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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