Question

Find the Additive inverse of $1-i$ is:
a). $0+0i$
b). $-1-i$
c). $-1+i$
d). None of these

Answer 1

Hint: Additive inverse is the number real or complex, that when added to the original number gives additive identity as the answer. Additive identity is zero. So, we will assign the additive inverse value to a general complex number $a+bi$ and we will find ‘a’ and ‘b’.

Complete step-by-step solution -
Let the additive inverse of the number $\left( 1-i \right)$ be a general complex number $a+bi$ .
Now, we know that when a number and its additive inverse are added, the result is the additive identity. The additive identity as we know is 0. Thus, adding the two and equation the sum to zero we get,
$1-i+a+bi=0$ .
Taking out I as common from the second term, we get $\left( 1+a \right)+\left( -i+bi \right)=0$ .
Rearranging the terms,
$\left( 1+a \right)+\left( b-1 \right)i=0$ .
Now since the RHS of the equation is zero, the LHS of the equation also must be 0.
Thus, the LHS of the equation is zero.
Now, if the LHS of the equation is zero, the real coefficients and the imaginary part, both must be zero.
Therefore, equating the real part to zero, we get,
$\begin{aligned} & 1+a=0 \\\ & \Rightarrow a=-1 \\\ \end{aligned}$
Thus we got our first value $a=-1$ .
Now, equating the complex part to zero, we get,
$\begin{aligned} & b-1=0 \\\ & \Rightarrow b=1 \\\ \end{aligned}$
Thus our additive inverse which is $a+bi$ is equal to $-1+1i$
Thus, the additive inverse of $1-i$ is $-1+i$ .
Therefore, the correct option is option (c).

Note: Essentially additive inverses are the numbers, real or complex, which when added to the original numbers give the result 0. Thus there is a shortcut in finding them. We can just change the signs of the coefficients of the real and imaginary part. The resulting number is the additive inverse of the original number.
Taking the number $1-i$ .
Changing signs of the real and imaginary parts, we get $-1+i$ which is the additive inverse and it matches our answer.