Mathematics FSc First Year FBISE NBF 2024

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

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

FBISE 2024 new course of mathematics FSc 11 national book foundation.

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

Unit 1: Complex Numbers

Exercise 1.1 (Solutions)

Question 4: To solve this question, we can separate the real and imaginary parts and then solve the resulting system of equations.

(i) (2 + 3i)x + (1 + 3i)y+2 = 0
Solution:

    \begin{align*} (2 + 3i)x + (1 + 3i)y + 2 &= 0 \qquad \text{by comparing} \\ 2x + y + 2 &= 0 \qquad \text{(i)} \\ 3x + 3y &= 0 \qquad \text{(ii)}\\ \text{from (ii)} \qquad \qquad y &= -x \\ \text{from (i)} \quad \qquad x + 2 &= 0 \\ x &= -2 \\ y &= 2 \end{align*}

(ii) \frac{x}{1+i} + \frac{y}{1-2i} = 1
Solution:

    \begin{align*} \frac{x}{1+i} + \frac{y}{1-2i} &= 1 \\ \frac{x(1-2i) + y(1+i)}{(1 + i)(1 - 2i)} &= 1 \\ \frac{x(1-2i) + y(1+i)}{3 - i} &= 1 \\ x(1-2i) + y(1+i) &= 3 - i \qquad \text{by comparing} \\ x + y &= 3 \quad \qquad \text{(i)} \\ -2x + y &= -1 \qquad \text{(ii)} \\ \text{by subtracting (ii) from (i)} x &= \frac{4}{3} \\ y &= \frac{5}{3} \end{align*}

(iii) \frac{x}{2+i} = \frac{1-5i}{3-2i} + \frac{y}{2-i}
Solution:

    \begin{align*}\frac{x}{2+i} - \frac{y}{2-i}&= \frac{1-5i}{3-2i} \\\frac{x(2-i)-y(2+i)}{(2+i)(2-i)}&= \frac{1-5i}{3-2i} \cdot \frac{3+2i}{3+2i} \\\frac{2x-2y+(-x-y)i}{2^2-i^2}&= \frac{3-10i^2+2i-15i}{3^2-(2i)^2}\\\frac{2x-2y+(-x-y)i}{5}&= \frac{13-13i}{13}\\2x-2y+(-x-y)i&= 5-5i\\\text{By comparing real and imaginary parts} \quad 2x-2y&=5 \qquad \text{(i)} \\-x-y=-5 \quad \text{or} \quad 2x+2y&=10 \qquad \text{(ii)}\\\text{By adding (i) and (ii) we will get} \quad 4x=15 \quad \text{or} \quad x&=\frac{15}{4}\\\text{By using this value of x in (i), we can easily get} \quad y & = \frac{5}{4}\end{align*}

(iv) x(1+i)^2+y(2-i)^2=3+10i
Solution:

    \begin{align*}x(1+i)^2+y(2-i)^2=3+10i \\x(1+2i+i^2) + y(4-4i+i^2) = 3 + 10i \\x(2i) + y(3-4i) = 3 + 10i \\2xi + 3y - 4yi = 3 + 10i \\\Rightarrow 3y = 3 \quad \text{(equating real parts)} \\y = 1 \\\Rightarrow 2x - 4y = 10 \quad \text{(equating imaginary parts)} \\2x - 4(1) = 10 \\2x = 14 \\x = 7\end{align*}

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

Question 5: Find z from 4z - 3\bar{z} = \frac{1-18i}{2-i}

Solution:

    \begin{align*}4z - 3\bar{z} =\frac{1-18i}{2-i} = \frac{(1-18i)(2+i)}{(2-i)(2+i)} = \frac{2+i-36i+18}{5} = 4 - 7i \\4(a + bi) - 3(a - bi) = 4 - 7i \\4a + 4bi - 3a + 3bi = 4 - 7i \\a + 7bi = 4 - 7i \\a = 4 \\7b = -7 \\b = -1 \\z = 4 - i\end{align*}

Question 6: Write conjugates

Solution:

    \begin{align*}&\text{(i)} \quad 4 - 3i \Rightarrow \quad \text{Conjugate:} \quad 4 + 3i\\&\text{(ii)} \quad 3i + 8 \Rightarrow \quad \text{Conjugate:} \quad -3i + 8\\&\text{(iii)} \quad 2 + \sqrt{-\frac{1}{5}} \Rightarrow \quad 2 +\frac{i}{\sqrt5} \quad \text{Conjugate:} \quad 2 - \frac{i}{\sqrt5} \\&\text{(iv)} \quad \frac{5i}{2} + \frac{7}{8} \Rightarrow \quad \text{Conjugate:} \quad -\frac{5i}{2} + \frac{7}{8}\end{align*}

Question 7: Find magnitudes

Solution:

    \begin{align*}&\text{(i)} \quad 11 + 12i \\&Solution: |11 + 12i| = \sqrt{11^2 + 12^2} = \sqrt{265}\\&\text{(ii)} \quad (2+3i) - (2 + 6i) \\&Solution: |(2+3i) - (2 + 6i)| = |-3i| = \sqrt{0^2+(-3)^2} = \sqrt{9} = 3 \\&\text{(iii)} \quad (2-i)(6 + 3i) \\&Solution: |(2-i)(6 + 3i)| = |15| = \sqrt{15^2} = 15 \\&\text{(iv)} \quad \frac{3-2i}{2+i} \\&Solution: |\frac{3-2i}{2+i}| = \frac{\sqrt{13}}{\sqrt{5}} = \frac{\sqrt{65}}{5} \\&\text{(v)} \quad (\sqrt{3} - \sqrt{-8}) + (\sqrt{3} + \sqrt{-8}) \\&Solution: |(\sqrt{3} - \sqrt{-8})(\sqrt{3} + \sqrt{-8})| = |11| = 11\end{align*}

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

Mathematics FSc First Year FBISE NBF 2024

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