MCQ
Additive inverse of $1 - i$is
  • A
    $0 + 0i$
  • B
    $ - 1 - i$
  • $ - 1 + i$
  • D
    None of these

Answer

Correct option: C.
$ - 1 + i$
c
(c) If $z = x + iy$ is the additive inverse of $1 - i$then $(x + iy) + (1 - i) = 0$ 

==> $x + 1 = 0$, $y - 1 = 0$

==> $x = - 1$, $y = 1$

$\therefore $ The additive inverse of $1 - i$is $z = - 1 + i$

Trick : Since $(1 - i) + ( - 1 + i) = 0$.

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Similar questions

$\text { If } \int_{0}^{100 \pi} \frac{\sin ^{2} x}{e^{\left(\frac{x}{\pi}-\left[\frac{x}{\pi}\right]\right)}} d x=\frac{\alpha \pi^{3}}{1+4 \pi^{2}}, \alpha \in R$

where $[x]$ is the greatest integer less than or equal to $x$, then the value of $\alpha$ is :

The foot of perpendicular of the point $(2,0,5)$ on the line $\frac{x+1}{2}=\frac{y-1}{5}=\frac{z+1}{-1}$ is $(\alpha, \beta, \gamma)$. Then. Which of the following is $NOT$ correct?
Let $f(x)=2+\cos x$ for all real $x$.

$STATEMENT -1$ : For each real $\mathrm{t}$, there exists a point $\mathrm{c}$ in $[\mathrm{t}, \mathrm{t}+\pi]$ such that $\mathrm{f}^{\prime}(\mathrm{c})=0$. because

$STATEMENT -2$: $f(t)=f(t+2 \pi)$ for each real $t$.

The numbers of solution $(s)$ of the equation $\left( {1 - \frac{1}{{2\,\sin x}}} \right){\cos ^2}\,2x\, = \,2\,\sin x\, - \,3\, + \,\frac{1}{{\sin x}}$ in $[0,4\pi ]$ is
In the expansion of the following expression $1 + (1 + x) + {(1 + x)^2} + ..... + {(1 + x)^n}$ the coefficient of ${x^k}(0 \le k \le n)$ is
The value of $\lambda$ for which $\int {\frac{{4{x^3} + \lambda {4^x}}}{{{4^x} + {x^4}}}} \,\,dx = \log ({4^x} + {x^4}) + c$ is
If $x = a{\cos ^4}\theta ,y = a{\sin ^4}\theta ,$ then ${{dy} \over {dx}}$, at $\theta = {{3\pi } \over 4}$, is
If $\frac{1}{1^{4}}+\frac{1}{2^{4}}+\frac{1}{3^{4}}+\ldots . . \infty=\frac{\pi^{4}}{90}$,
$\frac{1}{1^{4}}+\frac{1}{3^{4}}+\frac{1}{5^{4}}+\ldots \ldots \infty=\alpha$,
$\frac{1}{2^{4}}+\frac{1}{4^{4}}+\frac{1}{6^{4}}+\ldots . . \infty=\beta$,
then $\frac{\alpha}{\beta}$ is equal to
Sum of the series $1\cdot 2015 + 2\cdot 2014 + 3\cdot 2013 +.....2015\cdot 1$ is equal to :-
If $\{ p \}$ denotes the fractional part of the number $p$, then $\left\{\frac{3^{200}}{8}\right\},$ is equal to