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 ${\displaystyle \frac{1}{a+ib}}\times {\displaystyle \frac{a-ib}{a-ib}}$ . 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. ${\displaystyle \frac{1}{complexnumber}}$ .
Taking case (i): $i$ . Conjugate of this complex number $i$ is $-i$ . Reciprocal will be given as ${\displaystyle \frac{1}{i}}$ . To solve this, we will multiply this with i in numerator and denominator. So, we will get as
${\displaystyle \frac{1}{i}}\times {\displaystyle \frac{-i}{-i}}={\displaystyle \frac{-i}{-i^2}}$
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 $-(-i)=i$ . Reciprocal will be given as ${\displaystyle \frac{1}{-i}}$ . To solve this, we will multiply this with i in numerator and denominator. So, we will get as
${\displaystyle \frac{1}{-i}}\times {\displaystyle \frac{i}{i}}={\displaystyle \frac{i}{-i^2}}$
On solving, we will get
Reciprocal $={\displaystyle \frac{i}{-(-1)}}=i$ .
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 ${\displaystyle \frac{1}{3-i}}$ . So, on further solving and multiplying it with $3+i$ in numerator and denominator we will get as
${\displaystyle \frac{1}{3-i}}\times {\displaystyle \frac{3+i}{3+i}}={\displaystyle \frac{3+i}{9-i^2}}$
Using the formula $(a-b)(a+b)=a^2-b^2$ and $i^2=-1$ .
So, on further solving we will get
$={\displaystyle \frac{3+i}{9-(-1)}}={\displaystyle \frac{3+i}{10}}$
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 ${\displaystyle \frac{1}{i-3}}$ . So, we will multiply it with $-i-3$ in numerator and denominator. So, we will get as
${\displaystyle \frac{1}{i-3}}\times {\displaystyle \frac{-i-3}{-i-3}}={\displaystyle \frac{-i-3}{9-i^2}}$
Using the formula $(a-b)(a+b)=a^2-b^2$ and $i^2=-1$ .
So, on further solving we will get
$={\displaystyle \frac{-3-i}{9-(-1)}}={\displaystyle \frac{-3-i}{10}}$
Reciprocal $=-0.3-0.1i$
Taking case (v): $\sqrt{-9}-1$ . This can be simplified as $\sqrt{9\times (-1)}$ 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 ${\displaystyle \frac{1}{-1+3i}}$ . So, we will multiply it with conjugate in numerator and denominator. So, we will get as
${\displaystyle \frac{1}{-1+3i}}\times {\displaystyle \frac{-1-3i}{-1-3i}}={\displaystyle \frac{-1-3i}{1-9i^2}}$
Using the formula $(a-b)(a+b)=a^2-b^2$ and $i^2=-1$ .
So, on further solving we will get
$={\displaystyle \frac{-1-3i}{1-9(-1)}}={\displaystyle \frac{-1-3i}{10}}$
Reciprocal $=-0.1-0.3i$

Note: There is another method of finding reciprocal of complex number given as ${\displaystyle \frac{1}{z}}={\displaystyle \frac{\bar{z}}{{\left|z\right|}^2}}$ . let say we have $z=3-i$ . So, $\bar{z}=3+i$ 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 ${\displaystyle \frac{1}{z}}={\displaystyle \frac{3+i}{10}}=0.3+0.1i$ .
Thus, we get the same answer by this approach also.