Anand Classes offers comprehensive NCERT Solutions for the Miscellaneous Exercise of Complex Numbers & Quadratic Equations (Class 11 Maths), with step-by-step derivations, solved examples, and full explanations. This set includes all types of problems—expressing complex numbers in a + ib form, operations on complex numbers, roots of quadratic equations, nature of roots, discriminants, and relations between roots—in a coherent and student-friendly manner. All solutions align with the NCERT curriculum and are ideal for exam preparation and revision. Students can download the solutions in free PDF format to access the content offline. Click the print button to download study material and notes.
1. Evaluate: $$([i^{18} + \frac{1}{i^{25}}]^3)$$
Solution :
$$
([i^{18} + \frac{1}{i^{25}}]^3)
$$
Rewrite powers as multiples of 4:
$$
i^{18} = i^{4 \cdot 4 + 2} = (i^4)^4 \cdot i^2
$$
$$
(1/i)^{25} = 1/i^{25} = 1/(i^{4\cdot6 + 1}) = 1/((i^4)^6 \cdot i)
$$
Since ($i^4 = 1$) and ($i^2 = -1$):
$$
[i^{18} + (1/i)^{25}]^3 = [-1 + 1/i]^3
$$
Rationalize (1/i):
$$
\frac{1}{i} \cdot \frac{i}{i} = \frac{i}{i^2} = -i
$$
Hence:
$$
[-1 + 1/i]^3 = (-1 – i)^3
$$
Expand the cube:
$$
(-1 – i)^3 = (-1)^3 + 3(-1)^2(-i) + 3(-1)(-i)^2 + (-i)^3
$$
Compute powers of (i) (($i^2=-1, i^3=-i$)):
$$
(-1 – i)^3 = -1 – 3i + 3 + i = 2 – 2i
$$
Answer:
$$(\boxed{2 – 2i})$$
2. Prove that $$(\Re(z_1 z_2) = \Re z_1 \Re z_2 – \Im z_1 \Im z_2)$$
Solution
Let
$$
z_1 = x_1 + i y_1, \quad z_2 = x_2 + i y_2
$$
Multiply:
$$
z_1 z_2 = (x_1 + i y_1)(x_2 + i y_2)
$$
Step by step:
$$
z_1 z_2 = x_1 x_2 + i x_1 y_2 + i x_2 y_1 + i^2 y_1 y_2
$$
Since ($i^2=-1$):
$$
z_1 z_2 = (x_1 x_2 – y_1 y_2) + i(x_1 y_2 + x_2 y_1)
$$
Hence:
$$
Re(z_1 z_2) = x_1 x_2 – y_1 y_2
$$
Also:
$$
Re z_1 \;Re z_2 – Im z_1\; Im z_2 = x_1 x_2 – y_1 y_2
$$
Hence proved.
$$
Re(z_1 z_2) = Re z_1 \;Re z_2 – Im z_1\; Im z_2
$$
Note : Similar property :
$$
Im(z_1 z_2) = (x_1 y_2 + x_2 y_1)
$$
$$
Re z_1 \;Im z_2 + Re z_2\; Im z_1 = (x_1 y_2 + x_2 y_1)
$$
$$
Im(z_1 z_2) = Re z_1 \;Im z_2 + Re z_2\; Im z_1
$$
3. Reduce to standard form:
$$
\left(\frac{1}{1 – 4i} – \frac{2}{i + 1}\right) \frac{3 – 4i}{5 + i}
$$
Solution
Step 1: Compute the bracket:
$$
\frac{1}{1 – 4i} – \frac{2}{i + 1} = \frac{-1 + 9i}{5 – 3i}
$$
Step 2: Multiply by $(\frac{3 – 4i}{5 + i})$
$$
\frac{-1 + 9i}{5 – 3i} \cdot \frac{3 – 4i}{5 + i}
$$
Step 3: Split numerator and denominator:
Numerator:
$$
(-1 + 9i)(3 – 4i) = (-1 \cdot 3) + (-1 \cdot -4i) + (9i \cdot 3) + (9i \cdot -4i) = 33 + 31i
$$
Denominator:
$$
(5 – 3i)(5 + i) = (5 \cdot 5) + (5 \cdot i) + (-3i \cdot 5) + (-3i \cdot i) = 28 – 10i
$$
Step 4: Rationalize:
$$
\frac{33 + 31i}{28 – 10i} \cdot \frac{28 + 10i}{28 + 10i} = \frac{614 + 1198i}{884}
$$
Step 5: Simplify:
$$
\frac{614 + 1198i}{884} = \frac{307}{442} + i \frac{599}{442}
$$
Answer:
$$
\boxed{\frac{307}{442} + i \frac{599}{442}}
$$
