# 22. Complex Numbers Solutions Part 3

1. Since $$P, A, B$$ are collinear

$\therefore \begin{vmatrix}z & \overline{z} & 1\\a & \overline{a} & 1\\ b & \overline{b} & 1\end{vmatrix} = 0 \Rightarrow z(\overline{a} - \overline{b}) - \overline{z}(a - b) + (a\overline{b} - \overline{a}b) = 0$

Similarly, $$P, C, D$$ are collinear

$\Rightarrow z(\overline{c} - \overline{d}) - \overline{z}(c - d) + (c\overline{d} - \overline{c}d) = 0$

Multiplying the first equation with $$c - d$$ and second with $$a - b$$ and subtracting resulting equations we get

$z(\overline{a} - \overline{b})(c - d) - \overline{z}(\overline{c} - \overline{d})(a - b) = (c\overline{d} - \overline{c}d)(a - b) - (a\overline{b} - \overline{a}b)(c - d)$

Given that $$a, b, c, d$$ all lie on the circle $$|z| = r \therefore |z|^2 = r^2 \Rightarrow z\overline{z} = r^2 \Rightarrow \overline{z} = \frac{r^2}{z}$$

Substituting accordingly in previous equation

$z\left(\frac{r^2}{a} - \frac{r^2}{b}\right)(c - d) - z\left(\frac{r^2}{c} - \frac{r^2}{d}\right)(a - b) = \left(\frac{cr^2}{d} - \frac{dr^2}{c}\right)(a - b) - \left(\frac{ar^2}{b} - \frac{br^2}{a}\right)(c - d)$

Solving this will yield desired result.

2. We have,

$\frac{z + 1}{z - 2} = \frac{3 + t + i\sqrt{3 - t^2}}{1 + t + i\sqrt{3 - t^2}} \therefore \left|\frac{z + 1}{z - 1}\right|^2 = \frac{(3 + t)^2 + 3 - t^2}{(1 + t)^2 + 3 - t^2} = \frac{6(t + 2)}{2(t + 2)} = 3$

Thus modulus of required fraction is independent of $$t$$.

Also, $$z = x + iy = 2 + t + i\sqrt{3 - t^2} \Rightarrow y = 3 - (x - 2)^2$$ or $$(x - 2)^2 + y^2 = 3.$$

Thus locus of complex number represents a circle with center at $$(2, 0)$$ having radius 3.

3. Given,

$\begin{vmatrix}a & b & c\\b & c & a\\c & a & b\end{vmatrix} = 0 \Rightarrow a^3 + b^3 + c^3 - 3abc = 0 \Rightarrow (a + b + c)[a^2 + b^2 + c^2 - ab -bc - ca] = 0 \Rightarrow (a - b)^2 + (b - c)^2 + (c - a)^2 = 0 \Rightarrow a = b = c [\because a + b + c \ne 0 \because z_1 \ne = 0]$

The circle made by these is shown below:

Now OA = OB = OC where O is the origin and A, B and C are the points representing $$z_1, z_2$$ and $$z_3$$ respectively.

$$\therefore$$ O is the circumcenter of $$\triangle ABC.$$

Now

$arg\left(\frac{z_3}{z_2}\right) = \angle BOC = 2\angle BAC = 2arg\left(\frac{z_3 - z_1}{z_2 - z_1}\right) = arg\left(\frac{z_3 - z_1}{z_2 - z_1}\right)^2$

Hence, proven.

4. $z_2 = \frac{OQ}{OP}z_1e^{i\theta} = cos\theta z_1e^{i\theta}$

and

$z_3 = \frac{OR}{OP}z_1e^{i2\theta} = cos2\theta z_1e^{i2\theta}$

Hence,

$z_2^2cos2\theta = z_1z_3cos^2\theta.$
5. Given circles are $$|z| = 1$$ or $$x^2 + y^2 - 1 = 0$$ and $$|z - 1| = 4$$ or $$x^2 + y^2 -2x - 15 = 0.$$

Let the circles cut by these two circles orthogonally is

$x^2 + y^2 + 2gx + 2fy +c = 0$

Now since first two circles cut this third one orthogonally

$\therefore 2g.0 + 2f.0 = c - 1 \Rightarrow c = 1$

and

$\therefore 2g(-1) + 2f.0 = c - 15 \Rightarrow g = 7$

Therefore the required circles are

$x^2 + y^2 + 14x + 2fy + 1 = 0 |z + 7 + if| = \sqrt{48 + f^2}$
6. Given $$|z + 3| = t^2 - 2t + 6$$ represents a circle with center (-3, 0) and radius $$t^2 -2t + 6.$$ The inequality $$|z - 3\sqrt{3}i| < t^2$$ means $$z$$ lies in the interior of circle having center at $$(0, 3\sqrt{3})$$ having radius $$t^2.$$

Let A is center of first circle and B is center of second circle. Clearly when both the circles are disjoint or touching then no solution is possible.

Further solution is left as an exercise.

7. Let $$z = x + iy$$

$\frac{az + b}{cz + d} = \frac{ax + iay + b}{cx + icy + d} = \frac{[(ax + b) + iay][(cx + d) - icy]}{(cx + d)^2 + d^2} Im\left(\frac{az + b}{cz + d}\right) = \frac{ay(cx + d) - cy(ax + b)}{(cx + d)^2 + d^2} \Rightarrow \frac{ady - bcy}{(cx + d)^2 + d^2}$

Now since $$ad > bc$$ sign is same as $$y$$ i.e. positive. Hence, proven.

8. Given,

$z_1 = \frac{i(z_2 + 1)}{z_2 - 1} \Rightarrow x_1 + iy_1 = \frac{i(x_2 + iy_2 + 1)}{(x_2 - 1) + iy_2} = \frac{[-y_2 + i(x_2 + 1)][(x_2 - 1) + iy_2]}{(x_2 - 1)^2 + y_2^2}$

Now equating for real part and then evaluating the desired equation will yield the result.

9. $$sin25\theta + icos25\theta$$ This question is left as an exercise.

10. Let $$z = x + iy.$$ Now we have

$z^2 + |z| = x^2 - y^2 + i2xy + \sqrt{x^2 - y^2} = 0$

Equating imaginary parts we have $$2xy = 0$$ which means either $$x = 0$$ or $$y = 0.$$ Let $$y = 0$$ then we have

$x^2 + \sqrt{x^2} = 0$

Since $$x$$ is real the only possible solution is $$x = 0$$. So $$z = 0.$$

If $$x = 0$$ then we have

$y^2 + \sqrt{-y^2} = 0 y^4 + y^2 = 0 y^2 = -1 \Rightarrow y = \pm i$

Thus we have $$z = \pm i.$$

11. Problem no. 111 to 118 have been left as exercises to the reader.

1. Given,

$|1 - \overline{z_1}z_2|^2 - |z_1 - z_2|^2 = (1 - \overline{z_1}z_2)(1 - z_1\overline{z_2}) - (z_1 - z_2)(\overline{z_1} - \overline{z_2}) [\because |z|^2 = z\overline{z}] = (1 - \overline{z_1}z_2 - z_1\overline{z_2} + |z_1|^2|z_2|^2) - (|z_1|^2 - \overline{z_1}z_2 - z_1\overline{z_2} + |z_2|^2) = (1 - |z_1|^2 - |z_2|^2 + |z_1|^2|z_2|^2) = (1 - |z_1|^2)(1 - |z_2|^2)$
2. Consider two complex numbers $$z_1 = a_1 + ib_1$$ and $$z_2 = a_2 + ib_2.$$ Now we have to prove $$|z_1 + z_2| \le |z_1| + |z_2|$$ which can be further extended to prove the result.

$\sqrt{(a_1 + a_2)^2 + (b_2 + b_2)^2} \le \sqrt{a_1^2 + b_1^2} + \sqrt{a_2^2 + b_2^2}$

Squaring both sides and simplifying we get

$a_1a_2 + b_1b_2 \le \sqrt{(a_1^2 + b_1^2)(a_2^2 + b_2^2)} \Rightarrow (a_1a_2 + b_1b_2)^2 - (a_1^2 + b_1^2)(a_2^2 + b_2^2) \le 0 \Rightarrow -(a_1b_2 - a_2b_1)^2 \le 0$

which is true. Hence, proven.

3. We have,

$\left|\frac{\overline{z_1} - 2\overline{z_2}}{2 - z_1\overline{z_2}}\right| = 1 \Rightarrow |\overline{z_1} - 2\overline{z_2}|^2 = |2 - z_1\overline{z_2}|^2 \Rightarrow (\overline{z_1} - 2\overline{z_2})(z_1 - 2z_2) = (2 - z_1\overline{z_2})(2 - \overline{z_1}z_2) \Rightarrow |z_1|^2 - 2z_1\overline{z_2} - 2\overline{z_1}z_2 + 4|z_2|^2 = 4 - 2z_1\overline{z_2} - 2\overline{z_1}z_2 + |z_1|^2|z_2|^2 \Rightarrow |z_1|^2|z_2|^2 - 4|z_2|^2 - |z_1|^2 - 4 = 0 \Rightarrow \because |z_1| \ne 1 |z_2| = 2$
4. It can be solves similarly as 121 and is left as an exercise.

5. We have,

$\left|\frac{z_1 + z_2}{2} + \sqrt{z_1z_2}\right| + \left|\frac{z_1 + z_2}{2} - \sqrt{z_1z_2}\right| = \frac{1}{2}\left|(\sqrt{z_1} + \sqrt{z_2})^2\right| + \frac{1}{2}\left|(\sqrt{z_1} - \sqrt{z_2})^2\right| = |z_1| + |z_2|$
6. From problem no. 54 it follows that $$|a + \sqrt{a^2 - b^2}| + |a - \sqrt{a^2 - b^2}| = |a + b| + |a - b|.$$

Substituting $$a = \beta$$ and $$b = \sqrt{\alpha\gamma}$$ we have

$|\beta + \sqrt{\alpha\gamma}| + |\beta - \sqrt{\alpha\gamma}| = |\alpha|\left(|\frac{\beta}{\alpha} + \sqrt{\frac{\gamma}{\alpha}}| + |\frac{\beta}{\alpha} - \sqrt{\frac{\gamma}{\alpha}}|\right) = |alpha|\left(|-z_1 - z_2 + \sqrt{z_1z_2}| + |-z_1 - z_2 - \sqrt{z_1z_2}|\right) = |\alpha|(|z_1| + |z_2|)$
7. We have,

$|a| = 1 \Rightarrow |a|^2 = 1 \Rightarrow a\overline{a} = 1 \Rightarrow \overline{a} = \frac{1}{a}$

From this we can write

$\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = \overline{a} + \overline{b} + \overline{c} = 3$
8. Given,

$|z + 4| \le 3 \Rightarrow |z + 1 + 3| \le 3 \Rightarrow |z + 1| + 3 \le 3 [\because |z_1 + z_2| \le |z_1| + |z_2|] |z + 1| \le 0$

Following similarly

$|z + 4| = |z + 1 + 3| \ge |z + 1| - 3 \Rightarrow |z + 1| \ge 6$

So least value is 0 and greatest value is 6.

9. Let $$z_1 = r_1cos\theta_1 + isin\theta_1$$ and $$z_2 = r_2cos\theta_2 + isin\theta_2.$$ Now it can be easily shown that

$4|z_1 + z_2|^2 - (|z_1| + |z_2|)^2\left(\frac{z_1}{|z_2|} + \frac{z_2}{|z_2|}\right)^2 \ge 0$
10. Given equation is $$z^2 + az + b = 0.$$ Let $$p, q$$ are two of its roots. Then we have $$p + q = -a$$ and $$pq = b.$$ Taking modulus of both we have $$|p + q| = |a|$$ and $$|pq| = b.$$ Now it is required that $$|p| = |q| = 1.$$ Therefore we have $$|p + q| \le |p| + |q| = 2 \therefore |a| \le 2.$$ Similarly $$|b| = |pq| = |p||q| = 1.$$ Since $$p, q$$ have unit modulii we can have them as $$p = cos\theta_1 + isin\theta_1$$ and $$q = cos\theta_2 + isin\theta_2.$$

$pq = (cos\theta_1 + isin\theta_1)(cos\theta_2 + isin\theta_2) = cos(\theta_1 + \theta_2) + isin(\theta_1 + \theta_2) \therefore arg(b) = arg(pq) = \theta_1 + \theta_2 p + q = cos\theta_1 + isin\theta_1 + cos\theta_2 + isin\theta_2 = cos^2{\frac{\theta_1}{2}} + i^2sin^2{\frac{\theta_1}{2}} + i2sin{\frac{\theta_1}{2}}cos{\frac{\theta_1}{2}} + cos^2{\frac{\theta_2}{2}} + i^2sin^2{\frac{\theta_2}{2}} + i2sin{\frac{\theta_2}{2}}cos{\frac{\theta_2}{2}} = cos\frac{\theta_1 + \theta_2}{2} + isin\frac{\theta_1 + \theta_2}{2} \therefore arg(a) = arg(p + q) = \frac{\theta_1 + \theta_2}{2} \therefore argb = 2arga$
11. Let $$a = x + iy.$$ First we consider first two inequalities

$|z| \le |Re(z)| + |Im(z)| \Rightarrow \sqrt{x^2 + y^2} \le x + y$

Sqauting we have

$x^2 + y^2 \le x^2 + y^2 + 2xy \Rightarrow 2xy \ge 0$

which is true. Now we consider last two inequalities

$|Re(z)| + |Im(z)| \le \sqrt{2}|z| \Rightarrow x + y \le \sqrt{2(x^2 + y^2)}$

Squaring we get

$x^2 + y^2 + 2xy \le 2(x^2 + y^2) \Rightarrow (x - y)^2 \ge 0$

which is also true. Hence, proven.

12. Translating the given equation we have

$|z|^2 - 2|z| - 4 \ge 0$

The greatest root of this equation is $$\sqrt{5} + 1.$$ Hence proven.

13. Since $$\alpha, \beta, \gamma, \delta$$ are root of the equation.

$(x - \alpha)(x - \beta)(x - \gamma)(x - \delta) = ax^4 + bx^3 + cx^2 + dx + e$

Substituting $$x = i$$ we get following

$(i - \alpha)(i - \beta)(i - \gamma)(i - \delta) = ai^4 + bi^3 + ci^2 + di + e \Rightarrow (1 + i\alpha)(1 + i\beta)(1 + i\gamma)(1 + i\delta) = a - ib - c + id + e$

Taking modulus and squaring we get our desired result.

14. This is similar to 131 and is left as an exercise.

15. Let $$|z_1| = |z_2| = |z_3| = R.$$ $$\therefore$$ origin is the circumcenter of triangle. Since triangle is also equilateral circumcenter and origin coincide. Therefore, origin is also centroid. Thus

$\frac{z_1 + z_2 + z_3}{3} = 0 \Rightarrow z_1 + z_2 + z_3 = 0$
16. Similar to 133 it can be proven that it is an equilateral triangle. Now since $$|z_1| = |z_2| = |z_3| = 1$$ therefore it is an equilateral triangle inscribed in an unit circle.

17. Circumcenter of an equilateral triangle is given by $$z_0 = \frac{z_1 + z_2 + z_3}{3}$$ which is same as centroid. Now since this triangle is equilateral

$\sum z_1^2 = \sum z_1z_2 (\sum z_1)^2 = \sum z_1^2 + 2\sum z_1z_2 = 3\sum z_1^2$

Also,

$z_0 = \frac{\sum z_1}{3} \Rightarrow \sum z_1 = 3z_0 \Rightarrow 3\sum z_1^2 = 9z_0^2 \Rightarrow \sum z_1^2 = 3z_0^2$
18. Since $$z_1, z_2$$ and origin form an equilateral triangle we have

$z_1^2 + z_2^2 + 0^2 - z_1z_2 - z_2*0 - z_1*0 = 0$

Hence, proven.

19. From 136 $$z_1, z_2$$ and origin will form a triangle if $$z_1^2 + z_2^2 - z_1z_2 = 0.$$ Therefore,

$(z_1 + z_2)^2 = 3z_1z_2 \Rightarrow a^2 = 3b.$
20. Centroid of the triangle is given by $$\frac{z_1 + z_2 + z_3}{3}$$ i.e. $$\frac{-3\alpha}{3}$$ i.e. $$-\alpha.$$ Triangle will be equilateral if

$z_1^2 + z_2^2 + z_3^2 = z_1z_2 + z_2z_3 + z_3z_1 \Rightarrow (z_1 + z_2 + z_3)^2 = 3(z_1z_2 + z_2z_3 + z_3z_1) \Rightarrow 9\alpha^2 = 9\beta \Rightarrow \alpha^2 = \beta$
21. Given,

$z_2 = \frac{z_1 + z_3}{2}$

Clearly, from section formula we can deduce that $$z_2$$ divides line segment joining $$z_1$$ and $$z_3$$ in two equal segments hence the complex numbers are collinear.

22. $$z_3$$ will divide line segment joining $$z_1$$ and $$z_2$$ either externally or internally. Now section formula can be used to prove remaining.

23. We have,

$\frac{z - i}{ z + i}$

as a purely imaginary quantity. Let $$z = x + iy.$$

$\frac{[x + i(y - 1)][x - i(y + 1)])}{x^2 + (y + 1)^2}$

Equating real part to 0 we have

$\Rightarrow x^2 + y^2 - 1 = 0$

Therefore locus of z represents the circle $$x^2 + y^2 = 1.$$

24. $$z$$ represents the ring between the concentric circles whose center is at (3, 4i) having radii 1 and 2.

25. Let $$z = x+ iy.$$ Now we have

$\sqrt{(x - 1)^2 + y^2} + \sqrt{(x + 1)^2 + y^2} \le 4$

Let $$L + M = 4$$

$L^2 - M^2 = -4x \therefore L^2 - M^2 = -x \therefore 4L^2 = (4 - x)^2 4(x^2 + y^2 - 2x + 1) = 16 + x^2 - 8x 3x^2 + 4y^2 = 12 \frac{x^2}{4} + \frac{y^2}{3} = 1$

Hence it represent the above ellipse.

26. Let $$z = x + iy$$ then we have

$x = t + 5 \text{ and } y = \sqrt{4 -t^2} \Rightarrow (x - 5)^2 = t^2 \text{ and } y^2 = 4 -t^2$

Adding we get, $$(x - 5)^2 + y^2 = 4$$ which represents a circle with radius at (5, 0) with radius 2.

27. Given $$\frac{z^2}{z - 1}$$ is real i.e. its imaginary part is zero.

$\frac{(x^2 - y^2 + i2xy)((x - 1) - iy)}{(x - 1)^2 + y^2}$

Equating imaginary part to 0 we have

$x^2 + y^2 - 2x = 0 \therefore (x - 1)^2 + y^ = 1$

which represents a circle having center at (1, 0) and radius unity.

28. Given, $$|z^2 + (-1)| = |z|^2 + |(-1)| \Rightarrow \frac{z^2}{-1}$$ is non-negative real number. Thus $$z$$ is purely imaginary number. Thus locus of z is a straight line.

Question 147 to 149 are left as exercises.

1. Given,

$\log_{\sqrt{3}}\frac{|z|^2 - |z| + 1}{2 + |z|} < 2 \Rightarrow \frac{|z|^2 - |z| + 1}{2 + |z|} < (\sqrt{3})^2 \Rightarrow |z|^2 - 4|z| - 5 < 0 \Rightarrow |z| < 5$