Conjugate Properties of Complex Number, Multiplicative or Reciprocal, Additive Inverse of Complex Numbers

Conjugate, multiplicative inverse, reciprocal, and additive inverse are important concepts in complex numbers that help simplify algebraic operations and solve mathematical problems efficiently. The conjugate of a complex number changes the sign of its imaginary part, while the additive inverse gives a number whose sum with the original complex number is zero. Similarly, the multiplicative inverse or reciprocal of a complex number produces a value whose product with the original number equals one. These properties are widely used in simplifying expressions, rationalizing denominators, solving equations, and advanced applications of complex numbers in mathematics and engineering.

What is Conjugate of Complex Numbers ?

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 $\overline{z}$.

Similar topics for practice include Complex Numbers: Addition, Subtraction, Multiplication of Two Complex Numbers With Solved Examples

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What are the Properties of Complex Number Conjugate ?

(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 $x,y\in ℝ$.

$\therefore \overline{z}=x-iy.$

$\therefore \overline{\overline{z}}=x-iy=x+iy=z.$

(II) The sum and product of two conjugate complex numbers are purely real.

Let $z=x+iy$. Then $\overline{z}=x-iy$, where $x,y\in ℝ$.

(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
$z_1=x_1+i{y}_1$ and $z_2=x_2+i{y}_2$, $(x_1,x_2;y_1,y_2\in ℝ)$.

$\therefore \overline{z_1}=x_1-i{y}_1$ and $\overline{z_2}=x_2-i{y}_2.$

(i) $z_1+z_2=(x_1+i{y}_1)+(x_2+i{y}_2)=(x_1+x_2)+i(y_1+y_2).$

Hence, $\overline{z_1+z_2}=(x_1+x_2)-i(y_1+y_2)=(x_1-i{y}_1)+(x_2-i{y}_2)=\overline{z_1}+\overline{z_2}.$

(ii) $z_1{z}_2=(x_1+i{y}_1)(x_2+i{y}_2)=(x_1{x}_2-y_1{y}_2)+i(x_1{y}_2+y_1{x}_2).$

$\therefore \overline{z_1{z}_2}=(x_1{x}_2-y_1{y}_2)-i(x_1{y}_2+y_1{x}_2).$

Also $\overline{z_1}\cdot \overline{z_2}=(x_1-i{y}_1)(x_2-i{y}_2)=(x_1{x}_2-y_1{y}_2)-i(x_1{y}_2+y_1{x}_2).$

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) $\overline{\left({\displaystyle \frac{z_1}{z_2}}\right)}={\displaystyle \frac{\overline{z_1}}{\overline{z_2}}}$, $z_2\ne 0.$

Proof. Let
$z_1=x_1+i{y}_1$ and $z_2=x_2+i{y}_2$, $(x_1,x_2;y_1,y_2\in ℝ)$.

$\therefore \overline{z_1}=x_1-i{y}_1$ and $\overline{z_2}=x_2-i{y}_2.$

(i) $z_1-z_2=(x_1+i{y}_1)-(x_2+i{y}_2)=(x_1-x_2)+i(y_1-y_2).$

$\therefore \overline{z_1-z_2}=(x_1-x_2)-i(y_1-y_2)=(x_1-i{y}_1)-(x_2-i{y}_2)=\overline{z_1}-\overline{z_2}.$

(ii) ${\displaystyle \frac{z_1}{z_2}}={\displaystyle \frac{x_1+i{y}_1}{x_2+i{y}_2}}={\displaystyle \frac{x_1+i{y}_1}{x_2+i{y}_2}}\times {\displaystyle \frac{x_2-i{y}_2}{x_2-i{y}_2}}$

$={\displaystyle \frac{(x_1{x}_2+y_1{y}_2)+i(x_2{y}_1-x_1{y}_2)}{x_2^2+y_2^2}}={\displaystyle \frac{x_1{x}_2+y_1{y}_2}{x_2^2+y_2^2}}+i{\displaystyle \frac{x_2{y}_1-x_1{y}_2}{x_2^2+y_2^2}}.$

$\therefore \overline{\left({\displaystyle \frac{z_1}{z_2}}\right)}={\displaystyle \frac{x_1{x}_2+y_1{y}_2}{x_2^2+y_2^2}}-i{\displaystyle \frac{x_2{y}_1-x_1{y}_2}{x_2^2+y_2^2}}.$

Also

${\displaystyle \frac{\overline{z_1}}{\overline{z_2}}}={\displaystyle \frac{x_1-i{y}_1}{x_2-i{y}_2}}={\displaystyle \frac{x_1-i{y}_1}{x_2-i{y}_2}}\times {\displaystyle \frac{x_2+i{y}_2}{x_2+i{y}_2}}$

$={\displaystyle \frac{(x_1{x}_2+y_1{y}_2)+i(x_2{y}_2-x_1{y}_1)}{x_2^2+y_2^2}}={\displaystyle \frac{x_1{x}_2+y_1{y}_2}{x_2^2+y_2^2}}-i{\displaystyle \frac{x_2{y}_2-x_1{y}_2}{x_2^2+y_2^2}}.$

Hence,

$\overline{\left({\displaystyle \frac{z_1}{z_2}}\right)}={\displaystyle \frac{\overline{z_1}}{\overline{z_2}}},z_2\ne 0.$

Important concepts connected to this topic are What are Imaginary Numbers and iota (i), Powers of iota, Solved Examples

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Find the conjugate and modulus of complex number 7 – 24i.

Solution. Let z = 7 – 24i.

∴ Its conjugate, $\overline{z}=7+24i$

and its modulus $=|z|=\sqrt{(7{)}^2+(-24{)}^2}$

$|z|=\sqrt{49+576}=\sqrt{625}=25.$

Find the conjugate of ${\displaystyle \frac{(3-2i)(2+3i)}{(1+2i)(2-i)}}$. [N.C.E.R.T.]

Solution. ${\displaystyle \frac{(3-2i)(2+3i)}{(1+2i)(2-i)}}={\displaystyle \frac{6+5i-6i^2}{2+3i-2i^2}}$

${\displaystyle \frac{6+5i+6}{2+3i+2}}={\displaystyle \frac{12+5i}{4+3i}}$

$={\displaystyle \frac{12+5i}{4+3i}}\times {\displaystyle \frac{4-3i}{4-3i}}$

$={\displaystyle \frac{48-36i+20i-15i^2}{16-9i^2}}$

$={\displaystyle \frac{48-16i+15}{16+9}}={\displaystyle \frac{63-16i}{25}}.$

Thus $z={\displaystyle \frac{63}{25}}-{\displaystyle \frac{16}{25}}i$.

Hence, $\overline{z}={\displaystyle \frac{63}{25}}+{\displaystyle \frac{16}{25}}i$.

If $z_1=2-i$ and $z_2=-2+i$, find $\text{Re}\left({\displaystyle \frac{1}{z_1\overline{z_2}}}\right)$.

Solution. We have: $z_1=2-i$ and $z_2=-2+i$.

$\therefore \overline{z_2}=-2-i$.

$\therefore {\displaystyle \frac{1}{z_1\overline{z_2}}}={\displaystyle \frac{1}{(2-i)(-2-i)}}$

$={\displaystyle \frac{1}{(-4-2i+2i+i^2)}}={\displaystyle \frac{1}{-4-1}}=-{\displaystyle \frac{1}{5}}$

Hence, $\text{Re}\left({\displaystyle \frac{1}{z_1\overline{z_2}}}\right)=-{\displaystyle \frac{1}{5}}$,

$\text{Im}\left({\displaystyle \frac{1}{z_1\overline{z_2}}}\right)=0$.

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 $\bar{z}=x-iy$.

$$\therefore z\bar{z}=(x+iy)(x-iy)=x^2-i^2{y}^2=x^2+y^2.$$

Hence, the product $z\bar{z}$ is always a non-negative real number.

If $z={({\displaystyle \frac{\sqrt{5}}{2}}+{\displaystyle \frac{i}{2}})}^{105}+{({\displaystyle \frac{\sqrt{5}}{2}}-{\displaystyle \frac{i}{2}})}^{105}$, then prove that $\text{Im}(z)=0$.

Solution. We have $z={({\displaystyle \frac{\sqrt{5}}{2}}+{\displaystyle \frac{i}{2}})}^{105}+{({\displaystyle \frac{\sqrt{5}}{2}}-{\displaystyle \frac{i}{2}})}^{105}$.

$\overline{z}=\overline{{({\displaystyle \frac{\sqrt{5}}{2}}+{\displaystyle \frac{i}{2}})}^{105}+{({\displaystyle \frac{\sqrt{5}}{2}}-{\displaystyle \frac{i}{2}})}^{105}}$.

$\overline{z}=\overline{{({\displaystyle \frac{\sqrt{5}}{2}}+{\displaystyle \frac{i}{2}})}^{105}}+\overline{{({\displaystyle \frac{\sqrt{5}}{2}}-{\displaystyle \frac{i}{2}})}^{105}}$.

$\overline{z}={({\displaystyle \frac{\sqrt{5}}{2}}-{\displaystyle \frac{i}{2}})}^{105}+{({\displaystyle \frac{\sqrt{5}}{2}}+{\displaystyle \frac{i}{2}})}^{105}$.

Then $\bar{z}=z$, so $z$ is real. Hence $\text{Im}(z)=0$.

[Let $z=x+iy$ and $\bar{z}=x-iy$, if $\bar{z}=z$, that is $x+iy$ = $x-iy$ then 2$y=0$, $y=0$. Hence $\text{Im}(z)=0$ and $z$ is real]

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What is the Additive Inverse of Complex Numbers ?

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 ….. (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.

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What is the Multiplicative Inverse or Reciprocal of a Complex Numbers ?

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

(ax – by) + (bx + ay)i = 1 + 0i … (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 ${\displaystyle \frac{1}{a+ib}}$.

If (1) is true, then ax – by = 1 and bx + ay = 0 … (2).

Since $a+ib\ne 0$, $\therefore a^2+b^2\ne 0$.

Solving (2) simultaneously, we have :

$x={\displaystyle \frac{a}{a^2+b^2}}\text{and}y={\displaystyle \frac{-b}{a^2+b^2}}$

Hence, ${\displaystyle \frac{a}{a^2+b^2}}+{\displaystyle \frac{(-b)i}{a^2+b^2}}$ 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 :

${\displaystyle \frac{\text{Re}(z)}{|z{|}^2}}-{\displaystyle \frac{\text{Im}(z)}{|z{|}^2}}i={\displaystyle \frac{\bar{z}}{|z{|}^2}}$

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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.]

a + ib = 2 – 3i

Hence, the required additive inverse is $2-3i$.

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.]

$\Rightarrow a+ib=-3+4i.$

Hence, the additive inverse is $-3+4i$.

(ii) Reciprocal of $3-4i={\displaystyle \frac{1}{3-4i}}$

$={\displaystyle \frac{3+4i}{(3-4i)(3+4i)}}={\displaystyle \frac{3+4i}{9-16i^2}}$

$={\displaystyle \frac{3+4i}{9+16}}={\displaystyle \frac{3}{25}}+{\displaystyle \frac{4}{25}}i.$

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.]

$\Rightarrow a+ib={\displaystyle \frac{1}{3+4i}}={\displaystyle \frac{1}{3+4i}}\times {\displaystyle \frac{3-4i}{3-4i}}$

$a+ib={\displaystyle \frac{3-4i}{9-16i^2}}={\displaystyle \frac{3-4i}{9+16}}={\displaystyle \frac{3-4i}{25}}={\displaystyle \frac{3}{25}}-\frac{4}{25}i.$

Hence, the reqd. multiplicative inverse is ${\displaystyle \frac{3}{25}}-{\displaystyle \frac{4}{25}}i$.

(ii) $(5-7i{)}^2=25+49i^2-70i=25+49(-1)-70i=-24-70i$.

Let $(a+ib)$ be the multiplicative inverse.

Then by def., $(a+ib)(-24-70i)=1$

$a+ib={\displaystyle \frac{1}{-24-70i}}$

$a+ib={\displaystyle \frac{1}{-24-70i}}\times {\displaystyle \frac{-24+70i}{-24+70i}}$

$a+ib={\displaystyle \frac{-24+70i}{(-24{)}^2-(70i{)}^2}}$

$a+ib={\displaystyle \frac{-24+70i}{576-4900i^2}}$

$a+ib={\displaystyle \frac{-24+70i}{576+4900}}$

$a+ib={\displaystyle \frac{-24+70i}{5476}}$

$a+ib={\displaystyle \frac{-24}{5476}}+{\displaystyle \frac{70}{5476}}i$

$a+ib={\displaystyle \frac{-6}{1369}}+{\displaystyle \frac{35}{2738}}i.$

Hence, the required multiplicative inverse is :

${\displaystyle \frac{-6}{1369}}+{\displaystyle \frac{35}{2738}}i.$

Find the multiplicative inverse of ${\displaystyle \frac{3+4i}{4-5i}}$ and write it in the form $a+ib$.

Solution. ${\displaystyle \frac{3+4i}{4-5i}}={\displaystyle \frac{3+4i}{4-5i}}\times {\displaystyle \frac{4+5i}{4+5i}}$

$={\displaystyle \frac{12+15i+16i+20i^2}{(4{)}^2-(5i{)}^2}}$

$={\displaystyle \frac{12+31i-20}{16-25i^2}}$

$={\displaystyle \frac{-8+31i}{16+25}}$

$={\displaystyle \frac{-8+31i}{41}}.$

Let $(a+ib)$ be its multiplicative inverse.

Then, by def., $(a+ib)\left({\displaystyle \frac{-8+31i}{41}}\right)=1$

$a+ib={\displaystyle \frac{41}{-8+31i}}$

$a+ib={\displaystyle \frac{41}{-8+31i}}\times {\displaystyle \frac{-8-31i}{-8-31i}}$

$a+ib={\displaystyle \frac{-328-1271i}{(-8{)}^2-(31i{)}^2}}$

$a+ib={\displaystyle \frac{-328-1271i}{64+961}}$

$a+ib={\displaystyle \frac{-328}{1025}}-{\displaystyle \frac{1271}{1025}}i.$

Hence, the required multiplicative inverse is :

${\displaystyle \frac{-328}{1025}}-{\displaystyle \frac{1271}{1025}}i.$

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 :

$z=(1-i{)}^6+(1-i)\dots (1)$

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 $z={\displaystyle \frac{1}{1+7i}}$

$$={\displaystyle \frac{1-7i}{(1+7i)(1-7i)}}$$

$$={\displaystyle \frac{1-7i}{1-49i^2}}$$

$$={\displaystyle \frac{1-7i}{1+49}}$$

$$={\displaystyle \frac{1}{50}}(1-7i).$$

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