Question

Determine the conjugate and reciprocal of each complex number given below:
(i) $i$
(ii) ${{i}^{3}}$
(iii) $3-i$
(iv) $\sqrt{-1}-3$
(v) $\sqrt{-9}-1$

Answer 1

Hint: Here, to find conjugate of given complex number in the given form $a+ib$ , we have to just change sign of ib to get conjugate i.e. $a-ib$ . For reciprocal we have to multiply it with the conjugate form given as $\dfrac{1}{a+ib}\times \dfrac{a-ib}{a-ib}$ . On solving this we will get our answer. We will be using the formula ${{i}^{2}}=-1$ , $i=\sqrt{-1}$ .

Complete step-by-step answer:
Here, conjugate means we have to change the sign of the complex number given to us. While in reciprocal we write inverse number i.e. $\dfrac{1}{complex\text{ }number}$ .
Taking case (i): $i$ . Conjugate of this complex number $i$ is $-i$ . Reciprocal will be given as $\dfrac{1}{i}$ . To solve this, we will multiply this with i in numerator and denominator. So, we will get as
$\dfrac{1}{i}\times \dfrac{-i}{-i}=\dfrac{-i}{-{{i}^{2}}}$
We should know that ${{i}^{2}}=-1$ so, on putting this value, we will get
Reciprocal $=-i$
Taking case (ii): ${{i}^{3}}$ . We can write ${{i}^{3}}={{i}^{2}}\cdot i$ . Also, we know that ${{i}^{2}}=-1$ . So, we can write it as ${{i}^{3}}=-i$ . Conjugate of this complex number $-\left( -i \right)=i$ . Reciprocal will be given as $\dfrac{1}{-i}$ . To solve this, we will multiply this with i in numerator and denominator. So, we will get as
$\dfrac{1}{-i}\times \dfrac{i}{i}=\dfrac{i}{-{{i}^{2}}}$
On solving, we will get
Reciprocal $=\dfrac{i}{-\left( -1 \right)}=i$ .
Taking case (iii): $3-i$ . Conjugate of this complex number is just by changing the sign of the complex number only. So, we get $3+i$ . Reciprocal will be given as $\dfrac{1}{3-i}$ . So, on further solving and multiplying it with $3+i$ in numerator and denominator we will get as
$\dfrac{1}{3-i}\times \dfrac{3+i}{3+i}=\dfrac{3+i}{9-{{i}^{2}}}$
Using the formula $\left( a-b \right)\left( a+b \right)={{a}^{2}}-{{b}^{2}}$ and ${{i}^{2}}=-1$ .
So, on further solving we will get
$=\dfrac{3+i}{9-\left( -1 \right)}=\dfrac{3+i}{10}$
Reciprocal $=0.3+0.1i$
Taking case (iv): $\sqrt{-1}-3$ . We can write this as $i-3$ . As we know that ${{i}^{2}}=-1$ so, taking the square root on both the sides we get $i=\sqrt{-1}$ . So, conjugate of $i-3$ is just changing sign of complex number i so, we get $-i-3$ .
Reciprocal can be written as $\dfrac{1}{i-3}$ . So, we will multiply it with $-i-3$ in numerator and denominator. So, we will get as
$\dfrac{1}{i-3}\times \dfrac{-i-3}{-i-3}=\dfrac{-i-3}{9-{{i}^{2}}}$
Using the formula $\left( a-b \right)\left( a+b \right)={{a}^{2}}-{{b}^{2}}$ and ${{i}^{2}}=-1$ .
So, on further solving we will get
$=\dfrac{-3-i}{9-\left( -1 \right)}=\dfrac{-3-i}{10}$
Reciprocal $=-0.3-0.1i$
Taking case (v): $\sqrt{-9}-1$ . This can be simplified as $\sqrt{9\times \left( -1 \right)}$ i.e. $\sqrt{9}\cdot \sqrt{-1}$ . We know that $\sqrt{9}=3,\sqrt{-1}=i$ , so we can write it as $3i-1$ i.e. $-1+3i$ . Thus, conjugate can be written by changing the sign of complex numbers, so we get $-1-3i$ .
Reciprocal can be written as $\dfrac{1}{-1+3i}$ . So, we will multiply it with conjugate in numerator and denominator. So, we will get as
$\dfrac{1}{-1+3i}\times \dfrac{-1-3i}{-1-3i}=\dfrac{-1-3i}{1-9{{i}^{2}}}$
Using the formula $\left( a-b \right)\left( a+b \right)={{a}^{2}}-{{b}^{2}}$ and ${{i}^{2}}=-1$ .
So, on further solving we will get
$=\dfrac{-1-3i}{1-9\left( -1 \right)}=\dfrac{-1-3i}{10}$
Reciprocal $=-0.1-0.3i$

Note: There is another method of finding reciprocal of complex number given as $\dfrac{1}{z}=\dfrac{\overline{z}}{{{\left| z \right|}^{2}}}$ . let say we have $z=3-i$ . So, $\overline{z}=3+i$ which is a conjugate of complex numbers. Now $\left| z \right|$ can be find out using the formula $\left| z \right|=\sqrt{{{x}^{2}}+{{y}^{2}}}$ where x and y are coefficient of complex numbers. Here we have x as 3 and y as $-1$ . So, on putting values, we get $\left| z \right|=\sqrt{9+1}=\sqrt{10}$ So, reciprocal will be $\dfrac{1}{z}=\dfrac{3+i}{10}=0.3+0.1i$ .
Thus, we get the same answer by this approach also.