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If z=\frac{\sqrt{3}}{2}+\frac{i}{2}\left ( i=\sqrt{-1} \right ), then \left ( 1+iz+z^{5}+iz^{8} \right )^{9} is equal to : 

 

 

  • Option 1)

    0

  • Option 2)

    1

  • Option 3)

    \left ( -1+2i \right )^{9}

     

  • Option 4)

    -1

 
   (cube root of unity) Option 1) Option 2) Option 3)   Option 4)

If the fourth term of the binomial expansion \left ( \sqrt{\frac{1}{x^{1+\log_{10}x}}}+x^{\frac{1}{12}} \right )^{6} is equal to 200, and x>1, then the value of x is : 


 

  • Option 1)

    100

  • Option 2)

    10

  • Option 3)

    10^{3}

  • Option 4)

    10^{4}

 
Fourth term is given  So,  take log both side.   put  Option 1) Option 2) Option 3) Option 4)

If \alpha \; and\; \beta be the roots of the equation x^{2}-2x+2=0, then the least value of n for which \left ( \frac{\alpha }{\beta } \right )^{n}=1 is :

  • Option 1)

    2

  • Option 2)

    5

  • Option 3)

    4

  • Option 4)

    3

 
                                                                                                                    So   Option 1) Option 2) Option 3) Option 4)

The sum of the solutions of the equation \left | \sqrt{x}-2 \right |+\sqrt{x}\left (\sqrt{x}-4 \right )+2=0,(x>0) is equal to :

  • Option 1)

    12

  • Option 2)

    10

  • Option 3)

    9

  • Option 4)

    4

 
         Sum       Option 1) Option 2) Option 3) Option 4)
Let Z = x +iy Hence z will lie on imaginary axis for every x  
15519652398521476122459.jpg Solve it

The number of complex number z which satisfy z^2+2|z|^2=2

@mannika If you look at problem, , it is clear that  must be real. if you solve you get so, if we represent
15519582136261468787462.jpg The number of complex number z such that |z-i|=|z+i|=|z+1| is
@mannika We are given that, |z-i| = |z- (-i)| = |z- (-1)| we know that in a complex plane, |z-a| represents distance of z from complex number a. Now we are given, 3 points (0,1);(0,-1);(-1,0) from which distance of z is equal.But we know in a plane there exist only 1 point which are equidistant from these 3 points which is centroid of triangle made by these 3 points.
i have not understood anything in cube root of unity topic video

z3 = 1

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If the roots of the quadratic equationx^2 + px + q = 0 are tan30° and tan15°, respectively then the value of 2 + q − p is

  • Option 1)

    2

  • Option 2)

    3

  • Option 3)

    0

  • Option 4)

    1

 
Option 1) 2 Option 2) 3 Option 3) 0 Option 4) 1

All the values of m for which both roots of the equations x2 − 2mx + m2 − 1 = 0 are greater than −2 but less than 4, lie in the interval

 

  • Option 1)

    −2 < m < 0

  • Option 2)

    m > 3

  • Option 3)

    −1 < m < 3

  • Option 4)

    1 < m < 4

 
Option 1) −2 < m < 0 Option 2) m > 3 Option 3) −1 < m < 3 Option 4) 1 < m < 4
IMG_20190109_35438.jpg Sum of roots of the equation (x + — 41x + 31+ 3 = O is -
@Arvind 1.  
15470031512802109424855.jpg Number of negative integral value of x satisfying
@Vijay Q42
IMG_20190115_201103.jpg

The solution of the inequation |x2-2x-3|<|x2-x+5| is-

15475677420921088009151.jpg |f(x)|=|x2-2x-3| |g(x)|=|x2-x+5| |f(x)|<|g(x)| plot graph of f(x) and g(x)
Sir in |z-2+3i| how we get Centre as (2,3)
@Arvind
how do you get y is less than or equal to 2 from that (y-2)^2 =y (y-2)^2 is less than or equal to 2
@santhosh we are taking different cases (1) when y is less than or equal 2 (2) when 2<y less than or equal to 3 (3) y is greater than 3
Screenshot_716.png
santosh2.jpg for atleast one negative value of x (x-x_1)(x-x_2) is less than 0 and Discriminant is greater than 0
Screenshot_715.png
santosh.jpg put x-1=y^2 simplify use the concept that value under any square root is always greater then or equal to 0
rps20190122_174705.jpg

so roots are (x,y ) = (\frac{1}{\sqrt{2}},-\:\frac{1}{\sqrt{2}}\:) or (-\:\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\:)

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