Download App

Complex Numbers and Quadratic Equations — Maths STD 11 Science — Question

CBSE BoardEnglish MediumSTD 11 ScienceMathsComplex Numbers and Quadratic Equations1 Mark
MCQ
Choose the correct answer. The value of $(\text{z}+3)(\bar{\text{z}}+3)$ is equivalent to:
  • ✓
    $|z + 3|^2$
  • B
    $|z - 3|$
  • C
    $z^2 + 3$
  • D
    None of these.
✓

Answer

Correct option: A.
$|z + 3|^2$
Let $z = x + iy.$ Then
$(\text{z}+3)(\bar{\text{z}}+3)=(\text{x}+\text{iy}+3)(\text{x}-\text{iy}+3)$
$=(\text{x}+3)^2-(\text{iy})^2$
$=(\text{x}+3)^2+\text{y}^2$
$=|\text{x}+3+\text{iy}|^2$
$=|\text{z}+3|^2$

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 Complex Numbers and Quadratic Equations→Complex Numbers and Quadratic Equations — all question types→Maths STD 11 Science question papers→CBSE Board STD 11 Science question papers→CBSE Board question paper generator→

Similar questions

If $9$ times the $9^{th}$ term of an $A.P.$ is equal to $13$ times the $13^{th}$ term, then the $22^{nd} $ term of the $A.P.$ is:
→
$G(1, 1, -2)$ is the centroid of the triangle $\text{ABC}$ and $D$ is the mid point of $\text{BC.}$ If $A = (-1, 1, -4) D=$
→
If one of the diameters of the circle $x^{2}+y^{2}-2 \sqrt{2} x$ $-6 \sqrt{2} y+14=0$ is a chord of the circle $(x-2 \sqrt{2})^{2}$ $+(y-2 \sqrt{2})^{2}=r^{2}$, then the value of $r^{2}$ is equal to
→
If ${^\text{n+1}}\text{C}_{\text{3}}=2.{^\text{n}}\text{C}_{\text{2}},$ then $n:$
→
The mean and variance of $10$ observations were calculated as $15$ and $15$ respectively by a student who took by mistake $25$ instead of $15$ for one observation. Then, the correct standard deviation is$.....$
→
The eccentricity of the hyperbola whose latus$-$rectum is half of its transverse axis, is
→
${\cos ^2}\left( {\frac{\pi }{6} + \theta } \right) - {\sin ^2}\left( {\frac{\pi }{6} - \theta } \right) = $
→
The value of ${81^{(1/{{\log }_5}3)}} + {27^{{{\log }_{_9}}36}} + {3^{4/{{\log }_{_7}}9}}$ is equal to
→
What is the sum of all two digit numbers which give a remainder of $4$ when divided by $6$ ?
→
If $P \times Q$ has $10$ elements then which is not possible?
→