Test: Complex Numbers - JEE MCQ

(1/2)^{(x2 - 2x)} < (1/4) 
(1/2)^{(x2 - 2x)} < (1/2)²
x² − 2x > 2......(as after multiplicative inverse sign of inequality changes)
x² − 2x − 2 > 0
x²  -  2x + 1 - 3  >
(x-1)²  - 3  > 0
(x-1)²  > 3 
So for the above to hold good both the expression must be positive or both must be negative. After finding the solution the range of the solution will be
either x > 3
(3,¥)

Puting x=-1
a-b+c
but
a+c<b or a+c-b<0
hence f(x)<0 for all real values of x
therefore
putting x=-2
we get f(X)=4a+c-2b<0
or 4a+c<2b

As (1 – p) is root of the equation: x² + px + (1 – p) = 0

(1 – p)² + p(1 – p) + (1 – p) = 0

(1 – p)[1 – p + p + 1] = 0

(1 – p) = 0

p = 1

Therefore, given equation now becomes

x² + x = 0

x(x + 1) = 0

x = 0, -1

The triangle formed is an isosceles right angled triangle
Therefore it has an area

Hence
⇒|z|=20 units.

Since all roots are positive,
coeff. of x shld be negatuve i.e., b is negative
let roots be m(>0), n(>0), p(>0), q(>0)
m+n+p+q = 4
mn+np+pq+mp+mq+nq = a
mnp+mnq+npq+mpq = -b
mnpq = 1
For the objective case,
put m=n=p=q =1
so a = 1+1+1+1+1+1 = 6
    -b = 1+1+1+1 =4
so, a = 6 and b = -4

k+|k+z²|=|z|² 
Hence k and ∣∣z²∣∣ are collinear.
Now k∈R⁻ 
Since, k and z² is collinear, then z² has to be purely real.
Thus, arg(z²)=(2n−1)π
Now z=re^iθ 
z² = r²e^2iθ=me^2iθ
Hence arg(z²)=2arg(z)
Therefore, arg(z)=1/2arg(z²)

We have |z|=|a|<1, thus

But z²⁰¹⁵ = a²⁰¹⁵[cos(403π) + isin(403π)]=−a²⁰¹⁵ 

The numbers in the questions are not very clear.
z₁ = 2 + i
z₂ = 1+3i
z₁ – z₂ = (2 – 1) + i (1 – 3)
= 1 – 2i