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Math Notes For Class 11 Chapter 1 Ex 1.2 Solved & Quiz
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Math Notes For Class 11 Chapter 1 Ex 1.2 Solved & Quiz
A Complex Number is an expression of the form $z = a + bi$, where $a, b \in \mathbb{R}$ and $i = \sqrt{-1}$ represents the imaginary unit. Understanding the algebraic properties, conjugate relations, and modulus definitions is essential for solving Exercise 1.2.
| Concept / Operation | Mathematical Formula | Key Property / Rule |
|---|---|---|
| Powers of $i$ | $i^1 = i$, $i^2 = -1$, $i^3 = -i$, $i^4 = 1$ | $i^{4n + k} = i^k$ for $n \in \mathbb{Z}$ |
| Complex Conjugate | If $z = a + bi$, then $\bar{z} = a - bi$ | $z \cdot \bar{z} = a^2 + b^2 = |z|^2$ |
| Modulus of Complex Number | $|z| = \sqrt{a^2 + b^2}$ | $|z_1 \cdot z_2| = |z_1| \cdot |z_2|$ |
| Complex Division | $\frac{z_1}{z_2} = \frac{z_1 \cdot \bar{z}_2}{|z_2|^2}$ | Multiply numerator and denominator by $\bar{z}_2$ |
| Complex Factorization | $a^2 + b^2 = (a + bi)(a - bi)$ | Converts sum of squares to linear complex factors |
Question 1 (Part i): Simplify $i^{101}$.
Solution:
Divide the exponent $101$ by $4$: $101 = 4(25) + 1$.
Applying index laws:
$$i^{101} = i^{4(25) + 1} = (i^4)^{25} \cdot i^1$$
Since $i^4 = 1$, we get:
$$i^{101} = (1)^{25} \cdot i = 1 \cdot i = i$$
Answer: $$i$$
Question 1 (Part ii): Simplify $(-i)^{19}$.
Solution:
Separate the negative sign:
$$(-i)^{19} = (-1)^{19} \cdot i^{19} = -1 \cdot i^{19}$$
Express $i^{19}$ in terms of multiples of $4$: $19 = 4(4) + 3$.
$$i^{19} = (i^4)^4 \cdot i^3 = (1)^4 \cdot (-i) = -i$$
Substitute back into the expression:
$$(-i)^{19} = -1 \cdot (-i) = i$$
Answer: $$i$$
Question 1 (Part iii): Simplify $i^{-23}$.
Solution:
Rewrite with a positive exponent in the denominator:
$$i^{-23} = \frac{1}{i^{23}}$$
Express $23$ as $4(5) + 3$:
$$i^{23} = (i^4)^5 \cdot i^3 = (1)^5 \cdot (-i) = -i$$
Substitute back:
$$i^{-23} = \frac{1}{-i} = \frac{1 \cdot i}{-i \cdot i} = \frac{i}{-i^2} = \frac{i}{-(-1)} = \frac{i}{1} = i$$
Answer: $$i$$
Question 1 (Part iv): Simplify $(-1)^{-21/2}$.
Solution:
Recall that $\sqrt{-1} = i$, so $(-1)^{1/2} = i$.
$$(-1)^{-21/2} = \left((-1)^{1/2}\right)^{-21} = i^{-21}$$
Rewrite with a positive exponent:
$$i^{-21} = \frac{1}{i^{21}} = \frac{1}{(i^4)^5 \cdot i^1} = \frac{1}{(1)^5 \cdot i} = \frac{1}{i}$$
Multiply numerator and denominator by $i$:
$$\frac{1}{i} = \frac{1 \cdot i}{i \cdot i} = \frac{i}{i^2} = \frac{i}{-1} = -i$$
Answer: $$-i$$
Question 2 (Part i): Express in standard form $a + bi$: $(2 + 3i) + (4 - 5i)$.
Solution:
Group real parts and imaginary parts:
$$(2 + 3i) + (4 - 5i) = (2 + 4) + (3 - 5)i = 6 - 2i$$
Answer: $$6 - 2i$$
Question 2 (Part ii): Express in standard form $a + bi$: $(1 + 2i)(3 - 4i)$.
Solution:
Expand using the distributive property:
$$(1 + 2i)(3 - 4i) = 1(3) + 1(-4i) + 2i(3) + 2i(-4i)$$
$$= 3 - 4i + 6i - 8i^2$$
Since $i^2 = -1$:
$$= 3 + 2i - 8(-1) = 3 + 2i + 8 = 11 + 2i$$
Answer: $$11 + 2i$$
Question 2 (Part iii): Express in standard form $a + bi$: $\frac{2 + 3i}{1 - i}$.
Solution:
Multiply numerator and denominator by the conjugate of the denominator $(1 + i)$:
$$\frac{2 + 3i}{1 - i} = \frac{(2 + 3i)(1 + i)}{(1 - i)(1 + i)}$$
Numerator expansion:
$$(2 + 3i)(1 + i) = 2 + 2i + 3i + 3i^2 = 2 + 5i - 3 = -1 + 5i$$
Denominator expansion:
$$(1 - i)(1 + i) = 1^2 - i^2 = 1 - (-1) = 2$$
Combine and separate real and imaginary parts:
$$\frac{-1 + 5i}{2} = -\frac{1}{2} + \frac{5}{2}i$$
Answer: $$-\frac{1}{2} + \frac{5}{2}i$$
Question 2 (Part iv): Express in standard form $a + bi$: $\frac{(1 + i)^2}{2 - i}$.
Solution:
First, simplify the numerator $(1 + i)^2$:
$$(1 + i)^2 = 1 + 2i + i^2 = 1 + 2i - 1 = 2i$$
Now simplify the expression $\frac{2i}{2 - i}$ by multiplying by conjugate $(2 + i)$:
$$\frac{2i(2 + i)}{(2 - i)(2 + i)} = \frac{4i + 2i^2}{2^2 - i^2} = \frac{4i + 2(-1)}{4 - (-1)} = \frac{-2 + 4i}{5} = -\frac{2}{5} + \frac{4}{5}i$$
Answer: $$-\frac{2}{5} + \frac{4}{5}i$$
Question 3 (Part i): Find the conjugate and modulus of $z = 3 - 4i$.
Solution:
1. Conjugate ($\bar{z}$): Change sign of imaginary part.
$$\bar{z} = 3 + 4i$$
2. Modulus ($|z|$):
$$|z| = \sqrt{a^2 + b^2} = \sqrt{(3)^2 + (-4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5$$
Answer: $$\bar{z} = 3 + 4i, \quad |z| = 5$$
Question 3 (Part ii): Find the conjugate and modulus of $z = -2 + 5i$.
Solution:
1. Conjugate ($\bar{z}$):
$$\bar{z} = -2 - 5i$$
2. Modulus ($|z|$):
$$|z| = \sqrt{(-2)^2 + (5)^2} = \sqrt{4 + 25} = \sqrt{29}$$
Answer: $$\bar{z} = -2 - 5i, \quad |z| = \sqrt{29}$$
Question 3 (Part iii): Find the conjugate and modulus of $z = \frac{1 + i}{1 - i}$.
Solution:
First, simplify $z$ into $a + bi$ form:
$$z = \frac{(1 + i)(1 + i)}{(1 - i)(1 + i)} = \frac{1 + 2i + i^2}{1 - (-1)} = \frac{2i}{2} = i = 0 + 1i$$
1. Conjugate ($\bar{z}$):
$$\bar{z} = 0 - 1i = -i$$
2. Modulus ($|z|$):
$$|z| = \sqrt{0^2 + 1^2} = \sqrt{1} = 1$$
Answer: $$\bar{z} = -i, \quad |z| = 1$$
Question 4 (Part i): Find real numbers $x$ and $y$ if $(x + yi)(2 - 3i) = 4 + i$.
Solution:
Expand the left-hand side:
$$(x + yi)(2 - 3i) = 2x - 3xi + 2yi - 3yi^2 = (2x + 3y) + (-3x + 2y)i$$
Equate this to $4 + i$:
$$(2x + 3y) + (-3x + 2y)i = 4 + 1i$$
Equating real and imaginary parts gives a system of linear equations:
1) $2x + 3y = 4$
2) $-3x + 2y = 1$
Multiply (1) by 3 and (2) by 2:
$$6x + 9y = 12$$
$$-6x + 4y = 2$$
Add the two equations:
$$13y = 14 \implies y = \frac{14}{13}$$
Substitute $y = \frac{14}{13}$ into equation (1):
$$2x + 3\left(\frac{14}{13}\right) = 4 \implies 2x + \frac{42}{13} = 4$$
$$2x = 4 - \frac{42}{13} = \frac{52 - 42}{13} = \frac{10}{13} \implies x = \frac{5}{13}$$
Answer: $$x = \frac{5}{13}, \quad y = \frac{14}{13}$$