{2} \left( \cos \left(-\frac{\pi}{2}\right) + i \sin \left(-\frac{\pi}{2}\right) \right) \)
Now, convert to rectangular form:
\( \cos(-\frac{\pi}{2}) = 0 \)
\( \sin(-\frac{\pi}{2}) = -1 \)
So, \( z = \frac{1}{2} (0 + i(-1)) \)
\( z = -\frac{1}{2}i \)
This rectangular form is much simpler than the initial fractional form.
In simple words: When dividing complex numbers in polar form, you subtract their angles and divide their lengths. Then, turn this result into the \( x + iy \) form. Be careful with negative angles.

🎯 Exam Tip: Always remember that \( \cos(-\theta) = \cos(\theta) \) and \( \sin(-\theta) = -\sin(\theta) \), which is useful when dealing with negative angles.

Question 3. If \( (x_1+i y_1)(x_2+i y_2)(x_3+i y_3) \cdots (x_n+i y_n)=a+i b \), show that \( (x_1^2 + y_1^2)(x_2^2 + y_2^2)(x_3^2 + y_3^2)\cdots(x_n^2 + y_n^2) = a^2 + b^2 \).
Answer: Given the equation \( (x_1+i y_1)(x_2+i y_2)(x_3+i y_3) \cdots (x_n+i y_n)=a+i b \).
We need to show that \( (x_1^2 + y_1^2)(x_2^2 + y_2^2)(x_3^2 + y_3^2)\cdots(x_n^2 + y_n^2) = a^2 + b^2 \).
To do this, we can take the modulus (or absolute value) of both sides of the given equation.
The modulus of a complex number \( z = x + iy \) is \( |z| = \sqrt{x^2+y^2} \).
A key property of moduli is that the modulus of a product of complex numbers is equal to the product of their moduli: \( |z_1 z_2 \cdots z_n| = |z_1| |z_2| \cdots |z_n| \).
Applying this property to the given equation:
\( |(x_1+i y_1)(x_2+i y_2)(x_3+i y_3) \cdots (x_n+i y_n)| = |a+i b| \)
\( |x_1+i y_1| |x_2+i y_2| |x_3+i y_3| \cdots |x_n+i y_n| = |a+i b| \)
Now, substitute the definition of modulus for each complex number:
\( \sqrt{x_1^2+y_1^2} \sqrt{x_2^2+y_2^2} \sqrt{x_3^2+y_3^2} \cdots \sqrt{x_n^2+y_n^2} = \sqrt{a^2+b^2} \)
To remove the square roots, square both sides of the equation:
\( \left(\sqrt{x_1^2+y_1^2} \sqrt{x_2^2+y_2^2} \cdots \sqrt{x_n^2+y_n^2}\right)^2 = \left(\sqrt{a^2+b^2}\right)^2 \)
\( (x_1^2+y_1^2)(x_2^2+y_2^2)(x_3^2+y_3^2) \cdots (x_n^2+y_n^2) = a^2+b^2 \)
Hence proved. This method is very useful for relating products of sums of squares to the square of a sum.
In simple words: When you have a product of many complex numbers that equals another complex number, taking the 'length' (modulus) of both sides helps. The length of a product is the product of the lengths. If you square these lengths, you get the squares of the real and imaginary parts, which helps prove the statement.

🎯 Exam Tip: This type of proof often relies on the fundamental property of complex numbers: \( |z_1 z_2| = |z_1| |z_2| \) and \( |z|^2 = x^2+y^2 \).

Question 4. Given \( \frac{1+z}{1-z} = \cos 2\theta + i \sin 2\theta \), show that \( z = i \tan \theta \).
Answer: Given the equation \( \frac{1+z}{1-z} = \cos 2\theta + i \sin 2\theta \).
Let \( \frac{1+z}{1-z} = e^{i2\theta} \) (using Euler's formula \( e^{i\phi} = \cos \phi + i \sin \phi \)).
We need to solve for \( z \).
Multiply both sides by \( (1-z) \):
\( 1+z = e^{i2\theta}(1-z) \)
\( 1+z = e^{i2\theta} - z e^{i2\theta} \)
Now, gather all terms with \( z \) on one side:
\( z + z e^{i2\theta} = e^{i2\theta} - 1 \)
Factor out \( z \):
\( z(1 + e^{i2\theta}) = e^{i2\theta} - 1 \)
Now, solve for \( z \):
\( z = \frac{e^{i2\theta} - 1}{e^{i2\theta} + 1} \)
To simplify this expression, we can use a common trick: divide the numerator and denominator by \( e^{i\theta} \).
\( z = \frac{\frac{e^{i2\theta} - 1}{e^{i\theta}}}{\frac{e^{i2\theta} + 1}{e^{i\theta}}} \)
\( z = \frac{e^{i\theta} - e^{-i\theta}}{e^{i\theta} + e^{-i\theta}} \)
Now, use Euler's formulas again: \( \cos \phi = \frac{e^{i\phi} + e^{-i\phi}}{2} \) and \( \sin \phi = \frac{e^{i\phi} - e^{-i\phi}}{2i} \).
From these, we can see that \( e^{i\phi} - e^{-i\phi} = 2i \sin \phi \) and \( e^{i\phi} + e^{-i\phi} = 2 \cos \phi \).
Substitute these into the expression for \( z \):
\( z = \frac{2i \sin \theta}{2 \cos \theta} \)
\( z = i \frac{\sin \theta}{\cos \theta} \)
\( z = i \tan \theta \)
Hence proved. This method shows the power of Euler's formula in complex number manipulations.
In simple words: Start with the given equation. Rearrange it to solve for \( z \). Use Euler's formula \( e^{i\theta} = \cos \theta + i \sin \theta \) to replace the cosine and sine terms. Then, simplify the expression by dividing the top and bottom by \( e^{i\theta} \), and use other Euler identities to finally get \( i \tan \theta \).

🎯 Exam Tip: When simplifying expressions involving \( e^{i\theta} - 1 \) or \( e^{i\theta} + 1 \), a common trick is to factor out \( e^{i\theta/2} \) or divide by \( e^{i\theta/2} \) to create sine or cosine terms.

Question 5. If \( \cos \alpha + \cos \beta + \cos \gamma = \sin \alpha + \sin \beta + \sin \gamma = 0 \), then show that (i) \( \cos 3\alpha + \cos 3\beta + \cos 3\gamma = 3 \cos (\alpha + \beta + \gamma) \) and (ii) \( \sin 3\alpha + \sin 3\beta + \sin 3\gamma = 3 \sin (\alpha + \beta + \gamma) \).
Answer: Let's define three complex numbers:
\( a = \cos \alpha + i \sin \alpha = e^{i\alpha} \)
\( b = \cos \beta + i \sin \beta = e^{i\beta} \)
\( c = \cos \gamma + i \sin \gamma = e^{i\gamma} \)
Now, consider their sum:
\( a + b + c = (\cos \alpha + i \sin \alpha) + (\cos \beta + i \sin \beta) + (\cos \gamma + i \sin \gamma) \)
Group the real and imaginary parts:
\( a + b + c = (\cos \alpha + \cos \beta + \cos \gamma) + i (\sin \alpha + \sin \beta + \sin \gamma) \)
From the given conditions, \( \cos \alpha + \cos \beta + \cos \gamma = 0 \) and \( \sin \alpha + \sin \beta + \sin \gamma = 0 \).
So, \( a + b + c = 0 + i(0) \)
\( \implies a + b + c = 0 \)
A fundamental algebraic identity states that if \( a+b+c=0 \), then \( a^3+b^3+c^3 = 3abc \). This identity is very useful in complex number problems.
Substitute the exponential forms back into this identity:
\( (e^{i\alpha})^3 + (e^{i\beta})^3 + (e^{i\gamma})^3 = 3 (e^{i\alpha})(e^{i\beta})(e^{i\gamma}) \)
Using the property \( (e^{ix})^n = e^{inx} \) and \( e^x e^y e^z = e^{x+y+z} \):
\( e^{i3\alpha} + e^{i3\beta} + e^{i3\gamma} = 3 e^{i(\alpha+\beta+\gamma)} \)
Now, convert these exponential forms back to trigonometric (cosine and sine) forms:
\( (\cos 3\alpha + i \sin 3\alpha) + (\cos 3\beta + i \sin 3\beta) + (\cos 3\gamma + i \sin 3\gamma) = 3 (\cos (\alpha+\beta+\gamma) + i \sin (\alpha+\beta+\gamma)) \)
Group the real and imaginary parts on the left side:
\( (\cos 3\alpha + \cos 3\beta + \cos 3\gamma) + i (\sin 3\alpha + \sin 3\beta + \sin 3\gamma) = 3 \cos (\alpha+\beta+\gamma) + i 3\sin (\alpha+\beta+\gamma) \)
Finally, by equating the real parts and the imaginary parts on both sides of the equation:
(i) Equating Real Parts:
\( \cos 3\alpha + \cos 3\beta + \cos 3\gamma = 3 \cos (\alpha + \beta + \gamma) \)
(ii) Equating Imaginary Parts:
\( \sin 3\alpha + \sin 3\beta + \sin 3\gamma = 3 \sin (\alpha + \beta + \gamma) \)
Hence proved. This problem beautifully connects algebraic identities with complex number theory.
In simple words: We create three complex numbers using the given angles. Because the sums of their cosine parts and sine parts are zero, their total sum is zero. When three numbers add up to zero, a special math rule says their cubes add up to three times their product. Using this rule and converting back to cosine and sine forms, we can show the required relationships for \( 3\alpha, 3\beta, 3\gamma \).

🎯 Exam Tip: This is a standard result: if \( \sum \cos \alpha = 0 \) and \( \sum \sin \alpha = 0 \), then \( \sum \cos 3\alpha = 3 \cos (\sum \alpha) \) and \( \sum \sin 3\alpha = 3 \sin (\sum \alpha) \). Remember the identity \( a+b+c=0 \implies a^3+b^3+c^3 = 3abc \).

Question 6. If \( z = x + iy \) and \( \arg \left(\frac{z-i}{z+2}\right) = \frac{\pi}{4} \), then show that \( x^2 + y^2 + 3x - 3y + 2 = 0 \).
Answer: Given \( z = x + iy \) and \( \arg \left(\frac{z-i}{z+2}\right) = \frac{\pi}{4} \).
First, substitute \( z = x + iy \) into the expression \( \frac{z-i}{z+2} \):
\( \frac{z-i}{z+2} = \frac{(x+iy)-i}{(x+iy)+2} \)
\( = \frac{x + i(y-1)}{(x+2) + iy} \)
To simplify this complex fraction, multiply the numerator and denominator by the conjugate of the denominator, which is \( (x+2) - iy \):
\( = \frac{(x + i(y-1))((x+2) - iy)}{((x+2) + iy)((x+2) - iy)} \)
\( = \frac{x(x+2) - ixy + i(y-1)(x+2) - i^2 y(y-1)}{(x+2)^2 + y^2} \)
Since \( i^2 = -1 \), we have \( -i^2 y(y-1) = y(y-1) \).
\( = \frac{x(x+2) + y(y-1) + i(-(xy) + (y-1)(x+2))}{(x+2)^2 + y^2} \)
Now, expand the real and imaginary parts in the numerator:
Real Part: \( x^2 + 2x + y^2 - y \)
Imaginary Part: \( -xy + (xy + 2y - x - 2) = 2y - x - 2 \)
So, \( \frac{z-i}{z+2} = \frac{x^2+y^2+2x-y}{(x+2)^2+y^2} + i \frac{2y-x-2}{(x+2)^2+y^2} \)
We are given that \( \arg \left(\frac{z-i}{z+2}\right) = \frac{\pi}{4} \).
For a complex number \( A+iB \), its argument is \( \arg(A+iB) = \tan^{-1}\left(\frac{B}{A}\right) \).
Therefore, \( \tan^{-1}\left(\frac{\frac{2y-x-2}{(x+2)^2+y^2}}{\frac{x^2+y^2+2x-y}{(x+2)^2+y^2}}\right) = \frac{\pi}{4} \)
Since \( \tan^{-1} \) is \( \frac{\pi}{4} \), the ratio of the imaginary part to the real part must be \( \tan(\frac{\pi}{4}) = 1 \).
\( \frac{2y-x-2}{x^2+y^2+2x-y} = 1 \)
Cross-multiply:
\( 2y-x-2 = x^2+y^2+2x-y \)
Move all terms to one side to get the desired equation:
\( 0 = x^2+y^2+2x-y - (2y-x-2) \)
\( 0 = x^2+y^2+2x-y-2y+x+2 \)
\( x^2+y^2+3x-3y+2 = 0 \)
Hence proved. This demonstrates how geometric properties of complex numbers translate into algebraic equations.
In simple words: First, put \( z = x+iy \) into the given fraction and simplify it to a single complex number in the form \( A+iB \). Since the angle (argument) of this complex number is \( \frac{\pi}{4} \), it means its imaginary part \( B \) must be equal to its real part \( A \). Set \( A=B \) and then rearrange the equation to get the required result.

🎯 Exam Tip: When \( \arg(Z) = \frac{\pi}{4} \), it implies that the real part of \( Z \) is equal to its imaginary part (assuming the real part is positive). Use this shortcut after simplifying the complex number \( Z \).