CONJUGATE OF A COMPLEX NUMBER
When two complex numbers differ only in the sign of $i$, they are said to be conjugates of each other.
Thus $x + iy$ and $x – iy$ are two conjugate complex numbers.
The conjugate of a complex number $z$ is denoted by $bar{z}$.
PROPERTIES OF CONJUGATES
(I) The conjugate of the conjugate of a complex number is the complex number itself,
i.e. $\overline{\overline{z}} = z$.
Proof. Let
$z = x + iy$, where $x, y \in \mathbb{R}$.
$\therefore \overline{z} = x – iy.$
$\therefore \overline{\overline{z}} = x – iy = x + iy = z.$
(II) The sum and product of two conjugate complex numbers are purely real.
Let $z = x + iy$. Then $\overline{z} = x – iy$, where $x, y \in \mathbb{R}$.
(i) Sum $= z + \overline{z} = (x + iy) + (x – iy) = 2x$, which is purely real.
(ii) Product $= z \cdot \overline{z} = (x + iy)(x – iy) = x^2 – i^2 y^2 = x^2 – (-1)y^2 = x^2 + y^2$, which is purely real.
(III) The conjugate of the sum (product) of two complex numbers is the sum (product) of their conjugates,
i.e. $\overline{z_1 + z_2} = \overline{z_1} + \overline{z_2}$ and $\overline{z_1 z_2} = \overline{z_1} \cdot \overline{z_2}$
Proof. Let
$z_1 = x_1 + iy_1$ and $z_2 = x_2 + iy_2$, $(x_1, x_2; y_1, y_2 \in \mathbb{R})$.
$\therefore \overline{z_1} = x_1 – iy_1$ and $\overline{z_2} = x_2 – iy_2.$
(i) $z_1 + z_2 = (x_1 + iy_1) + (x_2 + iy_2) = (x_1 + x_2) + i(y_1 + y_2).$
Hence, $\overline{z_1 + z_2} = (x_1 + x_2) – i(y_1 + y_2) = (x_1 – iy_1) + (x_2 – iy_2) = \overline{z_1} + \overline{z_2}.$
(ii) $z_1 z_2 = (x_1 + iy_1)(x_2 + iy_2) = (x_1 x_2 – y_1 y_2) + i(x_1 y_2 + y_1 x_2).$
$\therefore \overline{z_1 z_2} = (x_1 x_2 – y_1 y_2) – i(x_1 y_2 + y_1 x_2).$
Also $\overline{z_1} \cdot \overline{z_2} = (x_1 – iy_1)(x_2 – iy_2) = (x_1 x_2 – y_1 y_2) – i(x_1 y_2 + y_1 x_2).$
Hence, $\overline{z_1 z_2} = \overline{z_1} \cdot \overline{z_2}$
(IV)
(i) $\overline{z_1 – z_2} = \overline{z_1} – \overline{z_2}$
(ii) $\overline{\left(\dfrac{z_1}{z_2}\right)} = \dfrac{\overline{z_1}}{\overline{z_2}}$, $z_2 \neq 0.$
Proof. Let
$z_1 = x_1 + iy_1$ and $z_2 = x_2 + iy_2$, $(x_1, x_2; y_1, y_2 \in \mathbb{R})$.
$\therefore \overline{z_1} = x_1 – iy_1$ and $\overline{z_2} = x_2 – iy_2.$
(i) $z_1 – z_2 = (x_1 + iy_1) – (x_2 + iy_2) = (x_1 – x_2) + i(y_1 – y_2).$
$\therefore \overline{z_1 – z_2} = (x_1 – x_2) – i(y_1 – y_2) = (x_1 – iy_1) – (x_2 – iy_2) = \overline{z_1} – \overline{z_2}.$
(ii) $\dfrac{z_1}{z_2} = \dfrac{x_1 + iy_1}{x_2 + iy_2} = \dfrac{x_1 + iy_1}{x_2 + iy_2} \times \dfrac{x_2 – iy_2}{x_2 – iy_2}$
$= \dfrac{(x_1x_2 + y_1y_2) + i(x_2y_1 – x_1y_2)}{x_2^2 + y_2^2} = \dfrac{x_1x_2 + y_1y_2}{x_2^2 + y_2^2} + i \dfrac{x_2y_1 – x_1y_2}{x_2^2 + y_2^2}.$
$\therefore \overline{\left(\dfrac{z_1}{z_2}\right)} = \dfrac{x_1x_2 + y_1y_2}{x_2^2 + y_2^2} – i \dfrac{x_2y_1 – x_1y_2}{x_2^2 + y_2^2}.$
Also
$\dfrac{\overline{z_1}}{\overline{z_2}} = \dfrac{x_1 – iy_1}{x_2 – iy_2} = \dfrac{x_1 – iy_1}{x_2 – iy_2} \times \dfrac{x_2 + iy_2}{x_2 + iy_2}$
$= \dfrac{(x_1x_2 + y_1y_2) + i(x_2y_2 – x_1y_1)}{x_2^2 + y_2^2} = \dfrac{x_1x_2 + y_1y_2}{x_2^2 + y_2^2} – i \dfrac{x_2y_2 – x_1y_2}{x_2^2 + y_2^2}.$
Hence,
$\overline{\left(\dfrac{z_1}{z_2}\right)} = \dfrac{\overline{z_1}}{\overline{z_2}}, z_2 \neq 0.$
Frequently Asked Questions
FAQs
Example 1. Find real and imaginary parts of $\dfrac{1-i}{1+i}$.
Solution.
$$
\frac{1-i}{1+i} = \frac{1-i}{1+i} \times \frac{1-i}{1-i}
$$
$$
= \frac{(1-i)^2}{1-i^2} = \frac{1+i^2 – 2i}{1-i^2}
$$
$$
= \frac{1-1-2i}{1-(-1)} = \frac{-2i}{2} = -i
$$
$$
= 0 – i.
$$
Hence, real part = 0 and imaginary part = -1.
Example 2. Write the following in the form $x + iy$ :
(i) $(3 + 2i)(2 – i)$
(ii) $\dfrac{3 + 4i}{2 – 4i}$
(iii) $2i^2 + 6i^3 + 3i^{16} – 6i^{19} + 4i^{25}$
(iv) $\dfrac{(3 – 2i)(2 + 3i)}{(1 + 2i)(2 – i)}$
(v) $\dfrac{1}{2 + \sqrt{-5}}$
(vi) $\dfrac{1}{1 – \cos \theta + 2i \sin \theta}$.
Solution.
(i) $(3 + 2i)(2 – i) = 6 – 3i + 4i – 2i^2$
$$
= 6 + i – 2(-1) = 6 + i + 2 = 8 + i,
$$
which is of the form $x + iy$. Here $x = 8$ and $y = 1$.
(ii) [Method. To get a real quantity in the denominator of the quotient of two complex numbers, we multiply the numerator and denominator by the conjugate of the denominator.]
$$
\frac{3 + 4i}{2 – 4i} = \frac{3 + 4i}{2 – 4i} \times \frac{2 + 4i}{2 + 4i}
$$
$$
= \frac{6 + 12i + 8i + 16i^2}{4 – 16i^2} = \frac{6 + 20i + 16(-1)}{4 – 16(-1)}
$$
$$
= \frac{6 + 20i – 16}{4 + 16} = \frac{-10 + 20i}{20} = -\frac{1}{2} + i,
$$
which is of the form $x + iy$. Here $x = -\dfrac{1}{2}$ and $y = 1$.
(iii) $2i^2 + 6i^3 + 3i^{16} – 6i^{19} + 4i^{25}$
$$
= 2(-1) + 6(-1)i + 3(1)^4 – 6(1)^4 \times (-1) \times i + 4(1)^6 \times i
$$
$$
= -2 – 6i + 3 + 6i + 4i = 1 + 4i,
$$
which is of the form $x + iy$. Here $x = 1$ and $y = 4$.
(iv) $\dfrac{(3-2i)(2+3i)}{(1+2i)(2-i)} = \dfrac{6+5i-6i^2}{2+3i-2i^2} = \dfrac{6+5i-6(-1)}{2+3i-2(-1)}$
$$
= \dfrac{12+5i}{4+3i} \times \dfrac{4-3i}{4-3i}
$$
$$
= \dfrac{48-16i-15i^2}{16-9i^2}
$$
$$
= \dfrac{48-16i-15(-1)}{16-9(-1)}
$$
$$
= \dfrac{63-16i}{25} = \dfrac{63}{25} – \dfrac{16}{25}i,
$$
which is of the form $x + iy$.
Here $x = \dfrac{63}{25}$ and $y = -\dfrac{16}{25}$.
(v) $\dfrac{1}{2 + \sqrt{-5}} = \dfrac{1}{2 + i\sqrt{5}}$
$$
= \dfrac{1}{2 + i\sqrt{5}} \times \dfrac{2 – i\sqrt{5}}{2 – i\sqrt{5}}
$$
$$
= \dfrac{2 – i\sqrt{5}}{(2)^2 – (i\sqrt{5})^2} = \dfrac{2 – i\sqrt{5}}{4 + 5} = \dfrac{2}{9} – \dfrac{\sqrt{5}}{9}i,
$$
which is of the form $x + iy$.
Here $x = \dfrac{2}{9}$ and $y = -\dfrac{\sqrt{5}}{9}$.
(vi) $\dfrac{1}{1 – \cos \theta + 2i \sin \theta}$
$$
= \dfrac{1}{(1 – \cos \theta) + 2i \sin \theta} \times \dfrac{(1 – \cos \theta) – 2i \sin \theta}{(1 – \cos \theta) – 2i \sin \theta}
$$
$$
= \dfrac{1 – \cos \theta – 2i \sin \theta}{(1 – \cos \theta)^2 – 4i^2 \sin^2 \theta} = \dfrac{1 – \cos \theta – 2i \sin \theta}{(1 – \cos \theta)^2 + 4 \sin^2 \theta}
$$
$$
= \dfrac{1 – \cos \theta – 2i \sin \theta}{1 – 2 \cos \theta + \cos^2 \theta + 4 \sin^2 \theta}
$$
$$
= \dfrac{1 – \cos \theta – 2i \sin \theta}{1 + (\cos^2 \theta + \sin^2 \theta) – 2 \cos \theta + 3 \sin^2 \theta}
$$
$$
= \dfrac{1 – \cos \theta – 2i \sin \theta}{2 – 2 \cos \theta + 3 \sin^2 \theta}
$$
$$
= \left( \dfrac{1 – \cos \theta}{2 – 2 \cos \theta + 3 \sin^2 \theta} \right) + i \left( \dfrac{-2 \sin \theta}{2 – 2 \cos \theta + 3 \sin^2 \theta} \right),
$$
which is of the form $x + iy$.
Here $x = \dfrac{1 – \cos \theta}{2 – 2 \cos \theta + 3 \sin^2 \theta}$ and $y = \dfrac{-2 \sin \theta}{2 – 2 \cos \theta + 3 \sin^2 \theta}$.
Example 3. (a) Express $\dfrac{1+7i}{(2-i)^2}$ in the form $a + ib$.
(b) Reduce $\dfrac{1}{1-4i} – \dfrac{2}{1+i}$ in the form $a + ib$.
Solution.
(a) $\dfrac{1+7i}{(2-i)^2} = \dfrac{1+7i}{4 + i^2 – 4i}$
$$
= \dfrac{1+7i}{4-1-4i} = \dfrac{1+7i}{3-4i}
$$
$$
= \dfrac{1+7i}{3-4i} \times \dfrac{3+4i}{3+4i} = \dfrac{3+4i+21i+28i^2}{9-16i^2}
$$
$$
= \dfrac{3+25i+28(-1)}{9-16(-1)} = \dfrac{3+25i-28}{9+16}
$$
$$
= \dfrac{-25+25i}{25} = -1 + i,
$$
which is of the form $a + ib$.
Here $a = -1$ and $b = 1$.
(b) $\dfrac{1}{1-4i} – \dfrac{2}{1+i} = \dfrac{1+i-2(1-4i)}{(1-4i)(1+i)}$
$$
= \dfrac{1+i-2+8i}{1+i-4i-4i^2} = \dfrac{-1+9i}{1-3i-4(-1)}
$$
$$
= \dfrac{-1+9i}{1-3i+4} = \dfrac{-1+9i}{5-3i} = \dfrac{-1+9i}{5-3i} \times \dfrac{5+3i}{5+3i}
$$
$$
= \dfrac{-5-3i+45i+27i^2}{25-9i^2} = \dfrac{-5+42i+27(-1)}{25-9(-1)}
$$
$$
= \dfrac{-5+42i-27}{25+9}
$$
$$
= \dfrac{-32+42i}{34} = -\dfrac{16}{17} + \dfrac{21}{17}i,
$$
which is of the form $a + ib$.
Here $a = -\dfrac{16}{17}$ and $b = \dfrac{21}{17}$.
Example 4. Perform the indicated operation and give your answer in the form $x + iy$, where $x$ and $y$ are real numbers and $i = \sqrt{-1}$:
(i) $\left(\dfrac{1}{2} + \dfrac{1}{4}i\right) \left(-\dfrac{2}{3} – \dfrac{1}{4}i\right)$
(ii) $\dfrac{5 + 2i}{1 + \sqrt{3}i}$
(iii) $(\sqrt{5} – 7i)(\sqrt{5} – 7i)^2 + (-2 + 7i)^2$
Solution.
(i) $\left(\dfrac{1}{2} + \dfrac{1}{4}i\right) \left(-\dfrac{2}{3} – \dfrac{1}{4}i\right)$
$= -\dfrac{1}{3} – \dfrac{1}{8}i – \dfrac{1}{6}i – \dfrac{1}{16}i^2 $
$= \left(-\dfrac{1}{3} + \dfrac{1}{16}\right) – \left(\dfrac{1}{8} + \dfrac{1}{6}\right)i $
$= -\dfrac{13}{48} – \dfrac{7}{24}i,$
which is of the form $x + iy$.
Here $x = -\dfrac{13}{48}$ and $y = -\dfrac{7}{24}$.
(ii) $\dfrac{5 + 2i}{1 + \sqrt{3}i} = \dfrac{5 + 2i}{1 + \sqrt{3}i} \times \dfrac{1 – \sqrt{3}i}{1 – \sqrt{3}i}$
$
= \dfrac{5 – 5\sqrt{3}i + 2i – 2\sqrt{3}i^2}{1 – 3i^2} $
$= \dfrac{-5 + 2\sqrt{3} + (-5\sqrt{3} + 2)i}{1 + 3} $
$= \dfrac{-5 + 2\sqrt{3}}{4} + \dfrac{-5\sqrt{3} + 2}{4}i,$
which is of the form $x + iy$.
Here $x = \dfrac{-5 + 2\sqrt{3}}{4}$ and $y = \dfrac{-5\sqrt{3} + 2}{4}$.
(iii) $(\sqrt{5} – 7i)(\sqrt{5} – 7i)^2 + (-2 + 7i)^2$
$$
= (\sqrt{5} – 7i)^3 + (-2 + 7i)^2
$$
$$
= [5\sqrt{5} – 3(\sqrt{5})^2(7i) + 3(\sqrt{5})(7i)^2 – (7i)^3] + [4 – 28i + 49i^2]
$$
$$
= (5\sqrt{5} – 105i + 147\sqrt{5}i^2 – 343i^3) + (4 – 28i – 49)
$$
$$
= (5\sqrt{5} – 105i – 147\sqrt{5} + 343i) + (-45 – 28i)
$$
$$
= (5\sqrt{5} – 147\sqrt{5} – 45) + (-105i + 343i – 28i)
$$
$$
= -(142\sqrt{5} + 45) + 210i,
$$
which is of the form $x + iy$.
Here $x = -(142\sqrt{5} + 45)$ and $y = 210$.
Example 5. Reduce $\left( \dfrac{1}{1-2i} + \dfrac{1}{1+i} \right) \dfrac{3+4i}{2-4i}$ to the standard form:
Solution.
Here $\dfrac{1}{1-2i} = \dfrac{1}{1-2i} \times \dfrac{1+2i}{1+2i} = \dfrac{1+2i}{1-4i^2}$
$$
= \dfrac{1+2i}{1+4} = \dfrac{1+2i}{5} = \dfrac{1}{5} + \dfrac{2}{5}i;
$$
$$
\dfrac{1}{1+i} = \dfrac{1}{1+i} \times \dfrac{1-i}{1-i} = \dfrac{1-i}{1-i^2}
$$
$$
= \dfrac{1-i}{2} = \dfrac{1}{2} – \dfrac{1}{2}i \quad \text{and}
$$
$$
\dfrac{3+4i}{2-4i} = \dfrac{3+4i}{2-4i} \times \dfrac{2+4i}{2+4i} = \dfrac{6+12i+8i+16i^2}{4-16i^2}
$$
$$
= \dfrac{-10 + 20i}{20} = \dfrac{-1}{2} + i.
$$
∴ Given expression
$$
= \left[ \left( \dfrac{1}{5} + \dfrac{2}{5}i \right) + \left( \dfrac{1}{2} – \dfrac{1}{2}i \right) \right] \left( \dfrac{-1}{2} + i \right)
$$
$$
= \left( \dfrac{17}{10} – \dfrac{11}{10}i \right) \left( \dfrac{-1 + 2i}{2} \right)
$$
$$
= \left( \dfrac{17-11i}{10} \right) \left( \dfrac{-1+2i}{2} \right)
$$
$$
= \frac{-17 + 34i + 11i – 22i^2}{20}
$$
$$
= \frac{(-17 + 22) + (34i + 11i)}{20}
$$
$$
= \frac{5 + 45i}{20} = \frac{5}{20} + \frac{45}{20}i
$$
$$
= \left( \frac{1}{4} + \frac{9}{4}i \right), \text{ which is of the standard form.}
$$
Example 6. Prove that :
(i) $\left[ \dfrac{(3 + 2i)}{(2 – 5i)} + \dfrac{(3 – 2i)}{(2 + 5i)} \right]$ is rational
(ii) $\left[ \dfrac{(2 – 3i)}{(3 – 4i)} \left( \dfrac{2 + 3i}{3 + 4i} \right) \right]$ is real.
Solution.
(i) $\dfrac{3 + 2i}{2 – 5i} = \dfrac{3 + 2i}{2 – 5i} \times \dfrac{2 + 5i}{2 + 5i}$
$$
= \dfrac{(3 + 2i)(2 + 5i)}{(2 – 5i)(2 + 5i)}
$$
$$
= \dfrac{6 + 15i + 4i + 10i^2}{4 – 25i^2}
$$
$$
= \dfrac{6 + 19i + 10(-1)}{4 – 25(-1)}
$$
$$
= \dfrac{-4 + 19i}{29}
$$
$$
= \dfrac{4}{29} + \dfrac{19}{29}i.
$$
Similarly $\dfrac{3 – 2i}{2 + 5i} = \dfrac{-4}{29} – \dfrac{19}{29}i$.
Adding $\left[ \dfrac{(3 + 2i)}{(2 – 5i)} + \dfrac{(3 – 2i)}{(2 + 5i)} \right] = \dfrac{-8}{29}$, which is rational.
(ii) $\left( \dfrac{2 – 3i}{3 – 4i} \right) \left( \dfrac{2 + 3i}{3 + 4i} \right)$
$$
= \dfrac{4 – 9i^2}{9 – 16i^2} = \dfrac{4 – 9(-1)}{9 – 16(-1)} = \dfrac{4 + 9}{9 + 16} = \dfrac{13}{25},
$$
which is real.
Example 7. If $z = 2 – 3i$, show that $z^2 – 4z + 13 = 0$ and hence find the value of $4z^3 – 3z^2 + 169$.
Solution. We have : $z = 2 – 3i$
$$
\Rightarrow \quad z – 2 = -3i.
$$
$$\text{Squaring}, (z – 2)^2 = 9i^2$$
$$
\Rightarrow \quad (z – 2)^2 = 9(-1)
$$
$$
\Rightarrow \quad z^2 – 4z + 4 = -9
$$
$$
\Rightarrow \quad z^2 – 4z + 13 = 0
$$
$$
\Rightarrow \quad z^2 = 4z – 13 \quad \dots(1)
$$
From (1),
$$
z^3 = 4z^2 – 13z
$$
$$
\Rightarrow \quad z^3 = 4(4z – 13) – 13z \quad \text{[Using (1)]}
$$
$$
\Rightarrow \quad z^3 = 16z – 52 – 13z = 3z – 52 \quad \dots (2)
$$
$$\text{Now}, 4z^3 – 3z^2 + 169$$
$$
= 4(3z – 52) – 3(4z – 13) + 169
$$
$$
= 12z – 208 – 12z + 39 + 169 = 0.
$$
Example 8. Show that a real value of ‘$x$’ will satisfy the equation $\dfrac{1 – ix}{1 + ix} = a – ib$ if $a^2 + b^2 = 1$, where $a, b$ are real.
Solution. We have : $\dfrac{1 – ix}{1 + ix} = a – ib$
By componendo and dividendo, we get :
$$
\dfrac{(1 – ix) + (1 + ix)}{(1 – ix) – (1 + ix)} = \dfrac{(a – ib) + 1}{(a – ib) – 1}
$$
$$
\Rightarrow \quad \dfrac{2}{-2ix} = \dfrac{1 + a – ib}{-1 + a – ib}
$$
$$
\Rightarrow \quad ix = \dfrac{1 – a + ib}{1 + a – ib}
$$
$$
\Rightarrow \quad ix = \dfrac{1 – a + ib}{1 + a – ib} \times \dfrac{1 + a + ib}{1 + a + ib}
$$
$$
\Rightarrow \quad ix = \dfrac{1 – a^2 – b^2 + 2ib}{(1 + a)^2 – i^2 b^2}
$$
$$
\Rightarrow \quad ix = \dfrac{1 – a^2 – b^2 + 2ib}{(1 + a)^2 + b^2}
$$
$$
\Rightarrow \quad ix = \dfrac{2ib}{(1 + a)^2 + b^2}
$$
$$
\Rightarrow \quad x = \dfrac{2b}{(1 + a)^2 + b^2},
$$
which is real.
Example 9. If $y = \sqrt{x^2 + 6x + 8}$, then show that
$(1+iy)^{1/2} + (1-iy)^{1/2} = \sqrt{2(x+4)}$.
Solution. Let $z = (1+iy)^{1/2} + (1-iy)^{1/2}$.
Squaring, $z^2 = (1+iy)^2 + (1-iy)^2$.
$z^2 = 2 + 2[(1+iy)(1-iy)]^{1/2} $
$ z^2 = 2 + 2(1 + y^2)^{1/2} $
$ z^2 = 2 + 2(x^2 + 6x + 8)^{1/2} $
$z^2 = 2 + 2(x^2 + 6x + 9)^{1/2} $
$ z^2 = 2 + 2[(x+3)^2]^{1/2} $
$ z^2 = 2 + 2(x+3) $
$z^2 = 2x + 8 $
$ z^2 = 2(x+4) $
$z = \sqrt{2(x+4)}. $
Hence, $(1+iy)^{1/2} + (1-iy)^{1/2} = \sqrt{2(x+4)}$.
Example 10. If $a + ib = \dfrac{c+i}{c-i}$, where ‘c’ is real, prove that $a^2 + b^2 = 1$ and $\dfrac{b}{a} = \dfrac{2c}{c^2 – 1}$.
Solution. We have : $a + ib = \dfrac{c+i}{c-i}$
$ a + ib = \dfrac{c+i}{c-i} \times \dfrac{c+i}{c+i} $
$ a + ib = \dfrac{c^2 + 2ic + i^2}{c^2 – i^2} $
$ a + ib = \dfrac{c^2 – 1}{c^2 + 1} + \dfrac{2c}{c^2 + 1}i. $
Comparing, $a = \dfrac{c^2 – 1}{c^2 + 1}$ and $b = \dfrac{2c}{c^2 + 1}$.
(i) Squaring and adding,
$ a^2 + b^2 = \left(\dfrac{c^2 – 1}{c^2 + 1}\right)^2 + \dfrac{4c^2}{(c^2 + 1)^2} = \dfrac{(c^2 + 1)^2}{(c^2 + 1)^2} = 1. $
(ii) Dividing, $\dfrac{b}{a} = \dfrac{2c}{c^2 – 1}$.
Example 11. If $\dfrac{a+ib}{c+id} = x + iy$, show that :
$$
x^2 + y^2 = \dfrac{a^2 + b^2}{c^2 + d^2}.
$$
Solution. We have :
$x + iy = \dfrac{a+ib}{c+id} $
$x + iy = \dfrac{a+ib}{c+id} \times \dfrac{c-id}{c-id} $
$x + iy = \dfrac{ac – iad + ibc – i^2bd}{c^2 – (id)^2} $
$x + iy = \dfrac{(ac + bd) + i(bc – ad)}{c^2 + d^2} $
$ x = \dfrac{ac + bd}{c^2 + d^2} \quad \text{and} \quad y = \dfrac{bc – ad}{c^2 + d^2}. $
$ \therefore \quad x^2 + y^2 = \dfrac{(ac + bd)^2 + (bc – ad)^2}{(c^2 + d^2)^2}$
$x^2 + y^2= \dfrac{a^2c^2 + b^2d^2 + 2abcd + b^2c^2 + a^2d^2 – 2abcd}{(c^2 + d^2)^2} $
$x^2 + y^2= \dfrac{(a^2c^2 + b^2c^2) + (b^2d^2 + a^2d^2)}{(c^2 + d^2)^2} $
$x^2 + y^2= \dfrac{c^2(a^2 + b^2) + d^2(a^2 + b^2)}{(c^2 + d^2)^2} = \dfrac{a^2 + b^2}{c^2 + d^2}, $
which is true.
Example 12. Find the conjugate and modulus of complex number $7 – 24i$.
Solution. Let $z = 7 – 24i$.
∴ Its conjugate, $\overline{z} = 7 + 24i$
and its modulus $= |z| = \sqrt{(7)^2 + (-24)^2}$
$$
|z| = \sqrt{49 + 576} = \sqrt{625} = 25.
$$
Example 13. Find the conjugate of $\dfrac{(3-2i)(2+3i)}{(1+2i)(2-i)}$.
[N.C.E.R.T.]
Solution. $\dfrac{(3-2i)(2+3i)}{(1+2i)(2-i)} = \dfrac{6+5i-6i^2}{2+3i-2i^2}$
$ \dfrac{6+5i+6}{2+3i+2} = \dfrac{12+5i}{4+3i} $
$= \dfrac{12+5i}{4+3i} \times \dfrac{4-3i}{4-3i} $
$= \dfrac{48-36i+20i-15i^2}{16-9i^2} $
$= \dfrac{48-16i+15}{16+9} = \dfrac{63-16i}{25}.$
Thus $z = \dfrac{63}{25} – \dfrac{16}{25}i$.
Hence, $\overline{z} = \dfrac{63}{25} + \dfrac{16}{25}i$.
Example 14. If $z_1 = 2 – i$ and $z_2 = -2 + i$, find $\text{Re}\left(\dfrac{1}{z_1 \overline{z_2}}\right)$.
Solution. We have: $z_1 = 2 – i$ and $z_2 = -2 + i$.
$\therefore \;\overline{z_2} = -2 – i$.
$\therefore \quad \dfrac{1}{z_1 \overline{z_2}} = \dfrac{1}{(2-i)(-2-i)} $
$= \dfrac{1}{(-4 -2i +2i +i^2)} = \dfrac{1}{-4 -1} = -\dfrac{1}{5} $
Hence, $\text{Re}\left(\dfrac{1}{z_1 \overline{z_2}}\right) = -\dfrac{1}{5}$,
$\text{Im}\left(\dfrac{1}{z_1 \overline{z_2}}\right) = 0$.
Example 15. Prove that for any complex number $z$, the product $z\bar{z}$ is always a non-negative real number.
Solution. Let $z = x + iy$. Then $\bar{z} = x – iy$.
$$\therefore \quad z\bar{z} = (x+iy)(x-iy) = x^2 – i^2 y^2 = x^2 + y^2.$$
Hence, the product $z\bar{z}$ is always a non-negative real number.
Example 16. If $z = \left(\dfrac{\sqrt{5}}{2} + \dfrac{i}{2}\right)^{105} + \left(\dfrac{\sqrt{5}}{2} – \dfrac{i}{2}\right)^{105}$, then prove that $\text{Im}(z) = 0$.
Solution. We have $z = \left(\dfrac{\sqrt{5}}{2} + \dfrac{i}{2}\right)^{105} + \left(\dfrac{\sqrt{5}}{2} – \dfrac{i}{2}\right)^{105}$.
$\overline{z} = \overline{\left(\dfrac{\sqrt{5}}{2} + \dfrac{i}{2}\right)^{105} + \left(\dfrac{\sqrt{5}}{2} – \dfrac{i}{2}\right)^{105}}$.
$\overline{z} = \overline{\left(\dfrac{\sqrt{5}}{2} + \dfrac{i}{2}\right)^{105}} + \overline{\left(\dfrac{\sqrt{5}}{2} – \dfrac{i}{2}\right)^{105}}$.
$\overline{z} =\left(\dfrac{\sqrt{5}}{2} – \dfrac{i}{2}\right)^{105} + \left(\dfrac{\sqrt{5}}{2} + \dfrac{i}{2}\right)^{105}$.
Then $\bar{z} = z$, so $z$ is real. Hence $\text{Im}(z) = 0$.
[Let $z = x + iy$ and $\bar{z} = x – i y $, if $\bar{z} = z$, that is $x + iy$ = $ x – i y $ then 2$y = 0$, $y=0$. Hence $\text{Im}(z) = 0$ and $z$ is real]
ADDITIVE INVERSE
Let $a + ib$ be a complex number.
The additive identity of the complex numbers is $0 + 0i$.
If we find $x + iy$ such that
$(a + ib) + (x + iy) = 0 + 0i $
$ (a + x) + (b + y)i = 0 + 0i \quad \dots (1)$
We call the complex number $x + iy$ so as to satisfy (1) as the additive inverse of $a + ib$ and is denoted by $-(a + ib)$.
If (1) is true, then $a + x = 0$ and $b + y = 0$.
Solving, $x = -a$ and $y = -b$.
Hence, $-a + (-b)i$ i.e. $-(a + ib)$ is the additive inverse of $a + ib$.
MULTIPLICATIVE INVERSE/RECIPROCAL
Let $a + ib$ be a non-zero complex number.
The multiplicative identity of the complex numbers is $1 + 0i$.
If we find $x + iy$ such that :
$$
(a + ib)(x + iy) = 1 + 0i
$$
$$
\Rightarrow \quad (ax – by) + (bx + ay)i = 1 + 0i \quad \dots (1)
$$
We call the complex number $x + iy$ so as to satisfy (1) as the multiplicative inverse (or reciprocal) of $a + ib$ and is denoted by $\dfrac{1}{a + ib}$.
If (1) is true, then $ax – by = 1$ and $bx + ay = 0 \quad \dots (2)$.
Since $a + ib \neq 0$, $\therefore a^2 + b^2 \neq 0$.
Solving (2) simultaneously, we have :
$$
x = \dfrac{a}{a^2 + b^2} \quad \text{and} \quad y = \dfrac{-b}{a^2 + b^2}
$$
Hence, $\dfrac{a}{a^2 + b^2} + \dfrac{(-b)i}{a^2 + b^2}$ is the multiplicative inverse (or reciprocal) of $a + ib$.
In Symbols, if $z = a + ib$ be a non-zero complex number, then its multiplicative inverse or its reciprocal is given by :
$$
\dfrac{\text{Re}(z)}{|z|^2} – \dfrac{\text{Im}(z)}{|z|^2} i = \dfrac{\bar{z}}{|z|^2}
$$
Frequently Asked Questions
Example 1. Write the additive inverse of the complex number $-2 + 3i$.
Solution. Let $(a + ib)$ be the additive inverse of $(-2 + 3i)$.
Then $(a + ib) + (-2 + 3i) = 0$ [Def.]
$$
\Rightarrow \quad a + ib = 2 – 3i.
$$
Hence, the reqd. additive inverse is $2 – 3i$.
Example 2. Find the additive inverse and reciprocal of complex number $3 – 4i$.
Solution.
(i) Let $(a + ib)$ be the additive inverse of $(3 – 4i)$.
Then $(a + ib) + (3 – 4i) = 0 + i0$ [Def.]
$$
\Rightarrow \quad a + ib = -3 + 4i.
$$
Hence, the additive inverse is $-3 + 4i$.
(ii) Reciprocal of $3 – 4i = \dfrac{1}{3 – 4i}$
$$
= \dfrac{3 + 4i}{(3 – 4i)(3 + 4i)} = \dfrac{3 + 4i}{9 – 16i^2}
$$
$$
= \dfrac{3 + 4i}{9 + 16} = \dfrac{3}{25} + \dfrac{4}{25}i.
$$
Example 3. Find the multiplicative inverse of the following :
(i) $3 + 4i$
(ii) $(5 – 7i)^2$.
Solution.
(i) Let $(a + ib)$ be the multiplicative inverse of $(3 + 4i)$.
Then $(a + ib)(3 + 4i) = 1$ [Def.]
$$
\Rightarrow \quad a + ib = \dfrac{1}{3 + 4i} = \dfrac{1}{3 + 4i} \times \dfrac{3 – 4i}{3 – 4i}
$$
$$
a + ib= \dfrac{3 – 4i}{9 – 16i^2} = \dfrac{3 – 4i}{9 + 16} = \dfrac{3 – 4i}{25} = \dfrac{3}{25} – \frac{4}{25}i.
$$
Hence, the reqd. multiplicative inverse is $\dfrac{3}{25} – \dfrac{4}{25}i$.
(ii) $(5 – 7i)^2 = 25 + 49i^2 – 70i = 25 + 49(-1) – 70i = -24 – 70i$.
Let $(a + ib)$ be the multiplicative inverse.
Then by def., $(a + ib)(-24 – 70i) = 1$
$a + ib = \dfrac{1}{-24 – 70i} $
$a + ib= \dfrac{1}{-24 – 70i} \times \dfrac{-24 + 70i}{-24 + 70i} $
$a + ib= \dfrac{-24 + 70i}{(-24)^2 – (70i)^2} $
$a + ib= \dfrac{-24 + 70i}{576 – 4900i^2} $
$a + ib= \dfrac{-24 + 70i}{576 + 4900} $
$a + ib= \dfrac{-24 + 70i}{5476} $
$a + ib= \dfrac{-24}{5476} + \dfrac{70}{5476}i $
$a + ib= \dfrac{-6}{1369} + \dfrac{35}{2738}i. $
Hence, the reqd. multiplicative inverse is :
$$
\dfrac{-6}{1369} + \dfrac{35}{2738}i.
$$
Example 4. Find the multiplicative inverse of $\dfrac{3+4i}{4-5i}$ and write it in the form $a + ib$.
Solution. $\dfrac{3+4i}{4-5i} = \dfrac{3+4i}{4-5i} \times \dfrac{4+5i}{4+5i}$
$= \dfrac{12 + 15i + 16i + 20i^2}{(4)^2 – (5i)^2} $
$= \dfrac{12 + 31i – 20}{16 – 25i^2} $
$= \dfrac{-8 + 31i}{16 + 25} $
$= \dfrac{-8 + 31i}{41}. $
Let $(a + ib)$ be its multiplicative inverse.
Then, by def., $(a + ib)\left(\dfrac{-8 + 31i}{41}\right) = 1$
$a + ib = \dfrac{41}{-8 + 31i} $
$a + ib= \dfrac{41}{-8 + 31i} \times \dfrac{-8 – 31i}{-8 – 31i} $
$a + ib= \dfrac{-328 – 1271i}{(-8)^2 – (31i)^2} $
$a + ib= \dfrac{-328 – 1271i}{64 + 961} $
$a + ib= \dfrac{-328}{1025} – \dfrac{1271}{1025}i. $
Hence, the reqd. multiplicative inverse is :
$$
\dfrac{-328}{1025} – \dfrac{1271}{1025}i.
$$
Example 5. If $z = (1 – i)^6 + (1 – i)$, then find the modulus of $z$ (i.e. $|z|$) and multiplicative inverse of $z$ (i.e. $z^{-1}$).
Solution : We have :
$$
z = (1 – i)^6 + (1 – i) \quad \dots(1)
$$
Now
$$(1 – i)^6 = ((1 – i)^2)^3$$
$$
(1 – i)^6= (1 + i^2 – 2i)^3 = (1 – 1 – 2i)^3
$$
$$
(1 – i)^6= (-2i)^3 = -8i^3
$$
$$
(1 – i)^6= -8(-i) = 8i.
$$
$\therefore$ From (1), $z = 8i + 1 – i = 1 + 7i$.
(i) $|z| = \sqrt{(1)^2 + (7)^2} = \sqrt{1+49} = \sqrt{50} = 5\sqrt{2}$.
(ii) Multiplicative inverse of $z = \dfrac{1}{1+7i}$
$$= \dfrac{1-7i}{(1+7i)(1-7i)} $$
$$= \dfrac{1-7i}{1-49i^2} $$
$$= \dfrac{1-7i}{1+49} $$
$$= \dfrac{1}{50} (1 – 7i). $$