Download App

STD 11 - 4.1 complex nubers — Maths STD 11 Science — Question

CBSE BoardEnglish MediumSTD 11 ScienceMathsSTD 11 - 4.1 complex nubers1 Mark
MCQ
If $\sqrt {x + iy} = \pm (a + ib),$ then $\sqrt { - x - iy} $ is equal to
  • A
    $ \pm (b + ia)$
  • B
    $ \pm (a - ib)$
  • ✓
    $ \pm (b - ia)$
  • D
    None of these
✓

Answer

Correct option: C.
$ \pm (b - ia)$
c
(c) $\sqrt {x + iy} = \pm (a + bi)$
==> $x + iy = {a^2} - {b^2} + 2iab$ ==> $x = {a^2} - {b^2},$$y = 2ab$
 $\sqrt { - x - iy} = \sqrt { - ({a^2} - {b^2}) - 2iab} = \sqrt {{b^2} - {a^2} - 2iab} $
$ = \sqrt {{{(b - ia)}^2}} = \pm (b - ia)$.

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

Explore more

More "M.C.Q (1 Marks)" from STD 11 - 4.1 complex nubers→STD 11 - 4.1 complex nubers — all question types→Maths STD 11 Science question papers→CBSE Board STD 11 Science question papers→CBSE Board question paper generator→

Similar questions

Let the sequence ${a_1},{a_2},{a_3},.............{a_{2n}}$ form an $A.P. $ Then $a_1^2 - a_2^2 + a_3^3 - ......... + a_{2n - 1}^2 - a_{2n}^2 = $
→
The value of $2\sin^2\text{B}+4\cos(\text{A+B})\sin\text{A}\sin\text{B}+\cos^2(\text{A+B})$ is:
→
If $a, b, c$ are digits, then the rotational number represeneted by $0.cababab ........ $is :-
→
Let $z$ be a complex number, then the equation ${z^4} + z + 2 = 0$ cannot have a root, such that
→
If $\alpha$ and $\beta $ are the roots of $9{x^2} + 6x + 1 = 0,$ then the equation with the roots $\frac{1}{\alpha },\,\frac{1}{\beta }$ is
→
The number of triangles that can be formed by choosing the vertices from a set of $12$ points, seven of which lie on the same straight line, is
→
If $\alpha \ne 1$ is any ${n^{th}}$ root of unity, then $S = 1 + 3\alpha + 5{\alpha ^2}$….. upto $n$ terms, is equal to
→
If $\frac{1}{(20-a)(40-a)}+\frac{1}{(40-a)(60-a)}+\ldots \ldots+$ $\frac{1}{(180-a)(200-a)}=\frac{1}{256}$, then the maximum value of $a$ is.
→
If $5\cos 2\theta + 2{\cos ^2}\frac{\theta }{2} + 1 = 0, - \pi < \theta < \pi $, then $\theta = $
→
If the tangents at the points $P$ and $Q$ on the circle $x ^2+ y ^2-2 x + y =5$ meet at the point $R \left(\frac{9}{4}, 2\right)$, then the area of the triangle $PQR$ is
→