OP Malhotra Class 11 Maths Solutions Chapter 9 Complex Numbers Exercise 9 (C)

S Chand Class 11 ICSE Maths Solutions Chapter 9 Complex Numbers Ex 9(c)

Question 1. If \( (- 2 + \sqrt{-3}) (- 3 + 2\sqrt{-3}) = a + bi \), find the real numbers a and b with values of a and b, also find the modulus of a + bi.
Answer: We are given the equation: \( (- 2 + \sqrt{-3}) (- 3 + 2\sqrt{-3}) = a + bi \).
First, we rewrite \( \sqrt{-3} \) as \( \sqrt{3}i \).
So, the equation becomes: \( (- 2 + \sqrt{3}i)(- 3 + 2\sqrt{3}i) = a + bi \).
Next, we multiply the terms on the left side:
\( 6 - 4\sqrt{3}i - 3\sqrt{3}i + 2(\sqrt{3}i)^2 = a + bi \)
\( 6 - 7\sqrt{3}i + 2(3i^2) = a + bi \)
Since \( i^2 = -1 \), we have:
\( 6 - 7\sqrt{3}i + 2(-3) = a + bi \)
\( 6 - 7\sqrt{3}i - 6 = a + bi \)
\( -7\sqrt{3}i = a + bi \)
Now, we compare the real and imaginary parts on both sides of the equation.
The real part on the left is 0, and the imaginary part is \( -7\sqrt{3} \).
So, \( a = 0 \) and \( b = -7\sqrt{3} \).
Finally, we need to find the modulus of \( a + bi \), which is \( |a + bi| \).
\( |a + bi| = |0 + (-7\sqrt{3})i| \)
\( |a + bi| = \sqrt{0^2 + (-7\sqrt{3})^2} \)
\( |a + bi| = \sqrt{0 + (49 \times 3)} \)
\( |a + bi| = \sqrt{147} \) This simplifies to \( 7\sqrt{3} \).
Therefore, the real numbers are \( a = 0 \) and \( b = -7\sqrt{3} \), and the modulus is \( 7\sqrt{3} \). It's helpful to remember that \( \sqrt{a \times b} = \sqrt{a} \times \sqrt{b} \).
In simple words: We first changed \( \sqrt{-3} \) to \( \sqrt{3}i \). Then, we multiplied the two complex numbers. After simplifying, we found that \( a \) is 0 and \( b \) is \( -7\sqrt{3} \). The size of this complex number, called its modulus, is \( 7\sqrt{3} \).

🎯 Exam Tip: When multiplying complex numbers, remember to substitute \( i^2 = -1 \) and then combine the real and imaginary parts to find a and b. Always simplify radicals at the end.

Question 2. Find the modulus of \( (1 - i)^{-2} + (1 + i)^{-2} \).
Answer: We need to find the modulus of the expression \( (1 - i)^{-2} + (1 + i)^{-2} \).
First, let's simplify the expression:
\( (1 - i)^{-2} + (1 + i)^{-2} = \frac{1}{(1-i)^2} + \frac{1}{(1+i)^2} \)
We know that \( (1-i)^2 = 1^2 - 2i + i^2 = 1 - 2i - 1 = -2i \).
And \( (1+i)^2 = 1^2 + 2i + i^2 = 1 + 2i - 1 = 2i \).
Substitute these values back into the expression:
\( = \frac{1}{-2i} + \frac{1}{2i} \)
\( = \frac{-1}{2i} + \frac{1}{2i} \)
\( = 0 \)
The expression simplifies to 0. Now, we find the modulus of 0.
The modulus of a complex number \( x + yi \) is \( \sqrt{x^2 + y^2} \). For 0, it is \( 0 + 0i \).
\( |0| = \sqrt{0^2 + 0^2} = 0 \)
So, the modulus of \( (1 - i)^{-2} + (1 + i)^{-2} \) is 0. Complex numbers with a modulus of zero are simply the number zero itself.
In simple words: We need to find the size of the given complex number. First, we simplified the expression by calculating \( (1-i)^2 \) and \( (1+i)^2 \). When we put these back, the whole expression became zero. The size of zero is just zero.

🎯 Exam Tip: Remember to simplify expressions involving powers of complex numbers before finding their modulus. Also, recall that \( i^2 = -1 \).

Question 3. If \( z = 6 + 8i \), verify that
(i) \( |z| = |\bar{z}| \)
(ii) \( -|z| < \text{Re}(z) \leq |z| \)
(iii) \( -|z| < \text{Im}(z) \leq |z| \)
(iv) \( z^{-1} = \frac{\bar{z}}{|z|^2} \)
Answer: Given \( z = 6 + 8i \).
First, let's find \( \bar{z} \) and \( |z| \).
The conjugate of \( z \), denoted as \( \bar{z} \), is obtained by changing the sign of the imaginary part: \( \bar{z} = 6 - 8i \).
The modulus of \( z \), denoted as \( |z| \), is \( \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10 \).

(i) To verify \( |z| = |\bar{z}| \):
We already found \( |z| = 10 \).
Now, let's find \( |\bar{z}| = |6 - 8i| = \sqrt{6^2 + (-8)^2} = \sqrt{36 + 64} = \sqrt{100} = 10 \).
Since \( 10 = 10 \), the statement \( |z| = |\bar{z}| \) is verified.

(ii) To verify \( -|z| < \text{Re}(z) \leq |z| \):
The real part of \( z \), \( \text{Re}(z) \), is 6.
So, we need to check if \( -10 < 6 \leq 10 \).
This statement is true because 6 is greater than -10 and less than or equal to 10. This property shows the real part of a complex number always lies between its modulus and negative modulus.

(iii) To verify \( -|z| < \text{Im}(z) \leq |z| \):
The imaginary part of \( z \), \( \text{Im}(z) \), is 8.
So, we need to check if \( -10 < 8 \leq 10 \).
This statement is true because 8 is greater than -10 and less than or equal to 10.

(iv) To verify \( z^{-1} = \frac{\bar{z}}{|z|^2} \):
First, let's find \( z^{-1} \).
\( z^{-1} = \frac{1}{z} = \frac{1}{6 + 8i} \)
To simplify this, multiply the numerator and denominator by the conjugate of the denominator:
\( z^{-1} = \frac{1}{6 + 8i} \times \frac{6 - 8i}{6 - 8i} = \frac{6 - 8i}{6^2 + 8^2} = \frac{6 - 8i}{36 + 64} = \frac{6 - 8i}{100} \)
Next, let's calculate \( \frac{\bar{z}}{|z|^2} \).
We know \( \bar{z} = 6 - 8i \) and \( |z|^2 = 10^2 = 100 \).
So, \( \frac{\bar{z}}{|z|^2} = \frac{6 - 8i}{100} \).
Since both sides are equal, the statement \( z^{-1} = \frac{\bar{z}}{|z|^2} \) is verified.
In simple words: We checked four rules for the complex number \( z=6+8i \). We first found its conjugate \( \bar{z}=6-8i \) and its size \( |z|=10 \). Then, we confirmed that the size of \( z \) is the same as the size of \( \bar{z} \). We also checked that the real and imaginary parts of \( z \) are always between -10 and 10. Finally, we showed that finding \( 1/z \) gives the same result as dividing \( \bar{z} \) by the square of \( |z| \).

🎯 Exam Tip: For verification problems, clearly state the given complex number and then compute each side of the equality or inequality separately to show they match. Remember to simplify complex fractions by multiplying by the conjugate.

Question 4. If \( Z_1 = 3 + 4i, z_2 = 8 - 15i \), verify that
(i) \( |-Z_1| = |Z_1| \)
(ii) \( |z_1^2| = |z_1|^2 \)
(iii) \( |Z_1Z_2| = |Z_1| |Z_2| \)
(iv) \( \left|\frac{z_1}{z_2}\right|=\frac{\left|z_1\right|}{\left|z_2\right|} \)
(v) \( |Z_1 + Z_2| < |Z_1| + |Z_2| \)
(vi) \( |Z_2 - Z_1| > ||z_2|-|z_1|| \)
(vii) \( |Z_1 + Z_2|^2 + |Z_1 - z_2|^2 = 2(|z_1|^2 + |z_2|^2) \)
Answer: Given \( Z_1 = 3 + 4i \) and \( Z_2 = 8 - 15i \).
First, let's find the moduli of \( Z_1 \) and \( Z_2 \).
\( |Z_1| = |3 + 4i| = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \).
\( |Z_2| = |8 - 15i| = \sqrt{8^2 + (-15)^2} = \sqrt{64 + 225} = \sqrt{289} = 17 \).

(i) To verify \( |-Z_1| = |Z_1| \):
\( -Z_1 = -(3 + 4i) = -3 - 4i \).
\( |-Z_1| = |-3 - 4i| = \sqrt{(-3)^2 + (-4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \).
Since \( 5 = 5 \), \( |-Z_1| = |Z_1| \) is verified. This means multiplying a complex number by -1 does not change its modulus.

(ii) To verify \( |z_1^2| = |z_1|^2 \):
First, find \( z_1^2 \):
\( z_1^2 = (3 + 4i)^2 = 3^2 + 2(3)(4i) + (4i)^2 = 9 + 24i + 16i^2 = 9 + 24i - 16 = -7 + 24i \).
Now, find \( |z_1^2| \):
\( |z_1^2| = |-7 + 24i| = \sqrt{(-7)^2 + 24^2} = \sqrt{49 + 576} = \sqrt{625} = 25 \).
We already know \( |z_1|^2 = 5^2 = 25 \).
Since \( 25 = 25 \), \( |z_1^2| = |z_1|^2 \) is verified.

(iii) To verify \( |Z_1Z_2| = |Z_1| |Z_2| \):
First, find \( Z_1Z_2 \):
\( Z_1Z_2 = (3 + 4i)(8 - 15i) = 3(8) + 3(-15i) + 4i(8) + 4i(-15i) \)
\( = 24 - 45i + 32i - 60i^2 = 24 - 13i + 60 = 84 - 13i \).
Now, find \( |Z_1Z_2| \):
\( |Z_1Z_2| = |84 - 13i| = \sqrt{84^2 + (-13)^2} = \sqrt{7056 + 169} = \sqrt{7225} = 85 \).
We already know \( |Z_1| |Z_2| = 5 \times 17 = 85 \).
Since \( 85 = 85 \), \( |Z_1Z_2| = |Z_1| |Z_2| \) is verified. This shows that the modulus of a product is the product of the moduli.

(iv) To verify \( \left|\frac{z_1}{z_2}\right|=\frac{\left|z_1\right|}{\left|z_2\right|} \):
First, find \( \frac{z_1}{z_2} \):
\( \frac{z_1}{z_2} = \frac{3 + 4i}{8 - 15i} \)
Multiply numerator and denominator by the conjugate of the denominator, \( 8 + 15i \):
\( = \frac{(3 + 4i)(8 + 15i)}{(8 - 15i)(8 + 15i)} = \frac{24 + 45i + 32i + 60i^2}{8^2 + 15^2} \)
\( = \frac{24 + 77i - 60}{64 + 225} = \frac{-36 + 77i}{289} \).
Now, find \( \left|\frac{z_1}{z_2}\right| \):
\( \left|\frac{-36 + 77i}{289}\right| = \frac{\sqrt{(-36)^2 + 77^2}}{289} = \frac{\sqrt{1296 + 5929}}{289} = \frac{\sqrt{7225}}{289} = \frac{85}{289} = \frac{5}{17} \).
We already know \( \frac{|Z_1|}{|Z_2|} = \frac{5}{17} \).
Since \( \frac{5}{17} = \frac{5}{17} \), \( \left|\frac{z_1}{z_2}\right|=\frac{\left|z_1\right|}{\left|z_2\right|} \) is verified.

(v) To verify \( |Z_1 + Z_2| < |Z_1| + |Z_2| \): (This is the triangle inequality, which holds unless \(Z_1\) and \(Z_2\) are in the same direction).
First, find \( Z_1 + Z_2 \):
\( Z_1 + Z_2 = (3 + 4i) + (8 - 15i) = (3 + 8) + (4 - 15)i = 11 - 11i \).
Now, find \( |Z_1 + Z_2| \):
\( |Z_1 + Z_2| = |11 - 11i| = \sqrt{11^2 + (-11)^2} = \sqrt{121 + 121} = \sqrt{242} = 11\sqrt{2} \).
We know \( |Z_1| + |Z_2| = 5 + 17 = 22 \).
Since \( 11\sqrt{2} \approx 11 \times 1.414 = 15.554 \) and \( 15.554 < 22 \), the statement \( |Z_1 + Z_2| < |Z_1| + |Z_2| \) is verified.

(vi) To verify \( |Z_2 - Z_1| > ||z_2|-|z_1|| \): (This is also a form of the triangle inequality).
First, find \( Z_2 - Z_1 \):
\( Z_2 - Z_1 = (8 - 15i) - (3 + 4i) = (8 - 3) + (-15 - 4)i = 5 - 19i \).
Now, find \( |Z_2 - Z_1| \):
\( |Z_2 - Z_1| = |5 - 19i| = \sqrt{5^2 + (-19)^2} = \sqrt{25 + 361} = \sqrt{386} \).
Next, find \( ||z_2|-|z_1|| \):
\( ||z_2|-|z_1|| = |17 - 5| = |12| = 12 \).
Since \( \sqrt{386} \approx 19.647 \) and \( 19.647 > 12 \), the statement \( |Z_2 - Z_1| > ||z_2|-|z_1|| \) is verified.

(vii) To verify \( |Z_1 + Z_2|^2 + |Z_1 - z_2|^2 = 2(|z_1|^2 + |z_2|^2) \): (This is the Parallelogram Law).
Left Hand Side (L.H.S.): \( |Z_1 + Z_2|^2 + |Z_1 - Z_2|^2 \).
We already found \( |Z_1 + Z_2| = 11\sqrt{2} \), so \( |Z_1 + Z_2|^2 = (11\sqrt{2})^2 = 121 \times 2 = 242 \).
We need \( Z_1 - Z_2 \):
\( Z_1 - Z_2 = (3 + 4i) - (8 - 15i) = (3 - 8) + (4 - (-15))i = -5 + 19i \).
Now, find \( |Z_1 - Z_2| \):
\( |Z_1 - Z_2| = |-5 + 19i| = \sqrt{(-5)^2 + 19^2} = \sqrt{25 + 361} = \sqrt{386} \).
So, \( |Z_1 - Z_2|^2 = (\sqrt{386})^2 = 386 \).
L.H.S. \( = 242 + 386 = 628 \).

Right Hand Side (R.H.S.): \( 2(|z_1|^2 + |z_2|^2) \).
We know \( |z_1|^2 = 5^2 = 25 \) and \( |z_2|^2 = 17^2 = 289 \).
R.H.S. \( = 2(25 + 289) = 2(314) = 628 \).
Since L.H.S. \( = \) R.H.S., the statement \( |Z_1 + Z_2|^2 + |Z_1 - z_2|^2 = 2(|z_1|^2 + |z_2|^2) \) is verified. This law is fundamental in understanding vector addition in complex numbers.
In simple words: We checked seven important rules about complex numbers using \( Z_1 = 3+4i \) and \( Z_2 = 8-15i \). We confirmed that the size of \( -Z_1 \) is the same as \( Z_1 \), and the size of \( Z_1 \) squared is the same as the square of \( Z_1 \)'s size. We also showed that when you multiply or divide complex numbers, you can just multiply or divide their sizes. We then confirmed two rules, called triangle inequalities, about the size of sums and differences of complex numbers. Finally, we proved the Parallelogram Law, which connects the sizes of the sum and difference of two complex numbers to the squares of their individual sizes.

🎯 Exam Tip: These properties are crucial for complex numbers. Remember to clearly state the values of \( Z_1 \), \( Z_2 \), their moduli, and then substitute them into each equation to perform the verification. The Parallelogram Law (vii) is often tested directly.

Question 5. Find the modulus of the following using the property of modulus.
(i) \( (3 + 4i) (8 - 6i) \)
(ii) \( \frac{8 + 15i}{8 - 6i} \)
(iii) \( \frac{3 + 2i}{2 - 5i} + \frac{3 - 2i}{2 + 5i} \)
(iv) \( \frac{(2 - 3i)(4 + 5i)}{(1 - 4i)(2 - i)} \)
Answer: We will find the modulus of each expression using the properties of moduli, such as \( |Z_1Z_2| = |Z_1||Z_2| \) and \( \left|\frac{Z_1}{Z_2}\right| = \frac{|Z_1|}{|Z_2|} \).

(i) For \( (3 + 4i) (8 - 6i) \):
Let \( Z = (3 + 4i)(8 - 6i) \).
Using the property \( |Z_1Z_2| = |Z_1||Z_2| \):
\( |Z| = |3 + 4i| \times |8 - 6i| \)
\( |3 + 4i| = \sqrt{3^2 + 4^2} = \sqrt{9 + 16} = \sqrt{25} = 5 \).
\( |8 - 6i| = \sqrt{8^2 + (-6)^2} = \sqrt{64 + 36} = \sqrt{100} = 10 \).
So, \( |Z| = 5 \times 10 = 50 \).

(ii) For \( \frac{8 + 15i}{8 - 6i} \):
Let \( Z = \frac{8 + 15i}{8 - 6i} \).
Using the property \( \left|\frac{Z_1}{Z_2}\right| = \frac{|Z_1|}{|Z_2|} \):
\( |Z| = \frac{|8 + 15i|}{|8 - 6i|} \)
\( |8 + 15i| = \sqrt{8^2 + 15^2} = \sqrt{64 + 225} = \sqrt{289} = 17 \).
\( |8 - 6i| = \sqrt{8^2 + (-6)^2} = \sqrt{64 + 36} = \sqrt{100} = 10 \).
So, \( |Z| = \frac{17}{10} \).

(iii) For \( \frac{3 + 2i}{2 - 5i} + \frac{3 - 2i}{2 + 5i} \):
Let \( Z = \frac{3 + 2i}{2 - 5i} + \frac{3 - 2i}{2 + 5i} \).
First, we need to combine the fractions by finding a common denominator:
\( Z = \frac{(3 + 2i)(2 + 5i) + (3 - 2i)(2 - 5i)}{(2 - 5i)(2 + 5i)} \)
\( = \frac{(6 + 15i + 4i + 10i^2) + (6 - 15i - 4i + 10i^2)}{2^2 + 5^2} \)
\( = \frac{(6 + 19i - 10) + (6 - 19i - 10)}{4 + 25} \)
\( = \frac{(-4 + 19i) + (-4 - 19i)}{29} \)
\( = \frac{-4 + 19i - 4 - 19i}{29} \)
\( = \frac{-8}{29} \).
Now, find the modulus of \( Z \):
\( |Z| = \left|\frac{-8}{29}\right| = \frac{8}{29} \).

(iv) For \( \frac{(2 - 3i)(4 + 5i)}{(1 - 4i)(2 - i)} \):
Let \( Z = \frac{(2 - 3i)(4 + 5i)}{(1 - 4i)(2 - i)} \).
Using the property \( \left|\frac{Z_1Z_2}{Z_3Z_4}\right| = \frac{|Z_1||Z_2|}{|Z_3||Z_4|} \):
\( |Z| = \frac{|2 - 3i||4 + 5i|}{|1 - 4i||2 - i|} \)
Calculate each modulus:
\( |2 - 3i| = \sqrt{2^2 + (-3)^2} = \sqrt{4 + 9} = \sqrt{13} \).
\( |4 + 5i| = \sqrt{4^2 + 5^2} = \sqrt{16 + 25} = \sqrt{41} \).
\( |1 - 4i| = \sqrt{1^2 + (-4)^2} = \sqrt{1 + 16} = \sqrt{17} \).
\( |2 - i| = \sqrt{2^2 + (-1)^2} = \sqrt{4 + 1} = \sqrt{5} \).
Now substitute these values:
\( |Z| = \frac{\sqrt{13} \times \sqrt{41}}{\sqrt{17} \times \sqrt{5}} = \frac{\sqrt{13 \times 41}}{\sqrt{17 \times 5}} = \frac{\sqrt{533}}{\sqrt{85}} \). This is the modulus of the complex number.
In simple words: To find the size (modulus) of these complex number expressions, we used rules that let us find the size of parts separately and then combine them. For multiplication, we multiplied the sizes. For division, we divided the sizes. For addition, we first added the numbers normally to get a single complex number, and then found its size. This helps simplify the calculations a lot.

🎯 Exam Tip: Always look for opportunities to use the properties \( |Z_1Z_2| = |Z_1||Z_2| \) and \( \left|\frac{Z_1}{Z_2}\right| = \frac{|Z_1|}{|Z_2|} \) as they can simplify calculations significantly. For addition or subtraction, first simplify the expression to a single complex number \( x + yi \) and then find its modulus \( \sqrt{x^2+y^2} \).

Question 6. If k number such that \( \left|\frac{z - 5i}{z + 5i}\right| = 1 \), then show that z is purely real.
Answer: We are given the condition \( \left|\frac{z - 5i}{z + 5i}\right| = 1 \). We need to show that \( z \) is a purely real number.
Let \( z = x + iy \), where \( x \) and \( y \) are real numbers.
Substitute \( z \) into the given equation:
\( \left|\frac{(x + iy) - 5i}{(x + iy) + 5i}\right| = 1 \)
\( \left|\frac{x + i(y - 5)}{x + i(y + 5)}\right| = 1 \)
Using the property \( \left|\frac{Z_1}{Z_2}\right| = \frac{|Z_1|}{|Z_2|} \):
\( \frac{|x + i(y - 5)|}{|x + i(y + 5)|} = 1 \)
This means \( |x + i(y - 5)| = |x + i(y + 5)| \).
Now, find the modulus on each side:
\( \sqrt{x^2 + (y - 5)^2} = \sqrt{x^2 + (y + 5)^2} \)
To remove the square roots, square both sides:
\( x^2 + (y - 5)^2 = x^2 + (y + 5)^2 \)
Expand the squared terms:
\( x^2 + (y^2 - 10y + 25) = x^2 + (y^2 + 10y + 25) \)
Subtract \( x^2 \) and \( y^2 \) from both sides:
\( -10y + 25 = 10y + 25 \)
Subtract 25 from both sides:
\( -10y = 10y \)
Add \( 10y \) to both sides:
\( 0 = 20y \)
Divide by 20:
\( y = 0 \).
Since \( z = x + iy \) and we found \( y = 0 \), it means \( z = x + i(0) = x \).
Therefore, \( z \) is a purely real number. This is a common method for finding loci in the complex plane.
In simple words: We are given that the distance from \( z \) to \( 5i \) is the same as the distance from \( z \) to \( -5i \). We let \( z \) be \( x + iy \). By using the definition of modulus and squaring both sides, we found that the imaginary part, \( y \), must be zero. If the imaginary part is zero, then \( z \) is a real number.

🎯 Exam Tip: When asked to prove a complex number is purely real (or purely imaginary), always start by assuming \( z = x + iy \). Use the modulus property \( |Z_1/Z_2| = |Z_1|/|Z_2| \) and then square both sides to eliminate square roots. The goal is to show that either \( x=0 \) or \( y=0 \).

Question 7. Find the complex number z satisfying the equation \( \left|\frac{z-12}{z-8 i}\right|=\frac{5}{3},\left|\frac{z-4}{z-8}\right| = 1 \).
Answer: We need to find the complex number \( z \) that satisfies two given equations.
Let \( z = x + iy \).

Equation 1: \( \left|\frac{z-12}{z-8 i}\right|=\frac{5}{3} \)
Substitute \( z = x + iy \):
\( \left|\frac{(x + iy) - 12}{(x + iy) - 8i}\right|=\frac{5}{3} \)
\( \left|\frac{(x - 12) + iy}{x + i(y - 8)}\right|=\frac{5}{3} \)
Using the property \( \left|\frac{Z_1}{Z_2}\right| = \frac{|Z_1|}{|Z_2|} \):
\( \frac{|(x - 12) + iy|}{|x + i(y - 8)|}=\frac{5}{3} \)
\( \frac{\sqrt{(x - 12)^2 + y^2}}{\sqrt{x^2 + (y - 8)^2}}=\frac{5}{3} \)
Square both sides to remove the square roots:
\( \frac{(x - 12)^2 + y^2}{x^2 + (y - 8)^2}=\frac{25}{9} \)
Cross-multiply:
\( 9[(x - 12)^2 + y^2] = 25[x^2 + (y - 8)^2] \)
Expand the squared terms:
\( 9[x^2 - 24x + 144 + y^2] = 25[x^2 + y^2 - 16y + 64] \)
\( 9x^2 - 216x + 1296 + 9y^2 = 25x^2 + 25y^2 - 400y + 1600 \)
Rearrange terms to one side:
\( (25x^2 - 9x^2) + (25y^2 - 9y^2) - 400y + 216x + 1600 - 1296 = 0 \)
\( 16x^2 + 16y^2 + 216x - 400y + 304 = 0 \)
This equation can be simplified by dividing by a common factor, but we'll keep it for now as (1). We can simplify later if needed.

Equation 2: \( \left|\frac{z-4}{z-8}\right| = 1 \)
Substitute \( z = x + iy \):
\( \left|\frac{(x + iy) - 4}{(x + iy) - 8}\right| = 1 \)
\( \left|\frac{(x - 4) + iy}{(x - 8) + iy}\right| = 1 \)
Using the property \( \left|\frac{Z_1}{Z_2}\right| = \frac{|Z_1|}{|Z_2|} \):
\( \frac{|(x - 4) + iy|}{|(x - 8) + iy|} = 1 \)
\( |(x - 4) + iy| = |(x - 8) + iy| \)
Find the modulus on each side and square both sides:
\( \sqrt{(x - 4)^2 + y^2} = \sqrt{(x - 8)^2 + y^2} \)
\( (x - 4)^2 + y^2 = (x - 8)^2 + y^2 \)
\( x^2 - 8x + 16 + y^2 = x^2 - 16x + 64 + y^2 \)
Cancel \( x^2 \) and \( y^2 \) from both sides:
\( -8x + 16 = -16x + 64 \)
Move \( -16x \) to the left and 16 to the right:
\( -8x + 16x = 64 - 16 \)
\( 8x = 48 \)
\( x = 6 \).

Now substitute \( x = 6 \) into Equation (1):
\( 16(6)^2 + 16y^2 + 216(6) - 400y + 304 = 0 \)
\( 16(36) + 16y^2 + 1296 - 400y + 304 = 0 \)
\( 576 + 16y^2 + 1296 - 400y + 304 = 0 \)
\( 16y^2 - 400y + (576 + 1296 + 304) = 0 \)
\( 16y^2 - 400y + 2176 = 0 \)
Divide the entire equation by 16 to simplify:
\( y^2 - 25y + 136 = 0 \)
This is a quadratic equation for \( y \). We can solve it using the quadratic formula \( y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a=1, b=-25, c=136 \).
\( y = \frac{-(-25) \pm \sqrt{(-25)^2 - 4(1)(136)}}{2(1)} \)
\( y = \frac{25 \pm \sqrt{625 - 544}}{2} \)
\( y = \frac{25 \pm \sqrt{81}}{2} \)
\( y = \frac{25 \pm 9}{2} \)
Two possible values for \( y \):
\( y_1 = \frac{25 + 9}{2} = \frac{34}{2} = 17 \)
\( y_2 = \frac{25 - 9}{2} = \frac{16}{2} = 8 \)
So, the possible complex numbers are \( z = x + iy \):
When \( x = 6 \) and \( y = 17 \), \( z = 6 + 17i \).
When \( x = 6 \) and \( y = 8 \), \( z = 6 + 8i \). These two complex numbers satisfy both given conditions. It's often helpful to sketch these solutions on an Argand diagram to visualize the conditions.
In simple words: We have two conditions for a complex number \( z \). We wrote \( z \) as \( x + iy \). From the first condition, we got a longer equation. From the second condition, we found that \( x \) must be 6. We put \( x=6 \) into the first equation and solved it to find two possible values for \( y \), which are 17 and 8. So, the complex numbers are \( 6 + 17i \) and \( 6 + 8i \).

🎯 Exam Tip: When solving simultaneous equations for complex numbers, always convert \( z \) to \( x + iy \) and separate the real and imaginary parts. Use the modulus properties carefully. Solving the quadratic equation for x or y is a common step, so be proficient in quadratic formula calculations.

Question 8. If \( |z - 1| = |z + 1| \), show that Re (z) = 0.
Answer: We are given the condition \( |z - 1| = |z + 1| \). We need to show that the real part of \( z \) is 0.
Let \( z = x + iy \), where \( x \) and \( y \) are real numbers.
Substitute \( z \) into the given equation:
\( |(x + iy) - 1| = |(x + iy) + 1| \)
Group the real and imaginary parts:
\( |(x - 1) + iy| = |(x + 1) + iy| \)
Now, find the modulus of each side:
\( \sqrt{(x - 1)^2 + y^2} = \sqrt{(x + 1)^2 + y^2} \)
Square both sides to eliminate the square roots:
\( (x - 1)^2 + y^2 = (x + 1)^2 + y^2 \)
Expand the squared terms:
\( x^2 - 2x + 1 + y^2 = x^2 + 2x + 1 + y^2 \)
Cancel \( x^2 \), \( y^2 \), and 1 from both sides:
\( -2x = 2x \)
Add \( 2x \) to both sides:
\( 0 = 4x \)
Divide by 4:
\( x = 0 \).
Since \( x \) is the real part of \( z \) (i.e., \( \text{Re}(z) = x \)), we have shown that \( \text{Re}(z) = 0 \). This condition geometrically means that z is equidistant from 1 and -1 on the complex plane, which describes the imaginary axis.
In simple words: We are given that the distance from \( z \) to 1 is the same as the distance from \( z \) to -1. When we write \( z \) as \( x + iy \) and use the formula for distance, we found that \( x \) must be zero. Since \( x \) is the real part of \( z \), this means the real part is 0.

🎯 Exam Tip: This problem is a common application of the distance interpretation of modulus. Geometrically, the condition \( |z - z_1| = |z - z_2| \) means \( z \) lies on the perpendicular bisector of the line segment joining \( z_1 \) and \( z_2 \). For \( z_1 = 1 \) and \( z_2 = -1 \), the perpendicular bisector is the imaginary axis, where the real part is always zero.

Question 9. Solve: \( |z| + z = 2 + i \), where z is a complex number.
Answer: We need to solve the equation \( |z| + z = 2 + i \) for \( z \).
Let \( z = x + iy \), where \( x \) and \( y \) are real numbers.
Then \( |z| = \sqrt{x^2 + y^2} \).
Substitute these into the given equation:
\( \sqrt{x^2 + y^2} + (x + iy) = 2 + i \)
Group the real and imaginary parts on the left side:
\( (\sqrt{x^2 + y^2} + x) + iy = 2 + 1i \)
Now, compare the real and imaginary parts on both sides of the equation.
Comparing the imaginary parts:
\( y = 1 \).
Comparing the real parts:
\( \sqrt{x^2 + y^2} + x = 2 \).
Substitute \( y = 1 \) into the real part equation:
\( \sqrt{x^2 + 1^2} + x = 2 \)
\( \sqrt{x^2 + 1} = 2 - x \).
To solve for \( x \), square both sides of the equation:
\( (\sqrt{x^2 + 1})^2 = (2 - x)^2 \)
\( x^2 + 1 = 4 - 4x + x^2 \).
Subtract \( x^2 \) from both sides:
\( 1 = 4 - 4x \).
Move the terms with \( x \) to one side and constants to the other:
\( 4x = 4 - 1 \)
\( 4x = 3 \)
\( x = \frac{3}{4} \).
We have \( x = \frac{3}{4} \) and \( y = 1 \).
Therefore, the complex number \( z \) is \( x + iy = \frac{3}{4} + 1i \). It's important to check that the solution satisfies the original equation, especially when squaring sides, which can introduce extraneous solutions. In this case, \( 2 - x \) must be non-negative, \( 2 - 3/4 = 5/4 \ge 0 \), so the solution is valid.
In simple words: We want to find the complex number \( z \) that fits the equation \( |z| + z = 2 + i \). We let \( z \) be \( x + iy \), where \( x \) is the real part and \( y \) is the imaginary part. By comparing the imaginary parts of the equation, we found \( y = 1 \). Then, by comparing the real parts and using \( y=1 \), we solved for \( x \) and found it to be \( \frac{3}{4} \). So, the number \( z \) is \( \frac{3}{4} + i \).

🎯 Exam Tip: When solving equations involving \( |z| \), always substitute \( z = x + iy \) and \( |z| = \sqrt{x^2 + y^2} \). Then, equate the real and imaginary parts separately. Remember to check for extraneous solutions after squaring both sides of an equation.