Week -2 | Complex Numbers – 4

Q-1) Show that $|z_1 \pm z_2|^2 = |z_1|^2+|z_2|^2 \pm 2\text{Re}(z_1\overline{z_2})$

Solution:

 

Q-2) If $(1+x)^n= P_0+P_1x+P_2x^2+…+P_nx^n$, then find the value of the following series:

$P_0-P_2+P_4+…$

Solution:

Substituting $x=i$ in the equation we have

We must remember that to represent a Complex Number in polar form, we take out the modulus from the algebraic form. On those lines we can say

Thus the equation boils down to the following:

Comparing the real and imaginary parts on LHS & RHS we conclude :

 

Q-3) Show that if $1,z_1,z_2,….,z_{n-1}$ be the nth roots of unity then

Solution:

We have from the concepts of nth root of unity:

$(z^n-1)=(z-1)(z-z_1)(z-z_2)..(z-z_{n-1})$. Thus $\frac{z_n-1}{z-1}=(z-z_1)(z-z_2)..(z-z_{n-1})$

Now using GP Sum formulae (We’ll cover this in our Progressions and Sequences Module) $\frac{z^n-1}{z-1}=1+z+z^2+..+z^{n-1}$. And thus we have $1+z^2+z^3+..+z^{n-1}=(z-z_1)(z-z_2)…(z-z_{n-1})$.

Finally putting $z=1$, we have

 

Q-4) The altitudes of $\triangle ABC$ meet its circumcircle at $D,E,F$ respectively. Let $z_1,z_2,z_3$ represent the complex numbers for points $D,E,F$ respectively. Given that $\frac{z_3-z_1}{z_2-z_1} \in R$, show that $\angle BAC=90$ deg.

Solution:

Consider the figure as shown

Fig 4.1

$\angle FDA = \angle FCA = 90^\circ -A$, since both of them lie on curve $FA$ and $CF \perp AB$.  We also have $\angle ADE = \angle ABE = 90^\circ – A$. Thus $\angle FDE = 180^\circ-2A$. Similarly we conclude $\angle FED = 180^\circ – 2B$ & $\angle DFE = 180^\circ – 2C$.

We also have $\arg \left(\frac{z_3-z_1}{z_2-z_1} \right ) =0$ or $\pi$. Frrom which we conclude $\pi -2A =0$ or $\pi$, which gives $A=\frac{\pi}{2}$ or $0$ – which is impossible.

 

Q-5)

Matrix Match Type

Column I	Column II
a) The set of points satisfying $|z-i|z||-|z+i|z||=0$ is contained in or equal to	p) An ellipse with eccentricity $\frac{4}{5}$
b) The set of points $z$ satisfying $|z+4|+|z-4|=10$ is contained in or equal to	q) The set of points satisfying $\text{Img}(z)=0$
c) If $|\omega |=2$, then the set of points $z$ satisfying $z=\omega – \frac{1}{\omega}$ is contained in or equal to	r) The set of points $z$ satisfying $|\text{Img}(z)| \le 1$
d) If $|\omega|=1$The set of points $z$ satisfying$z=\omega + \frac{1}{\omega}$ is contained in or equal to	s) The set of points $z$ satisfying $|\text{Re}(z)| \le 1$
	t) The set of points $z$ satisfying $|z| \le 3$.

Solution:

a) We have $\left|\frac{z}{|z|}-i \right| = \left|\frac{z}{|z|} +i \right|$, which means from our locus concepts that $\frac{z}{|z|}$ lies on the perpendicular bisector of $i$ & $-i$. This implies $\frac{z}{|z|} \in R \iff z \in R$. Thus $\text{Img}(z) = 0$.

b) From our locus concepts, we can conclude that the locus of $z$ is an ellipse. Also we know $2|z| \le |z+4|+|z-4| = 10$, thus giving $\text{Re}(z) \le |z| \le 5$. The distance between the 2 focal points of the ellipse is $8$. Thus eccentricity = $\frac{10}{8} = \frac{4}{5}$.

c) $|\omega| = 2$. which implies we can have $\omega = 2(\cos \theta + i\sin \theta)$. Assuming $z=x+iy$, we have $z= 2(\cos \theta + i\sin \theta) – \frac{1}{2}\left( \cos \theta -i\sin \theta\right)$. COmparing the real and imaginary parts we finally have

$x=\frac{3}{2}\cos\theta$ & $y=\frac{5}{2}\sin\theta$.

We thus can say

This is the equation of an ellipse, eccentrivity of which is $e^2=1-\frac{\frac{9}{4}}{\frac{25}{4}} = \frac{16}{25}$. Thus $e=\frac{4}{5}$. Also $|z|^2=x^2+y^2 = \frac{34}{4} <9$. Thus $|z|<3$.

d) Assuming $\omega = \cos \theta +i\sin \theta$, we have

$z=x+iy = \cos\theta + i\sin\theta + \cos\theta -i\sin \theta = 2\cos\theta$. Thus $|z|<3$ and $\text{Img}(z)=0$.

Thus a->q, b->p, c->p,t  d->q,t.

 

Q-6)Let $z=x+iy$ be a complex number with $x$ and $y$ as integers. Find the area of the rectangle whose vertices are roots of the equation  $\overline{z}z^3+z\overline{z}^3=350$

Solution:

As we have discussed in our editorials, the best way to solve complex equations is to make use of the algebraic form of the complex numbers. Thus

$z\overline{z}(\overline{z}^2+z^2)=350$, which boils down to

$(x^2+y^2)(x^2-y^2)=175=5*5*7$. Thus we have 2 equations

$x^2+y^2=25$ and $x^2-y^2=7$. Solving them we get$x=\pm 4$ & $y=\pm 3$. Thus area = $8*6=48$.

 

Q-7) If $z=\lambda +3+i\sqrt{5-\lambda^2}$, find the locus of $z$ if $\lambda \in R$

Solution:

Let $z=x+iy$. Comparing real and imaginary parts we have

$x=\lambda+3$ & $y=\sqrt{5-\lambda^2}$. Thus we can say $(x-3)^2+y^2=5$, which is the locus of a circle.

 

Q-8) If $1,\omega , \omega^2,…\omega^{n-1}$ be the nth roots of unity then find the value of $(9-\omega)(9-\omega^2)…(9-\omega^{n-1})$

Solution : We know that $(x^n-1)=(x-1)(x-\omega )…(x-\omega^{n-1})$. Thus we can say that $\frac{x^n-1}{x-1} = (x-\omega )(x-\omega^2)…(x-\omega^{n-1})$. Plugging $x=9$, we have $(9-\omega)(9-\omega^2)…(9-\omega^{n-1})=\frac{9^n-1}{8}$.

 

Q-9)

Comprehension for the following 3 problems

Let A,B,C be 3 sets of complex numbers as defined

Q-9.1) The number of elements in the set $A \cap B \cap C$ is

a)0    b)1    c)2    d)$\infty$

Q-9.2) Let $z \in A\cap B \cap C$. Then $|z+1-i|^2+|z-5-i|^2$ lies between

a)25 & 29    b)30 & 34    c)35 & 39    d)40 & 44

Q-9.3) Let $z \in A\cap B \cap C$. Let $\omega$ be any point satisfying $|\omega -2-i| <3$. Then $|z|-|\omega|+3$ lies between

a)-6 & 3    b)-3 & 6    c)-6 & 6    d)-3 & 9

Solution:

9.1) The best solution would be to geometrically view this. {A} signifies the region above the line $y=1$. {B} signifies the circle $(x-2)^2+(y-1)^2=9$ and finally {C} signifies $\text{Re}((1-i)z) =\text{Re}((1-i)(x+iy)) = x+y=\sqrt{2}$. We see that this chord represented by {C} cuts the circle represented by {B} at exactly 1 point above the line $y=1$, which is represented by {A}. Thus  9.1->b.

9.2) The points $(-1,1)$ & $(5,1)$ represent the extremities of the diameter of the circle represented by {B}. Let us recollect that the sum of the squares of 2 perpendicular sides of a right angled triangle inscribed in a  circle equals to $4r^2$, where $r$ is the radius of the circle (Refer to Ex-3 of this post). Thus the sum = $36$.

9.3) We have already shown that $z$ is a fixed point.  $||z|-|\omega||<|z-\omega|$, and the maximum distance in a circle can be between 2 diametrically opposite points. Thus $|z-\omega|<6$, which gives $-6<|z|-|\omega|<6$. Thus $-3<|z|-|\omega|+3<9$.

 

Q-10) Let $s,t,r$ be non-zero complex numbers and $L$ be the set of solutions $z=x+iy$, with $x,y \in R$ of the equation $sz+t\overline{z}+r=0$. Which of the following statement(s) is(are) true?

a) If $L$ has exactly 1 element, then $|s| \ne |t|$

b) If |s|=|t|, then $L$ has infinitely many elements

c) The number of elements in $L \cap \left \{ z:|z-1+i|=5 \right \}$ is at most 2

d) If  $L$ has more than 1 element, then $L$ has infinitely many elements.

Solution:

$(1)+(2)$ gives

$(1)-(2)$ gives

For unique solution we must have $\frac{t+\overline{s}}{t-\overline{s}} \ne \frac{s+\overline{t}}{s-\overline{t}}$. Cross multiplication and further simplification gives us $|s| \ne |t|$.

Thus (A) is correct. Note that if $L$ has more than 1 element then $|s|=|t|$. For (B) to be true we need the lines to overlap, which means

Coupled with the fact that $|s|=|t|$, we can say that the lines may be parallel but no guarantee that the lines will overlap. Thus (B) may not be true.

(C) is trivial since $|z-1+i|=5$ represents a circle and if the elements of {L} represent a line we can have maximum 2 points of intersection.

With $|s| \ne |t|$, we can conclude that there can be infinitely many elements in L.

Thus A,C,D are correct

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