Question 1
True and False
(i) If three complex numbers are in A.P, then they lie on a circle in the complex plane
(ii) The cube root of the unit when represented on the Argand plane form the vertices of an equilateral triangle
(iii)Region represented by $|z + 7| \leq 5$ is On or inside the circle having centre (–7, 0) and radius 5 units
(iv) Region represented by $|z + 7i| \leq 5$ is On or inside the circle having centre (0, -7) and radius 5 units
(v) if the complex numbers $z_1$, $z_2$ and $z_3$ represent the vertices of the equilateral triangle such that $|z_1|=|z_2|=|z_3|$ then $z_1 + z_2 + z_3=0$
(vi) $|z_1 + z_2 + z_3 ...+ z_n| \leq |z_1| + |z_2| + |z_3| +.... + |z_n|$
(vii) If n is a positive integer, then the value of $i^n+ (i)^{n+1} + (i)^{n+2} +(i)^{n+3}$ is 0 Solution
(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
$i^n+ (i)^{n+1} + (i)^{n+2} +(i)^{n+3}= i^n(1 + i + i^2+ i^3) = i^n(1 + i – 1 – i)=0$
Question 2
Fill in the blanks
(i) For any two complex numbers $z_1$ and $z_2$ and real numbers a and b. then $|az_1 -bz_2|^2 + |bz_1 +a z_2|^2$ is ____________
(ii) if arg z < 0, then arg(-z) - arg (z) is _____
(iii) If $z_1$ and $z_2$ are two non-zero complex number such that $|z_1 + z_2| = |z_1| + |z_2|$, then $arg z_1 -arg z_2$ is ____ Solution
(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
Question 3
Let A ,B and C be the set of complex number defined as
A={z :||z+2| -|z-2|| =2}
B={z: $arg \frac {z-1}{z} = \frac {\pi}{2}$}
C={z : arg (z-1) =π }
Then find (A ∩ B ∩ C) Solution
$\phi$
Question 4
Let A ,B and C be the set of complex number defined as
A = {z : $Im(z) \geq 1$}
B={z: |z - 2 - i| = 3}
C={z : $Re((1- i)z) =\sqrt 2$ }
Then find n(A ∩ B ∩ C) Solution
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.
Question 5
Find the complex numbers z which simlutanously satisfy the equations
$|\frac {z-12}{z-8i}| = \frac {5}{3}$
$|\frac {z-4}{z-8}| = 1$ Solution
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
Question 6
The complex numbers $z_1$,$z_2$ and the origin form an equilateral triangle only if $z_1^2 + z_2^2 -z_1 z_2=0$
Solution
Question 7
Let $A_1,A_2, A_3,...., A_n$ be the vertices of the n-sided polygon such that
$\frac {1}{A_1A_2} = \frac {1}{A_1A_3} + \frac {1}{A_1A_4}$
Find the value of n Solution
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
Question 8
What is the locus of z, if amplitude of z – 2 – 3i is $\frac {\pi}{4}$ ? Solution
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
Question 9
Let a complex number $\alpha$, $\alpha \ne 1$, be a root of the equation $z^{p+q} -z^p -z^q +1 =0$ where p and q are distint primes. Show that either
$1 + \alpha + \alpha ^2 +... +\alpha ^{p-1}=0$
or
$1 + \alpha + \alpha ^2 +... +\alpha ^{q-1}=0$
But not both together
Question 10
Find the locus of the complex number
(i) |z+i|=|z+2|
(ii) $ |\frac {z-2}{z+3}| =1$
(iii)$z^2 + |z| =0$
(iv) $| \frac {z-1}{z+1}| < 1$ Solution
(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)
