Mathematics FSc First Year FBISE NBF 2024 Ex 1.2 (Q1 to 3)

Math11 FBISE 2024 Ex1.2 FSc

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Math11 FBISE 2024 Ex1.2 FSc

Mathematics FSc First Year FBISE NBF 2024 Ex 1.1 (Q1 to 3)

Mathematics FSc First Year FBISE NBF 2024 Ex 1.1 (Q4 to 7)

Math11 FBISE 2024 Ex1.2 FSc

Mathematics new book 2024 followed by the Federal Board of Intermediate and Secondary Education, Islamabad, Pakistan.

Unit 1: Complex Numbers

Exercise 1.2 (Solutions)

Question 1: Prove (i) Re(iz)&=-Im(z) and (ii) Im(iz)&=Re(z)

Solution: (i) Let

    \begin{align*} z&=x+iy \\iz&=ix+i^2y \\iz&=ix-y \\\Rightarrow \quad Re(iz)&=-y \\\Rightarrow \quad Re(iz)&=-Im(z)\end{align*}

(ii) Let

    \begin{align*} z&=x+iy \\iz&=ix+i^2y \\iz&=ix-y \\\Rightarrow \quad Im(iz)&= x \\\Rightarrow \quad Im(iz)&=Re(z)\end{align*}

Question 2: Prove that (z_1z_2)(z_3z_4)= (z_1z_3)(z_2z_4)= z_3(z_1z_2)z_4

Solution:

    \begin{align*} (z_1z_2)(z_3z_4) &= z_1(z_2(z_3z_4)) \quad \text{by associative property of multiplication} \\&= z_1((z_2z_3)z_4) \quad \text{by associative property of multiplication} \\&= (z_1(z_3z_2)z_4) \quad \text{by commutative property of multiplication} \\&=(z_1z_3)(z_2z_4) \quad \text{by associative property of multiplication} \end{align*}

Now

    \begin{align*} (z_1z_3)(z_2z_4)&= (z_3z_1)(z_2z_4)\quad \text{by commutative property of multiplication} \\&= z_3(z_1z_2)z_4 \quad \text{by associative property of multiplication}\end{align*}

Question 3: Prove following

{i) z is real iff z =\bar{z}
Solution: If z is real, then z = a, where a ∈ ℝ. Since the conjugate of a real number is itself, \bar{z} = a. Therefore, z =\bar{z}.

If z = \bar{z}, let z = a + bi where a, b ∈ ℝ. Then, \bar{z}= a - bi. Since z =\bar{z}, we have a + bi = a - bi. This implies that b = 0, so z =a, which is real.

(ii) \frac{z-\bar{z}}{z+\bar{z}}=i \frac{Im z}{Re z}
Solution: If z=x+iy then z=x-iy.
Now \frac{z-\bar{z}}{z+\bar{z}}=\frac{x+iy-(x-iy)}{x+iy+x-iy}=\frac{2iy}{2x}=i \frac{y}{x}=i \frac{Im z}{Re z}

Math11 FBISE 2024 Ex1.2 FSc

(iii) z is either real or pure imaginary iff (\bar{z})^2 = z^2
Solution:

Math11 FBISE 2024 Ex1.2 FSc

If z is either real or pure imaginary:
Case 1: z is real. \Rightarrow z = a \Rightarrow \bar{z} = a \Rightarrow \bar{z}^2 = a^2 = z^2

Case 2: z is pure imaginary. \Rightarrow z = bi \Rightarrow \bar{z} = -bi \Rightarrow \bar{z}^2 = b^2 = z^2, so (\bar{z})^2 = z^2.

If (\bar{z})^2 = z^2:
z = a + bi \Rightarrow \bar{z}^2= a^2 - 2abi + b^2 = a^2 + 2abi + b^2 = z^2

\Rightarrow -4abi = 0 \Rightarrow a = 0 or b = 0

Question 4: Find |z_2| if |z_1z_2|=16 and z_1=2-3i

Solution: Given z_1=2-3i
|z_1|=\sqrt{2^2+3^2}=\sqrt{13}

Then |z_2| = \frac{|z_1z_2|}{|z_1|} = \frac{16}{\sqrt{13}}

Question 5: Show that |z_1 + z_2|^2 - |z_1 - z_2|^2 = 4Re(z_1) Re(z_2)+ 4Im(z_1)Im(z_2)

Solution: Let z_1 = a + bi and z_2 = c + di, where a, b, c, d ∈ ℝ. Then, z_1 + z_2 = (a + c) + (b + d)i and z_1 - z_2 = (a - c) + (b - d)i.

|z_1 + z_2|^2= (a + c)^2 + (b + d)^2= a^2 + c^2 +2ac + b^2 + d^2 + 2bd
and
|z_1 - z_2|^2 = (a - c)^2 + (b - d)^2 = a^2 + c^2 - 2ac + b^2 + d^2 - 2bd
Subtracting these two expressions, we get:
|z_1 + z_2|^2 - |z_1 - z_2|^2 = 4ac + 4bd

Since Re(z_1) = a, Re(z_2) = c, Im(z_1)=b and Im(z_2)=d, so
|z_1 + z_2|^2 - |z_1 - z_2|^2 = 4Re(z_1) Re(z_2) + 4Im(z_1)Im(z_2)

Problem 6: Find the value of λ if |(z₁/z₂) + λ| = √(λ+2), where z₁ = 3 + i and z₂ = 1 + i.

Solution:

Calculate z₁/z₂: z₁/z₂ = (3 + i) / (1 + i) = (3 + i)(1 – i) / ((1 + i)(1 – i)) = (3 – 3i + i – i²) / (1 – i²) = (4 – 2i) / 2 = 2 – i

Substitute z₁/z₂ into the given equation:
|(2 – i) + λ| = √(λ+2)
| (2 + λ) – i | =√(λ+2)
√((2 + λ)² + (-1)²)=√(λ+2)
√((2 + λ)² + 1) = √(λ+2)
Taking squares on both sides:
(2 + λ)² + 1 = λ + 2
4 + 4λ + λ² + 1 – λ – 2 = 0
λ² + 3λ + 3 = 0
Using the quadratic formula:
λ = (-3 ± √(3² – 4(1)(3))) / (2(1)) = (-3 ± √(-3)) / 2 = (-3 ± i √3) / 2

Math11 FBISE 2024 Ex1.2 FSc

Problem 7: Prove √2 |z| ≥ (Re(z) + Im(z))

Solution: Let z = a + bi, where a, b ∈ ℝ. Then, Re(z) = a and Im(z) = b.

Step 1: Start with the inequality:

(|a| – |b|)^2 ≥ 0

Expanding the left side:

a² – 2|ab| + b² ≥ 0

Step 2: Add a² + b² to both sides:

a² + b² + a² – 2|ab| + b² ≥ a² + b²

Simplifying:

2(a² + b²) ≥ 2|ab| + a² + b²
then
2(a² + b²) ≥ (a + b)²
2 |z|² ≥ (Re(z) + Im(z))²
√2 |z| ≥ |Re(z)| + |Im(z)|

Math11 FBISE 2024 Ex1.2 FSc

Problem 8: Write in x, y form

i) |z – i| = 4

Solution: |2z – i| = 4,
|2(x + iy) – i|=4,
|2x + (2y – 1)i|=4,
√((2x)² + (2y – 1)²) = 4,
4x² + 4y² – 4y + 1 = 16
4x² + 4y² – 4y – 15= 0

ii) |z – 1| = |z + i|

Solution:

|(x + iy) – 1| = |(x + iy) + i|

√((x – 1)² + y²) = √((x)² + (y + 1)²)

Squaring both sides:

(x – 1)² + y² = x² + (y + 1)²

Simplifying:

-2x + 1 = 2y + 1

Rearranging:

y = -x

This is the equation of a line with slope -1 passing through the origin.

iii) |z – 4i| + |z + 4i| = 10

Solution:

|(x + iy) – 4i| + |(x + iy) + 4i| = 10

√(x² + (y – 4)²) + √(x² + (y + 4)²) = 10

This equation represents an ellipse.

iv) Re(z²) = 4

Solution:

z² = (x + iy)² = x² – y² + 2xyi

Re(z²) = x² – y²

So, x² – y² = 4

This is the equation of a hyperbola.

v) Im(z²) = -5

Solution:

Im(z²) = 2xy

So, 2xy = -5

This is the equation of a hyperbola.

vi) -2 ≤ Im(z + i) ≤ 0

Solution:

Im(z + i) = y + 1

So, -2 ≤ y + 1 ≤ 0

Subtracting 1 from all sides:

-3 ≤ y ≤ -1

Math11 FBISE 2024 Ex1.2 FSc

Mathematics FSc First Year FBISE NBF 2024 Ex 1.1 (Q1 to 3)

Mathematics FSc First Year FBISE NBF 2024 Ex 1.1 (Q4 to 7)

Mathematics FSc First Year FBISE NBF 2024 Ex 1.2 (Q1 to 3)

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