(i) False
False, because then $z_3 = \frac {z_1 + z_2}{2}$ which means $z_3$ is mid point of $z_1$ and $z_2$, So they collinear
(ii) True
.The cube roots are 1, $ \frac {-1+i \sqrt {3}}{2}$ and $ \frac {-1-i \sqrt {3}}{2}$
. We can see that distance between them and they are all equal
(iii) True
(iv) True
(v) True
Let O be the origin,then OA=OB=OC. So, O is the circumcenter
Now circumcenter is given by $\frac {z_1 + z_2 + z_3}{2}$.
Therfore
$z_1 + z_2 + z_3=0$
(vi) True
$i^n+ (i)^{n+1} + (i)^{n+2} +(i)^{n+3}= i^n(1 + i + i^2+ i^3) = i^n(1 + i – 1 – i)=0$
(i) $(a^2 + b^2)(|z_1|^2 + |z_2|^2$
$|az_1 -bz_2|^2 + |bz_1 +a z_2|^2 = (az_1 -bz_2)(a \bar {z_1} -b\bar {z_2}) + (bz_1 +a z_2)(b\bar {z_1} +a \bar {z_2})$
$=a^2 z_1 \bar {z_1} -ab z_1\bar {z_2} -ab z_2\bar {z_1} + b^2 z_2 \bar {z_2} + b^2 z_1 \bar {z_1} + ab z_1\bar {z_2} + ab z_2\bar {z_1} + a^2 z_2 \bar {z_2}$
$=(a^2 + b^2)(|z_1|^2 + |z_2|^2$
(ii) $\pi$
Let
$z= r (cos \theta + i sin \theta)$
$\theta < 0$
$-z = - r(cos \theta + i sin \theta)$
$-z = r[ cos (\pi + \theta) + i sin (\pi + \theta)]$
$arg(-z) = \pi + \theta$
So, $arg(-z) - arg (z) =\pi + \theta - \theta = \pi$
(iii) 0
$\phi$
A = set of points on and above the line y = 1 in the argand plane.
B = set of points on the circle $(x - 2)^2 + (y -1)^2 = 9$
C = set of point lies on the line $x + y = \sqrt 2$
We can draw this on the Argand plane and we can see there is only point of intersection
Hence n(A ∩ B ∩ C) = 1.
Let z=x + iy
Then
$|\frac {z-4}{z-8}| = 1$
$(x-4)^2 +y^2 =(x-8)^2 + y^2$
x=6
Now
$|\frac {z-12}{z-8i}| = \frac {5}{3}$
$|\frac {-6 +iy}{^+i(y-8)}) =\frac {5}{3}$
or
y^2 -25t + 136=0
y=17 or 8
For equilateral triangle
$|z_1|=|z_2|=|z_1 -z_2|=k$
$|z_1|=k$
$|z_1|^2=k^2$
$z_1 \bar{z_1)} =k^2$
$\bar{z_1}= \frac {k^2}{z_1}$ -(1)
Similarly
$\bar{z_2}= \frac {k^2}{z_2}$ --(2)
$\bar{z_1} -\bar{z_2}= \frac {k^2}{z_1 -z_2 }$ --(3)
Adding (1) ,(2) and (3)
$0= k^2 [ \frac {1}{z_1}+ \frac {1}{z_2} + \frac {1}{z_1 -z_2 }]$
or
$z_1^2 + z_2^2 -z_1 z_2=0$
Let $z_0$ be the center of the polygon and $z_1,z_2,z_3,.....,z_n$ be the complex numbers for $A_1,A_2, A_3,...., A_n$
Now angle subtended by a side at the center of the polygon =$\frac {2\pi}{n}$
Then
$z_2 -z_0 = (z_1 -z_0)e^ {2i \pi/n}$
or
$z_2 = z_0 + (z_1 -z_0)e^ {2i \pi/n}$
Similarly
$z_3=z_0 + (z_1 -z_0)e^ {4i \pi/n}$
$z_r=z_0 + (z_1 -z_0)e^ {2(r-1)i \pi/n}$
Now
$A_1A_r = |z_r -z_1|=|z_0 + (z_1 -z_0)e^ {2(r-1)i \pi/n} - z_1| = |z_1 -z_0|| 1 -e^ {2(r-1)i \pi/n}|$
$A_1A_r =|z_1 -z_0| |1 -cos {2(r-1) \pi/n} + i sin {2(r-1) \pi/n}|$
$A_1A_r =|z_1 -z_0| \sqrt {(1 -cos {2(r-1)\pi/n})^2 + sin^2 {2(r-1) \pi/n}}$
$A_1A_r =|z_1 -z_0|\sqrt { 2 -2 cos {2(r-1) \pi/n}}= 2|z_1 -z_0| sin ((r-1) \pi/n)$
Now
$A_1A_2 =2|z_1 -z_0| sin ( \pi/n)$
$A_1A_3 =2|z_1 -z_0| sin ( 2\pi/n)$
$A_1A_4 =2|z_1 -z_0| sin ( 3\pi/n)$
Now as per question
$\frac {1}{A_1A_2} = \frac {1}{A_1A_3} + \frac {1}{A_1A_4}$
Substituting the values
$ \frac {1}{sin ( \pi/n)} =\frac {1}{sin ( 2\pi/n)} + \frac {1}{sin ( 3\pi/n)}$
$ sin ( 2\pi/n) sin ( 3\pi/n)= sin ( \pi/n) [ sin ( 2\pi/n) + sin ( 3\pi/n)]$
$2 cos ( \pi/n) sin ( 3\pi/n)=sin ( 2\pi/n) + sin ( 3\pi/n)$
$ sin (4\pi/n) + sin ( 2\pi/n)=sin ( 2\pi/n) + sin ( 3\pi/n)$
$sin (4\pi/n)=sin ( 3\pi/n)$
$sin (4\pi/n)=sin (\pi - 3\pi/n)$
$\frac {4\pi}{n} = \pi -\frac {3\pi}{n}$
n=7
Let z=x+iy
Then
z – 2 – 3i =x+ iy -2 -3i = (x-2) +i(y-3)
Now $tan \theta = \frac {y-3}{x-2}$
Now Given $\theta=\frac {\pi}{4}$
Therefore
$1=\frac {y-3}{x-2}$
x – y + 1 = 0
Hence locus is a straight line
(i) |z+i|=|z+2|
$x^2 + (y+1)^2 = (x+2)^2 + y^2$
$2y+1 =4x + 4$
$y = 2x + \frac {3}{2}$
So point on the straight line
(ii) $ |\frac {z-2}{z+3}| =1$
$\frac {|z-2|}{|z+3} =1$
$|z-2|= |z+3|$
$(x-2)^2 + y^2 = (x-3)^2 + y^2$
$x= -1/2$
So point on the straight line
(iii) $z^2 + |z| =0$
$x^2 -y^2 + 2ixy + \sqrt {x^2 + y^2}=0$
$x^2 -y^2 + \sqrt {x^2 + y^2} + 2ixy=0$
Now xy=0
for x=0,
$-y^2 + \sqrt {y^2}=0$
$y^2 = \pm y$
y=0, 1,-1
So (0,0) ,(0,1) and (0,-1)
Now take y=0
$x^2 + \sqrt {x^2}=0$
$x^2 +x=0$
x=0 ,x=-1,1
So, (0,0) but x=1 and -1 does not satisfy
So Answe (0) ,(i) , (i)
(iv)$| \frac {z-1}{z+1}| < 1$
$\frac {|z-1|}{|z+1|} < 1$
$|z-1| < |z+1|$
$(x-1)^2 + y^2 < (x+1)^2 + y^2$
$-2x < 2x$
$x > 0$
