Class 11 Mathematics Chapter 4 Complex Numbers And Quadratics Equations

Key Concept: Complex Number Division

b) $\frac{3}{2} + \frac{5}{2}i$

[Solution Description] To divide two complex numbers, multiply numerator and denominator by the conjugate of the denominator:

$$\frac{4+i}{1-i} \cdot \frac{1+i}{1+i} = \frac{(4+i)(1+i)}{(1-i)(1+i)}$$

Calculate the denominator:

$$(1-i)(1+i) = 1^2 – (-i^2) = 1+1 = 2$$

Now, calculate the numerator:

$$(4+i)(1+i) = 4\cdot1 + 4\cdoti + i\cdot1 + i\cdoti = 4 + 4i + i – 1 = 3 + 5i$$

Therefore, the division simplifies to:

$$\frac{3+5i}{2} = \frac{3}{2} + \frac{5}{2}i$$

Consequently, the result of the division is $$\frac{3}{2} + \frac{5}{2}i$$.

Key Concept: Basic Addition

b) $7 – i$

[Solution Description] To find the sum, add the real parts and the imaginary parts separately:

The real part is $5 + 2 = 7$.

The imaginary part is $3 + (-4) = -1$.

Therefore, the sum is $7 – i$.

Key Concept: Complex Number Equations, Complex Conjugate Properties

a) $2 \pm 3i$

[Solution Description] The given quadratic equation is $$x^2 – 4x + 13 = 0$$. To find the roots, we calculate the discriminant $$D = b^2 – 4ac = (-4)^2 – 4(1)(13) = 16 – 52 = -36$$. Since $D < 0$, the roots are complex and can be determined by using the formula $$x = \frac{-b \pm \sqrt{D}}{2a}.$$ Substituting the values, we get $$x = \frac{-(-4) \pm \sqrt{-36}}{2 \times 1} = \frac{4 \pm 6i}{2}.$$ Thus, the roots are $2 \pm 3i$.

Key Concept: Square Roots

c) $3i$

[Solution Description] The square root of -9 can be expressed using $$\sqrt{-1} = i$$. Therefore, $$\sqrt{-9} = \sqrt{9} \cdot \sqrt{-1} = 3i$$.

Key Concept: Complex Multiplication

b) $10 – 5i$

[Solution Description] To find the product of two complex numbers, we use the formula:

$$z_1 z_2 = (ac – bd) + i(ad + bc)$$

where $z_1 = a + ib$ and $z_2 = c + id$. Here, $a = 4$, $b = 3$, $c = 1$, and $d = -2$.

Substitute these values into the formula:

$$ac – bd = (4)(1) – (3)(-2) = 4 + 6 = 10$$

$$ad + bc = (4)(-2) + (3)(1) = -8 + 3 = -5$$

Thus, the product is $10 – 5i$.

Key Concept: Modulus Formula

c) 10

[Solution Description] The modulus of a complex number $$z = a + ib$$ is given by $$|z| = \sqrt{a^2 + b^2}$$. For $$z = 6 + 8i$$, we have:

$$|z| = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10$$

Key Concept: Multiplication Basics

b) $23 + i2$

[Solution Description] The formula for multiplication of complex numbers $z_1 z_2 = (ac – bd) + i(ad + bc)$ is used here. Given $z_1 = 2 + i3$ and $z_2 = 4 – i5$, we identify $a = 2$, $b = 3$, $c = 4$, and $d = -5$.

$$ac – bd = (2)(4) – (3)(-5) = 8 + 15 = 23$$,

$$ad + bc = (2)(-5) + (3)(4) = -10 + 12 = 2$$.

Thus, the product is $23 + i2$.

Key Concept: Conjugate Identification

c) $5 + 7i$

[Solution Description] If $$z = a + ib$$, then the conjugate $$\overline{z} = a – ib$$. Here, $$z = 5 – 7i$$, so $$\overline{z} = 5 + 7i$$.

Key Concept: Basic Addition

c) $6 + 2i$

[Solution Description] To add the complex numbers, we simply add the real parts together and the imaginary parts together:

$$(4 + 2) + (3i – i) = 6 + 2i.$$

Key Concept: Basic Division

a) $1 + 3i$

[Solution Description] To divide $(4 + 2i)$ by $(1 – i)$, multiply both numerator and denominator by the conjugate of the denominator:

The conjugate of $(1 – i)$ is $(1 + i)$.

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

Simplify the denominator:

$$(1-i)(1+i) = 1^2 – i^2 = 1 – (-1) = 2$$

Simplify the numerator:

$$(4 + 2i)(1 + i) = 4 \times 1 + 4 \times i + 2i \times 1 + 2i \times i = 4 + 4i + 2i – 2 = 2 + 6i$$

Therefore,

$$\frac{2 + 6i}{2} = 1 + 3i$$

So, the answer is $1 + 3i$.

Key Concept: Basic Equality

c) $(4 – 5i)$ and $(4 – 5i)$

[Solution Description] To check if two complex numbers $z_1 = a + ib$ and $z_2 = c + id$ are equal, we compare their real and imaginary parts: $a = c$ and $b = d$. For pair C, the real and imaginary parts match: $4 = 4$ and $-5 = -5$. Therefore, they are equal.

Key Concept: Modulus Calculation

b) 13

[Solution Description] The modulus of a complex number $z = a + ib$ is calculated using:

$$|z| = \sqrt{a^2 + b^2}$$

Here, $a = 5$ and $b = -12$, so:

$$|z| = \sqrt{5^2 + (-12)^2} = \sqrt{25 + 144} = \sqrt{169} = 13$$

Key Concept: Real and Imaginary Axis

a) Real axis

[Solution Description] A complex number with an imaginary part equal to zero lies on the real axis. Since $-7 + 0i$ has no imaginary component, it is located on the real axis.

Key Concept: Modulus Calculation

d) $25$

[Solution Description] The modulus of a complex number $a + bi$ is calculated using the formula $$|z| = \sqrt{a^2 + b^2}$$.

In this case, $a = 7$ and $b = 24$. Thus,

$$|z| = \sqrt{7^2 + 24^2} = \sqrt{49 + 576} = \sqrt{625} = 25$$

The modulus is $25$.

Key Concept: Discriminant Calculation

c) 12, Two distinct real roots

[Solution Description] To find the discriminant, use the formula $\Delta = b^2 – 4ac$. Here, $a = 3$, $b = -6$, and $c = 2$.

$$\Delta = (-6)^2 – 4 \cdot 3 \cdot 2$$

$$= 36 – 24$$

$$= 12$$

Since $\Delta > 0$, the quadratic equation has two distinct real roots.

Key Concept: Roots Calculation

a) $x = 3$

[Solution Description] First, we find the discriminant $D$ with $a = 1$, $b = -6$, $c = 9$:

$$D = b^2 – 4ac = (-6)^2 – 4 \times 1 \times 9 = 36 – 36 = 0$$

As $D = 0$, the equation has two equal real roots. The roots are calculated as follows:

$$x = \frac{-b \pm \sqrt{D}}{2a} = \frac{6 \pm \sqrt{0}}{2 \times 1} = \frac{6}{2} = 3$$

Both roots are $x = 3$.

Key Concept: Identify Commutative Property

a) Addition

[Solution Description] An operation is considered commutative if changing the order of the operands does not affect the result. For complex numbers, both addition and multiplication are commutative operations. Hence:

– Addition: $$z_1 + z_2 = z_2 + z_1$$

– Multiplication: $$z_1 \cdot z_2 = z_2 \cdot z_1$$

However, checking between given options here:

The correct answer among these is only addition.

Key Concept: Additive Inverse Check

b) $0$

[Solution Description] The additive inverse of a complex number $z = a + ib$ is $-a – ib$. Adding the additive inverse to the original number gives zero: $$(8 + 2i) + (-8 – 2i) = 0$$.

Key Concept: Mirror Image

a) (3, -4)

[Solution Description] The mirror image of a point $(x, y)$ for a complex number $z = x + iy$ on the real axis is given by $(x, -y)$. Here, the point for $z = 3 + 4i$ is $(3, 4)$. Thus, its mirror image on the real axis will be $(3, -4)$.

Key Concept: Complex Number Addition

c) $2 + 3i$

[Solution Description] To find the sum of two complex numbers, we add the real parts together and the imaginary parts together:

Real part: $5 + (-3) = 2$

Imaginary part: $7i + (-4i) = 3i$

Therefore, the sum is $2 + 3i$.

Key Concept: Basic Definition

b) 7

[Solution Description] In a complex number of the form $a + ib$, the imaginary part is represented by $b$. Therefore, for the complex number $5 + 7i$, the imaginary part is $7$.

Key Concept: Basic Definition

a) $6 + 5i$

[Solution Description] The complex conjugate of a complex number $z = a + ib$ is given by $\overline{z} = a – ib$. Therefore, for $z = 6 – 5i$, the complex conjugate is $\overline{z} = 6 + 5i$.

Key Concept: Conceptual Understanding

b) The grouping of numbers does not change the sum.

[Solution Description] The associative law for addition states that the grouping of numbers does not change the sum: $$(z1 + z2) + z3 = z1 + (z2 + z3)$$.

Key Concept: Basic Roots

c) 2i and -2i

[Solution Description] The square roots of a negative number like $-4$ can be found using the formula $$\sqrt{-a} = \sqrt{a} \cdot i$$. So, for $-4$, we have:

$$\sqrt{-4} = \sqrt{4} \cdot i = 2i$$

Therefore, the two square roots are $2i$ and $-2i$.

Key Concept: Real and Imaginary Parts

a) Real: 4, Imaginary: 8

[Solution Description] The addition of two complex numbers $z_1 = a + ib$ and $z_2 = c + id$ results in:

$$z_1 + z_2 = (a + c) + i(b + d)$$

Given $z_1 = 7 + 2i$ and $z_2 = -3 + 6i$, we have:

$$a = 7, \, b = 2, \, c = -3, \, d = 6$$

Substitute these values:

$$z_1 + z_2 = (7 + (-3)) + i(2 + 6) = 4 + 8i$$

Therefore, the real part is 4, and the imaginary part is 8.

Key Concept: Addition of Complex Numbers

a) 10 + i

[Solution Description] To add two complex numbers $z_1 = a + ib$ and $z_2 = c + id$, we use the formula: $z_1 + z_2 = (a + c) + i(b + d)$. So, $(4 + 3i) + (6 – 2i) = (4 + 6) + i(3 – 2) = 10 + i$.

Key Concept: Complex Conjugate Product, Real World Application

a) $200e^{i0}$

[Solution Description] The complex impedance is $Z = 10 – 10i$. Its conjugate is $\overline{Z} = 10 + 10i$.

The product of $Z$ and its conjugate is:

$$Z \cdot \overline{Z} = (10 – 10i)(10 + 10i) = 10^2 – (10i)^2 = 100 + 100 = 200$$

Expressing this real number in polar form, the modulus is $200$ and the argument is $0$ radians.

Hence, the result in polar form is $200e^{i0}$.

Key Concept: Modulus Formula

c) 10

[Solution Description] The modulus of a complex number $z = a + ib$ is given by $|z| = \sqrt{a^2 + b^2}$. Here, $a = 6$ and $b = 8$. Therefore:

$$|6 + 8i| = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10$$

So, the modulus is 10.

Key Concept: Basic Definition

b) 5

[Solution Description] The imaginary part of a complex number $a + ib$ is represented by $b$. For the complex number $7 + 5i$, the imaginary part is $5$.

Key Concept: Complex Number Form

a) $5 + 0i$

[Solution Description] A complex number on the x-axis is of the form $a + i0$. Among the options, $5 + 0i$ fits this description.

Key Concept: Multiplicative Inverse

a) $\frac{4}{25} + \frac{3}{25}i$

[Solution Description] To find the multiplicative inverse of a complex number $$z = a + ib$$, we use the formula $$\frac{a}{a^2 + b^2} – \frac{b}{a^2 + b^2}i$$. Here, $$a = 4$$ and $$b = -3$$, so:

$$a^2 + b^2 = 4^2 + (-3)^2 = 16 + 9 = 25$$

The multiplicative inverse is:

$$\frac{4}{25} + \frac{3}{25}i$$

Therefore, the multiplicative inverse of $4 – 3i$ is $$\frac{4}{25} + \frac{3}{25}i$$.

Key Concept: Conjugate Identification

c) $6 – 2i$

[Solution Description] The conjugate of a complex number $z = x + yi$ is given by $z^* = x – yi$. For the complex number $6 + 2i$, the real part $x = 6$ and the imaginary part $y = 2$. Therefore, the conjugate is:

$z^* = 6 – 2i.$

Thus, the conjugate of $6 + 2i$ is $6 – 2i$.

Key Concept: Identity Concept

d) 1

[Solution Description] The multiplicative identity for complex numbers is a number that, when multiplied with any complex number, does not change its value. For complex numbers, this identity is $1$.

Key Concept: Real and Imaginary Parts

a) Real: 7, Imaginary: -4

[Solution Description] For a complex number $z = a + ib$, the real part is $a$ and the imaginary part is $b$. Here, $z = 7 – i4$, so the real part is $7$ and the imaginary part is $-4$.

Key Concept: Multiplication of Complex Numbers

b) $11 + 2i$

[Solution Description] To find the product, we use the distributive property:

$$(3 – 4i)(1 + 2i) = 3(1) + 3(2i) – 4i(1) – 4i(2i)$$

Simplifying each term gives:

$$= 3 + 6i – 4i – 8(i^2)$$

Since $i^2 = -1$, this becomes:

$$= 3 + 6i – 4i + 8$$

Combining like terms, we obtain: $= 11 + 2i$

Key Concept: Addition of Complex Numbers

c) $3 + 4i$

[Solution Description] To add two complex numbers, add their real parts and imaginary parts separately.

Given $$z_1 = 1 + i$$ and $$z_2 = 2 + 3i$$,

$$\text{Real part: }  1 + 2 = 3$$

$$\text{Imaginary part: }  1 + 3 = 4$$

Therefore,

$$z_1 + z_2 = 3 + 4i$$

Hence, the correct answer is $3 + 4i$.

Key Concept: Modulus Identification

c) 5

[Solution Description] The modulus of a complex number $z = x + iy$ is given by $|z| = \sqrt{x^2 + y^2}$. For $4 + 3i$, $x=4$ and $y=3$. So,

$$|z| = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5.$$

Key Concept: Additive Inverse

c) $2 – 5i$, sum is $0$

[Solution Description] The additive inverse of a complex number $(-2 + 5i)$ is found by changing the signs of both real and imaginary parts, which gives $(2 – 5i)$. Their sum is calculated as follows:

$$(-2 + 5i) + (2 – 5i) = (-2 + 2) + (5i – 5i) = 0 + 0i$$

Thus, the sum is $0$.

Key Concept: Simple Addition

a) 6 + 2i

[Solution Description] To add two complex numbers $(a+ib)$ and $(c+id)$, we use the rule: $$(a + ib) + (c + id) = (a+c) + i(b+d)$$.

Therefore, $$(4 + 3i) + (2 – i) = (4+2) + i(3-1) = 6 + 2i$$.

Key Concept: Subtraction with Variables

d) $x=7, y=4$

[Solution Description] To solve for $x$ and $y$, equate the real and imaginary parts from both sides of the equation:

From the real part:

$$x – 4 = 3$$

Solving for $x$:

$$x = 7$$

From the imaginary part:

$$y + 1 = 5$$

Solving for $y$:

$$y = 4$$

Thus, $x = 7$ and $y = 4$.