Mathematics FSc First Year FBISE NBF 2024

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

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

The solutions for the textbook mathematics were followed by the Federal Board of Intermediate and Secondary Education, Islamabad, Pakistan. The book is published by the National Book Foundation Islamabad.

Keep in mind the exam will be SLO-based. This means you will practice the questions from the book. But that exam doesn’t need to be exactly from this textbook. FBISE has provided the course outlines you must follow those outlines and prepare your exam accordingly.

If you found any error in the solution feel free to comment at the end of this post. We will update the solution within 48 hours.

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)

Unit 1: Complex Numbers

Complex numbers are a mathematical concept that extends the real number system to include imaginary numbers. They are essential in various fields, including electrical engineering, physics, and signal processing.

One of the key reasons for the importance of complex numbers is their ability to represent periodic phenomena. For example, in electrical engineering, alternating current (AC) can be represented using complex numbers. This allows for easier analysis and manipulation of AC circuits.

Exercise 1.1 (Solutions)

Question 1: Evaluate:

(i) i^{31}
Solution: i^{31} = (i)^{30} \cdot i = (i^2)^{15} \cdot i =(-1)^{15} \cdot i = -1 \cdot i = -i

(ii) (-i)^6
Solution: (-i)^6 =(-i)^{2 \cdot 3} = (i^2)^3 = (-1)^3 = -1

(iii) (-1)^{-\frac{13}{2}}
Solution: (-1)^{-\frac{13}{2}} = (i^2)^{-\frac{13}{2}} = i^{-13} = \frac{1}{i^{13}} = \frac{1}{i^{12}\cdot i} = \frac{1}{(i^2)^{6}\cdot i} = \frac{1}{(-1)^{6}\cdot i} = \frac{1}{(1) \cdot i} = \frac{1}{i} = \frac{1}{i} \cdt \frac{-i}{-i} = \frac{-i}{-i^2} = \frac{-i}{-(-1)} =-i

Keep in mind -1=i^2

(iv) \frac{2}{(-1)^{\frac{3}{2}}}
Solution: \frac{2}{(-1)^{\frac{3}{2}}} = \frac{2}{((-1)^3)^{\frac{1}{2}}} = \frac{2}{(-1)^{\frac{1}{2}}} = \frac{2}{-i} = \frac{2}{-i} \cdot \frac{i}{i} = \frac{2i}{-i^2} = 2i

(v) i^{23} + i^{58} + i^{21}
Solution: i^{23} + i^{58} + i^{21} = i^{22} \cdot i + (i^2)^{27} + i^{20} \cdot i
=(i^2)^{11} \cdot i+ (-1)^{27} + (i^2)^{10} \cdot i
= (-1)^{11} \cdot i-1+ (-1)^{10} \cdot i = -i - 1 + i = -1

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

Question 2: Write in simplified form

(i) (3 + 2i) + (2 + 4i)
Solution:

    \begin{align*} (3 + 2i) + (2 + 4i) &= (3 + 2) + (2i + 4i) \\ &=5 + 6i. \end{align*}

(ii) (4 + 3i) - (2 + 5i)
Solution:

    \begin{align*} (4 + 3i) - (2 + 5i) &= (4 - 2) + (3i - 5i) \\ &=2 - 2i. \end{align*}

(iii) (4 + 7i) +(4 - 7i)
Solution:

    \begin{align*} (4 + 7i) + (4 - 7i) &= 4 + 7i + 4 - 7i \\ &= 4 + 4 - 7i + 7i \\ &= 8 + 0i . \end{align*}

(iv) {2 + 5i}-{2 - 5i}
Solution:

    \begin{align*} {2 + 5i}-{2 - 5i} &= {2 - 2} + {5i + 5i} \\ &= 0 + 10i. \end{align*}

(v) (3+2i)(4- 3i)
Solution:

    \begin{align*} (3+2i)(4- 3i) &= 3 \cdot 4 + 3 \cdot (-3i) + 2i \cdot 4 + 2i \cdot (-3i) \\ &= 12 - 9i + 8i - 6i^2 \\ &= 12 - i + 6 \\ &= 18 - i \end{align*}

(vi) (3,2)\div (3,-1)
Solution:

    \begin{align*} \frac{3 + 2i}{3 - i} &= \frac{3 + 2i}{3 - i} \cdot \frac{3 + i}{3 + i} \\ &= \frac{(3 \cdot 3 + 3 \cdot i) + (2i \cdot 3 + 2i \cdot i)}{3^2 + 1^2} \\ &= \frac{9 + 3i + 6i + 2i^2}{9 + 1} \\ &= \frac{9 + 9i - 2}{10} \\ &= \frac{7 + 9i}{10} \\ &= \frac{7}{10} + \frac{9}{10}i \end{align*}

(vii) (1 + i)(1 - i)(2+i)
Solution:

    \begin{align*} (1 + i)(1 - i)(2+i) &= (1 \cdot 1 + 1 \cdot (-i) + i \cdot 1 + i \cdot (-i))(2 + i) \\ &= (1 - i^2)(2 + i) \\ &= (1 + 1)(2 + i) \\ &= 2(2 + i) \\ &= 4 + 2i \end{align*}

(viii) \frac{1}{2+3i} 
Solution:

    \begin{align*} \frac{1}{2+3i} &= \frac{1}{2+3i} \cdot \frac{2-3i}{2-3i} \\ &= \frac{1 \cdot 2 - 1 \cdot 3i}{2^2 + 3^2} \\ &= \frac{2 - 3i}{4 + 9} \\ &= \frac{2 - 3i}{13} \\ & = \frac{2}{13}- \frac{3i}{13} \end{align*}

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

Question 3: Simplified form

(i) \frac{(2 + i)\cdot (3 - 2i)}{1+i}
Solution:

    \begin{align*} \frac{(2 + i)(3 - 2i)}{1+i} &= \frac{8 - i}{1 + i} \\ &= \frac{8 - i}{1 + i} \cdot \frac{1 - i}{1 - i} \\ &= \frac{7 - 9i}{2} \\ &= \frac{7}{2} - \frac{9}{2}i \end{align*}

(ii) \frac{(1 + i)}{(2+i)^2}
Solution:

    \begin{align*} \frac{(1 + i)}{(2+i)^2} &= \frac{1 + i}{(2^2 + 2 \cdot 2 \cdot i + i^2)} \\ &= \frac{1 + i}{(4 + 4i - 1)} \\ &= \frac{1 + i}{3 + 4i} \cdot \frac{3 - 4i}{3 - 4i} \\ &= \frac{(1 \cdot 3 + 1 \cdot (-4i)) + (i \cdot 3 + i \cdot (-4i))}{3^2 + 4^2} \\ &= \frac{3 - 4i + 3i - 4i^2}{9 + 16} \\ &= \frac{3 - i + 4}{25} \\ &= \frac{7 - i}{25} \\ &= \frac{7}{25} - \frac{1}{25}i \end{align*}

(iii) \frac{1}{3 + i}-\frac{1}{3 - i}
Solution:

    \begin{align*} \frac{1}{3 + i}-\frac{1}{3 - i} &= \frac{3 - i}{(3 + i)(3 - i)} - \frac{3 + i}{(3 + i)(3 - i)} \\ &= \frac{-2i}{(3 + i)(3 - i)} \\ &= \frac{-2i}{10} \\ &= -\frac{1}{5}i \end{align*}

(iv) (1 + i)^{-2}+(1 - i)^{-2}
Solution:

    \begin{align*} (1 + i)^{-2}+(1 - i)^{-2} &= \left(\frac{1}{1 + i}\right)^2 + \left(\frac{1}{1 - i}\right)^2 \\ &= \left(\frac{1}{1 + i} \cdot \frac{1 - i}{1 - i}\right)^2 + \left(\frac{1}{1 - i} \cdot \frac{1 + i}{1 + i}\right)^2 \\ &= \left(\frac{1 - i}{1^2 + 1^2}\right)^2 + \left(\frac{1 + i}{1^2 + 1^2}\right)^2 \\ &= \left(\frac{1 - i}{2}\right)^2 + \left(\frac{1 + i}{2}\right)^2 \\ &= \frac{(1 - i)^2 + (1 + i)^2}{4} \\ &= \frac{1^2 - 2i + i^2 + 1^2 + 2i + i^2}{4} \\ &= \frac{2 + 2i^2}{4} \\ &= \frac{2 - 2}{4} \\ &= 0 \end{align*}

(v) (2 + i)^2 + \frac{7-4i}{2+i}
Solution:

    \begin{align*} \frac{7-4i}{2+i} + (2 + i)^2 &= \frac{7-4i}{2+i} \cdot \frac{2-i}{2-i} + (2 + i)(2 + i) \\ &= \frac{14 - 7i - 8i + 4i^2}{4 - i^2} + 4 + 4i + i^2 \\ &= \frac{10 - 15i}{5} + 3 + 4i \\ &= 2 - 3i + 3 + 4i \\ &= 5 + i \end{align*}

For the remaining questions of this exercise 1.1 please visit the link given below.

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

FBISE 2024 new course of mathematics FSc 11 national book foundation.
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Mathematics FSc First Year FBISE NBF 2024

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