Math Notes For Class 11 Chapter 1 Ex 1.1 Solved & Quiz

Question 1 (Part i): Simplify $i^7$.

Solution:
Express $i^7$ in terms of $i^2$:
$$i^7 = i^6 \cdot i = (i^2)^3 \cdot i$$ Substitute $i^2 = -1$:
$$i^7 = (-1)^3 \cdot i = -1 \cdot i = -i$$
Answer: $$-i$$

Question 1 (Part ii): Simplify $i^{50}$.

Solution:
Express $i^{50}$ using power of $i^2$:
$$i^{50} = (i^2)^{25}$$ Since $i^2 = -1$:
$$i^{50} = (-1)^{25} = -1$$
Answer: $$-1$$

Question 1 (Part iii): Simplify $i^{-13}$.

Solution:
Write with a positive exponent in denominator:
$$i^{-13} = \frac{1}{i^{13}} = \frac{1}{i^{12} \cdot i} = \frac{1}{(i^2)^6 \cdot i}$$ Since $i^2 = -1$:
$$i^{-13} = \frac{1}{(-1)^6 \cdot i} = \frac{1}{1 \cdot i} = \frac{1}{i}$$ Multiply numerator and denominator by $i$:
$$\frac{1}{i} = \frac{i}{i^2} = \frac{i}{-1} = -i$$
Answer: $$-i$$

Question 1 (Part iv): Simplify $(-i)^{19}$.

Solution:
Factor out $-1$:
$$(-i)^{19} = (-1)^{19} \cdot i^{19} = -1 \cdot i^{18} \cdot i$$ Substitute $i^{18} = (i^2)^9 = (-1)^9 = -1$:
$$(-i)^{19} = -1 \cdot (-1) \cdot i = 1 \cdot i = i$$
Answer: $$i$$

Question 2 (Part i): Express $(2 + 3i) + (4 - 5i)$ in standard form $a + bi$.

Solution:
Group the real parts and imaginary parts separately:
$$(2 + 3i) + (4 - 5i) = (2 + 4) + (3 - 5)i = 6 + (-2)i = 6 - 2i$$
Answer: $$6 - 2i$$

Question 2 (Part ii): Express $(1 - 2i)(3 + 4i)$ in standard form $a + bi$.

Solution:
Expand using the distributive property:
$$(1 - 2i)(3 + 4i) = 1(3 + 4i) - 2i(3 + 4i) = 3 + 4i - 6i - 8i^2$$ Substitute $i^2 = -1$:
$$= 3 - 2i - 8(-1) = 3 - 2i + 8 = 11 - 2i$$
Answer: $$11 - 2i$$

Question 2 (Part iii): Express $\frac{2 + i}{3 - 2i}$ in standard form $a + bi$.

Solution:
Multiply numerator and denominator by the complex conjugate of denominator $(3 + 2i)$:
$$\frac{2 + i}{3 - 2i} = \frac{(2 + i)(3 + 2i)}{(3 - 2i)(3 + 2i)}$$ Numerator expansion:
$$(2 + i)(3 + 2i) = 6 + 4i + 3i + 2i^2 = 6 + 7i + 2(-1) = 4 + 7i$$ Denominator expansion:
$$(3 - 2i)(3 + 2i) = 3^2 - (2i)^2 = 9 - 4(-1) = 9 + 4 = 13$$ Divide real and imaginary parts:
$$\frac{4 + 7i}{13} = \frac{4}{13} + \frac{7}{13}i$$
Answer: $$\frac{4}{13} + \frac{7}{13}i$$

Question 3 (Part i): Find the real and imaginary parts of $(3 - 4i)^2$.

Solution:
Expand using identity $(u - v)^2 = u^2 - 2uv + v^2$:
$$(3 - 4i)^2 = 3^2 - 2(3)(4i) + (4i)^2 = 9 - 24i + 16i^2$$ Substitute $i^2 = -1$:
$$= 9 - 24i + 16(-1) = 9 - 24i - 16 = -7 - 24i$$ Therefore, $\text{Re}(z) = -7$ and $\text{Im}(z) = -24$.
Answer: $$\text{Re}(z) = -7, \quad \text{Im}(z) = -24$$

Question 3 (Part ii): Find the real and imaginary parts of $\frac{1 + 2i}{1 - i}$.

Solution:
Rationalize the expression by multiplying numerator and denominator by $(1 + i)$:
$$\frac{1 + 2i}{1 - i} = \frac{(1 + 2i)(1 + i)}{(1 - i)(1 + i)}$$ Numerator:
$$(1 + 2i)(1 + i) = 1 + i + 2i + 2i^2 = 1 + 3i - 2 = -1 + 3i$$ Denominator:
$$(1 - i)(1 + i) = 1^2 - i^2 = 1 - (-1) = 2$$ Separate parts:
$$\frac{-1 + 3i}{2} = -\frac{1}{2} + \frac{3}{2}i$$ Hence, $\text{Re}(z) = -\frac{1}{2}$ and $\text{Im}(z) = \frac{3}{2}$.
Answer: $$\text{Re}(z) = -\frac{1}{2}, \quad \text{Im}(z) = \frac{3}{2}$$

Question 4 (Part i): Find the conjugate $\bar{z}$ and modulus $|z|$ of $z = -2 + 5i$.

Solution:
1. Conjugate ($\bar{z}$): Change sign of imaginary part:
$$\bar{z} = -2 - 5i$$ 2. Modulus ($|z|$): Apply formula $|z| = \sqrt{a^2 + b^2}$ where $a = -2, b = 5$:
$$|z| = \sqrt{(-2)^2 + 5^2} = \sqrt{4 + 25} = \sqrt{29}$$
Answer: $$\bar{z} = -2 - 5i, \quad |z| = \sqrt{29}$$

Question 4 (Part ii): Find the conjugate $\bar{z}$ and modulus $|z|$ of $z = \frac{4 + i}{1 - i}$.

Solution:
First, simplify $z$ into standard form $a + bi$:
$$z = \frac{(4 + i)(1 + i)}{(1 - i)(1 + i)} = \frac{4 + 4i + i + i^2}{1^2 - i^2} = \frac{4 + 5i - 1}{1 - (-1)} = \frac{3 + 5i}{2} = \frac{3}{2} + \frac{5}{2}i$$ 1. Conjugate ($\bar{z}$):
$$\bar{z} = \frac{3}{2} - \frac{5}{2}i$$ 2. Modulus ($|z|$):
$$|z| = \sqrt{\left(\frac{3}{2}\right)^2 + \left(\frac{5}{2}\right)^2} = \sqrt{\frac{9}{4} + \frac{25}{4}} = \sqrt{\frac{34}{4}} = \frac{\sqrt{34}}{2}$$
Answer: $$\bar{z} = \frac{3}{2} - \frac{5}{2}i, \quad |z| = \frac{\sqrt{34}}{2}$$

Question 5 (Part i): Solve for real numbers $x$ and $y$ if $(x + iy)(2 - 3i) = 4 + i$.

Solution:
Expand the left-hand side:
$$(x + iy)(2 - 3i) = 2x - 3xi + 2yi - 3yi^2 = (2x + 3y) + (-3x + 2y)i$$ Set equal 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 Equation (1) by 3 and Equation (2) by 2:
$$6x + 9y = 12$$ $$-6x + 4y = 2$$ Adding these 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: $$x = \frac{5}{13}, \quad y = \frac{14}{13}$$

Question 5 (Part ii): Solve for real numbers $x$ and $y$ if $(x + 2i)^2 = 3 + 4yi$.

Solution:
Expand LHS:
$$(x + 2i)^2 = x^2 + 4xi + 4i^2 = (x^2 - 4) + 4xi$$ Equate to $3 + 4yi$:
$$(x^2 - 4) + 4xi = 3 + 4yi$$ Equate real parts:
$$x^2 - 4 = 3 \implies x^2 = 7 \implies x = \pm \sqrt{7}$$ Equate imaginary parts:
$$4x = 4y \implies y = x$$ Therefore, when $x = \sqrt{7}, y = \sqrt{7}$ and when $x = -\sqrt{7}, y = -\sqrt{7}$.
Answer: $$x = \pm \sqrt{7}, \quad y = \pm \sqrt{7} \quad (\text{with } x = y)$$

Question 6 (Part i): Prove that $\overline{z_1 + z_2} = \bar{z}_1 + \bar{z}_2$ for $z_1 = 3 + 2i$ and $z_2 = 1 - 4i$.

Solution:
Left Hand Side (LHS):
$$z_1 + z_2 = (3 + 2i) + (1 - 4i) = (3 + 1) + (2 - 4)i = 4 -