Complex Numbers and Quadratic Equations — NCERT Solutions

Given: $\left[i^{18} + \left(\dfrac{1}{i}\right)^{25}\right]^3$

Step 1: Simplify $i^{18}$:
$$i^{18} = i^{4\times4+2} = (i^4)^4 \cdot i^2 = 1 \cdot (-1) = -1$$

Step 2: Simplify $\left(\dfrac{1}{i}\right)^{25} = i^{-25}$:
$$i^{-25} = \frac{1}{i^{25}}, \quad i^{25} = i^{4\times6+1} = i$$
$$i^{-25} = \frac{1}{i} = \frac{-i}{-i^2} = \frac{-i}{1} \cdot \frac{i}{i}\Rightarrow \frac{1}{i}\times\frac{i}{i}=\frac{i}{i^2}=\frac{i}{-1}=-i$$

Step 3: Add and cube:
$$i^{18} + i^{-25} = -1 + (-i) = -1 - i$$
$$(-1-i)^3 = -(1+i)^3$$
$$(1+i)^2 = 1 + 2i + i^2 = 1 + 2i - 1 = 2i$$
$$(1+i)^3 = (1+i)(2i) = 2i + 2i^2 = 2i - 2 = -2 + 2i$$
$$-(1+i)^3 = -(-2+2i) = 2 - 2i$$

Answer: $2 - 2i$.

Given: $z_1$ and $z_2$ are any two complex numbers.

Let $z_1 = a + ib$ and $z_2 = c + id$, where $a,b,c,d \in \mathbb{R}$.

So $\operatorname{Re}z_1 = a,\ \operatorname{Im}z_1 = b,\ \operatorname{Re}z_2 = c,\ \operatorname{Im}z_2 = d$.

Compute $z_1 z_2$:
$$z_1 z_2 = (a+ib)(c+id) = ac + iad + ibc + i^2 bd = (ac - bd) + i(ad + bc)$$

Therefore:
$$\operatorname{Re}(z_1 z_2) = ac - bd = \operatorname{Re}z_1 \cdot \operatorname{Re}z_2 - \operatorname{Im}z_1 \cdot \operatorname{Im}z_2$$

Hence proved.

Step 1: Simplify $\dfrac{1}{1-4i}$:
$$\frac{1}{1-4i} = \frac{1+4i}{(1-4i)(1+4i)} = \frac{1+4i}{1+16} = \frac{1+4i}{17}$$

Step 2: Simplify $\dfrac{2}{1+i}$:
$$\frac{2}{1+i} = \frac{2(1-i)}{(1+i)(1-i)} = \frac{2(1-i)}{2} = 1-i$$

Step 3: Subtract:
$$\frac{1+4i}{17} - (1-i) = \frac{1+4i - 17(1-i)}{17} = \frac{1+4i-17+17i}{17} = \frac{-16+21i}{17}$$

Step 4: Simplify $\dfrac{3-4i}{5+i}$:
$$\frac{3-4i}{5+i} = \frac{(3-4i)(5-i)}{(5+i)(5-i)} = \frac{15-3i-20i+4i^2}{25+1} = \frac{15-23i-4}{26} = \frac{11-23i}{26}$$

Step 5: Multiply:
$$\frac{-16+21i}{17} \times \frac{11-23i}{26} = \frac{(-16+21i)(11-23i)}{442}$$
$$(-16+21i)(11-23i) = -176 + 368i + 231i - 483i^2 = -176 + 599i + 483 = 307 + 599i$$
$$= \frac{307 + 599i}{442} = \frac{307}{442} + \frac{599}{442}i$$

Answer: $\dfrac{307}{442} + \dfrac{599}{442}i$.

Given: $x - iy = \sqrt{\dfrac{a-ib}{c-id}}$

Step 1: Square both sides:
$$(x-iy)^2 = \frac{a-ib}{c-id}$$
$$x^2 - y^2 - 2xyi = \frac{a-ib}{c-id}$$

Step 2: Take modulus of both sides. Recall $|z^2| = |z|^2$ and $\left|\dfrac{z_1}{z_2}\right| = \dfrac{|z_1|}{|z_2|}$:
$$|(x-iy)^2| = \left|\frac{a-ib}{c-id}\right|$$
$$|x-iy|^2 = \frac{|a-ib|}{|c-id|}$$
$$(x^2+y^2) = \frac{\sqrt{a^2+b^2}}{\sqrt{c^2+d^2}}$$

Step 3: Square both sides:
$$(x^2+y^2)^2 = \frac{a^2+b^2}{c^2+d^2}$$

Hence proved.

Given: $z_1 = 2-i,\ z_2 = 1+i$

Step 1: Compute numerator $z_1+z_2+1$:
$$(2-i)+(1+i)+1 = 4+0i = 4$$

Step 2: Compute denominator $z_1-z_2+1$:
$$(2-i)-(1+i)+1 = 2-i-1-i+1 = 2-2i$$

Step 3: Compute the modulus:
$$\left|\frac{4}{2-2i}\right| = \frac{|4|}{|2-2i|} = \frac{4}{\sqrt{4+4}} = \frac{4}{2\sqrt{2}} = \frac{2}{\sqrt{2}} = \sqrt{2}$$

Answer: $\sqrt{2}$.

Given: $a + ib = \dfrac{(x+i)^2}{2x^2+1}$

Step 1: Expand the numerator:
$$(x+i)^2 = x^2 + 2xi + i^2 = x^2 - 1 + 2xi$$
$$a + ib = \frac{x^2-1}{2x^2+1} + \frac{2x}{2x^2+1}i$$

So $a = \dfrac{x^2-1}{2x^2+1}$ and $b = \dfrac{2x}{2x^2+1}$.

Step 2: Compute $a^2+b^2$:
$$a^2+b^2 = \frac{(x^2-1)^2 + 4x^2}{(2x^2+1)^2} = \frac{x^4 - 2x^2 + 1 + 4x^2}{(2x^2+1)^2} = \frac{x^4+2x^2+1}{(2x^2+1)^2} = \frac{(x^2+1)^2}{(2x^2+1)^2}$$

Hence proved.

Given: $z_1 = 2-i,\ z_2 = -2+i$

So $\bar{z}_1 = 2+i$.

(i) Find $\operatorname{Re}\left(\dfrac{z_1 z_2}{\bar{z}_1}\right)$:

Step 1: Compute $z_1 z_2$:
$$(2-i)(-2+i) = -4+2i+2i-i^2 = -4+4i+1 = -3+4i$$

Step 2: Divide by $\bar{z}_1 = 2+i$:
$$\frac{-3+4i}{2+i} = \frac{(-3+4i)(2-i)}{(2+i)(2-i)} = \frac{-6+3i+8i-4i^2}{4+1} = \frac{-6+11i+4}{5} = \frac{-2+11i}{5}$$

$$\operatorname{Re}\left(\frac{z_1 z_2}{\bar{z}_1}\right) = -\frac{2}{5}$$

(ii) Find $\operatorname{Im}\left(\dfrac{1}{z_1\bar{z}_1}\right)$:

Step 1: $z_1\bar{z}_1 = |z_1|^2 = (2)^2+(-1)^2 = 5$ (a real number).

Step 2: $\dfrac{1}{z_1\bar{z}_1} = \dfrac{1}{5}$, which is purely real.

$$\operatorname{Im}\left(\frac{1}{z_1\bar{z}_1}\right) = 0$$

Answers: (i) $-\dfrac{2}{5}$, (ii) $0$.

Given: $(x-iy)(3+5i) = \overline{(-6-24i)} = -6+24i$

Step 1: Expand the left side:
$$(x-iy)(3+5i) = 3x + 5xi - 3yi - 5yi^2 = 3x + 5y + (5x-3y)i$$

Step 2: Equate real and imaginary parts:
$$3x + 5y = -6 \quad \cdots (1)$$
$$5x - 3y = 24 \quad \cdots (2)$$

Step 3: Solve the system:

Multiply (1) by 3 and (2) by 5:
$$9x + 15y = -18$$
$$25x - 15y = 120$$
Adding: $34x = 102 \Rightarrow x = 3$

Substitute in (1): $9 + 5y = -6 \Rightarrow y = -3$

Answer: $x = 3,\ y = -3$.

Step 1: Simplify $\dfrac{1+i}{1-i}$:
$$\frac{1+i}{1-i} = \frac{(1+i)^2}{(1-i)(1+i)} = \frac{1+2i-1}{2} = \frac{2i}{2} = i$$

Step 2: Simplify $\dfrac{1-i}{1+i}$:
$$\frac{1-i}{1+i} = \frac{(1-i)^2}{(1+i)(1-i)} = \frac{1-2i-1}{2} = \frac{-2i}{2} = -i$$

Step 3: Subtract:
$$i - (-i) = 2i$$

Step 4: Find modulus:
$$|2i| = 2$$

Answer: $2$.

Given: $(x+iy)^3 = u+iv$

Step 1: Expand $(x+iy)^3$ using $(a+b)^3 = a^3+3a^2b+3ab^2+b^3$:
$$(x+iy)^3 = x^3 + 3x^2(iy) + 3x(iy)^2 + (iy)^3$$
$$= x^3 + 3x^2 yi + 3x(-y^2) + (-iy^3)$$
$$= (x^3 - 3xy^2) + i(3x^2y - y^3)$$

Step 2: Equate real and imaginary parts:
$$u = x^3 - 3xy^2, \quad v = 3x^2y - y^3$$

Step 3: Compute $\dfrac{u}{x} + \dfrac{v}{y}$:
$$\frac{u}{x} = \frac{x^3-3xy^2}{x} = x^2 - 3y^2$$
$$\frac{v}{y} = \frac{3x^2y - y^3}{y} = 3x^2 - y^2$$
$$\frac{u}{x} + \frac{v}{y} = x^2 - 3y^2 + 3x^2 - y^2 = 4x^2 - 4y^2 = 4(x^2-y^2)$$

Hence proved.

Given: $|\beta|=1$, $\alpha \neq \beta$.

Step 1: Compute $\left|\dfrac{\beta-\alpha}{1-\bar{\alpha}\beta}\right|^2$:
$$\left|\frac{\beta-\alpha}{1-\bar{\alpha}\beta}\right|^2 = \frac{|\beta-\alpha|^2}{|1-\bar{\alpha}\beta|^2}$$

Step 2: Expand $|\beta-\alpha|^2 = (\beta-\alpha)\overline{(\beta-\alpha)} = (\beta-\alpha)(\bar{\beta}-\bar{\alpha})$

Step 3: Expand $|1-\bar{\alpha}\beta|^2 = (1-\bar{\alpha}\beta)\overline{(1-\bar{\alpha}\beta)} = (1-\bar{\alpha}\beta)(1-\alpha\bar{\beta})$

Step 4: Expand both:
$$|\beta-\alpha|^2 = |\beta|^2 - \beta\bar{\alpha} - \bar{\beta}\alpha + |\alpha|^2 = 1 - \beta\bar{\alpha} - \bar{\beta}\alpha + |\alpha|^2$$
$$|1-\bar{\alpha}\beta|^2 = 1 - \bar{\alpha}\beta - \alpha\bar{\beta} + |\alpha|^2|\beta|^2 = 1 - \bar{\alpha}\beta - \alpha\bar{\beta} + |\alpha|^2$$

Since $|\beta|^2 = 1$, both expressions are equal.

$$\left|\frac{\beta-\alpha}{1-\bar{\alpha}\beta}\right|^2 = 1 \implies \left|\frac{\beta-\alpha}{1-\bar{\alpha}\beta}\right| = 1$$

Answer: $1$.

Given: $|1-i|^x = 2^x$

Step 1: Compute $|1-i|$:
$$|1-i| = \sqrt{1^2+(-1)^2} = \sqrt{2}$$

Step 2: Substitute:
$$(\sqrt{2})^x = 2^x$$
$$2^{x/2} = 2^x$$

Step 3: Equate exponents:
$$\frac{x}{2} = x \implies x = 2x \implies x = 0$$

The only solution is $x = 0$, which is not a non-zero integer.

Answer: There is no non-zero integral solution (the number of non-zero integral solutions is $0$).

Given: $(a+ib)(c+id)(e+if)(g+ih) = A+iB$

Step 1: Take modulus of both sides:
$$|(a+ib)(c+id)(e+if)(g+ih)| = |A+iB|$$

Step 2: Use the property $|z_1 z_2| = |z_1||z_2|$:
$$|a+ib|\cdot|c+id|\cdot|e+if|\cdot|g+ih| = |A+iB|$$
$$\sqrt{a^2+b^2}\cdot\sqrt{c^2+d^2}\cdot\sqrt{e^2+f^2}\cdot\sqrt{g^2+h^2} = \sqrt{A^2+B^2}$$

Step 3: Square both sides:
$$(a^2+b^2)(c^2+d^2)(e^2+f^2)(g^2+h^2) = A^2+B^2$$

Hence proved.

Step 1: Simplify $\dfrac{1+i}{1-i}$:
$$\frac{1+i}{1-i} = \frac{(1+i)(1+i)}{(1-i)(1+i)} = \frac{(1+i)^2}{1+1} = \frac{1+2i-1}{2} = \frac{2i}{2} = i$$

Step 2: The equation becomes:
$$i^m = 1$$

Step 3: Since $i^4 = 1$ and the powers of $i$ cycle with period 4, the smallest positive integer $m$ for which $i^m = 1$ is:
$$m = 4$$

Answer: The least positive integral value of $m$ is $\boxed{4}$.