Скачать или смотреть Class 9th unit 01 Real and complex numbers exercise 1.6 Q2 | Sindh board |

Exercise 1.6, Question 2 from Unit 1: Real and Complex Numbers (Class 9, Sindh Board). You can adjust wording or details as per your style or the exact question text.


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Chapter: Real and Complex Numbers

Exercise 1.6 — Question 2

Topic: Simplification of Complex Expressions, writing in the form


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What the Question Asks

In Question 2 of Exercise 1.6, you are required to simplify given complex-number expressions and present your answer in the standard form:

a + i b

where and are real numbers. This involves performing algebraic operations (addition, subtraction, multiplication, division) on complex numbers, and sometimes rationalizing denominators (i.e. removing from the denominator by multiplying numerator & denominator by the conjugate) when needed.


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Key Concepts & Techniques Needed

To solve this kind of question, you should recall:

1. Basic operations on complex numbers

2. Conjugate of a complex number

The conjugate of is .

Use it especially when you have a complex number in the denominator.



3. Division involving complex numbers

To divide , multiply numerator and denominator by the conjugate of the denominator:




\frac{p + i q}{r + i s} \cdot \frac{r - i s}{r - i s} = \frac{(p + i q)(r - i s)}{r^2 + s^2}

4. Simplification and combining like terms

After expansion, collect all real parts together and imaginary parts together.

Sometimes you may get , so replace that.





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A Possible Worked Outline (Generic)

Suppose the expression in Q2 is something like:

\frac{2 + 3i}{1 - 2i} + (4 + 5i)

Here’s how you might approach:

1. Simplify the fraction

\frac{2 + 3i}{1 - 2i} \cdot \frac{1 + 2i}{1 + 2i}
= \frac{(2 + 3i)(1 + 2i)}{1^2 + (2)^2}
= \frac{2 + 4i + 3i + 6i^2}{1 + 4}
= \frac{2 + 7i + 6(-1)}{5}
= \frac{2 - 6 + 7i}{5}
= \frac{-4}{5} + \frac{7}{5} i

2. Add the other complex number



\frac{-4}{5} + \frac{7}{5} i + (4 + 5i)
= \left(\frac{-4}{5} + 4\right) + i\left(\frac{7}{5} + 5\right)
= \left(\frac{-4}{5} + \frac{20}{5}\right) + i\left(\frac{7}{5} + \frac{25}{5}\right)
= \frac{16}{5} + \frac{32}{5} i
= \frac{16}{5} + \frac{32}{5} i

3. Final answer in form



\boxed{\frac{16}{5} + \frac{32}{5} i}

Note: The actual numbers in your question 2 may differ, but the same method applies.

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