10 Most SAQ’s of Complex Numbers Chapter in Inter 2nd Year Maths-2A (TS/AP)
4 Marks
SAQ-1 : Show that the points in the Argand diagram represented by the complex numbers 2+2i, -2-2i, -2√3 + 2√3 i are the vertices of an equilateral triangle
Given complex numbers are taken as $$A(2,2)$$ $$B(-2,-2)$$ $$C(-2\sqrt{3},2\sqrt{3})$$
Now,
$$AB = \sqrt{(2+2)^2 + (2+2)^2} = \sqrt{4^2 + 4^2}$$
$$= \sqrt{16 + 16} = \sqrt{32}$$
$$BC = \sqrt{(-2 + 2\sqrt{3})^2 + (-2-2\sqrt{3})^2}$$
$$= \sqrt{2[(-2)^2 + (-2\sqrt{3})^2]} = \sqrt{2(4 + 12)}$$
$$= \sqrt{2(16)} = \sqrt{32}$$
$$\text{Using the identity } (a+b)^2 + (a-b)^2 = 2(a^2+b^2)$$
$$CA = \sqrt{(2 + 2\sqrt{3})^2 + (2 – 2\sqrt{3})^2}$$
$$= \sqrt{2[(2)^2 + (2\sqrt{3})^2]} = \sqrt{2(4+12)}$$
$$= \sqrt{2(16)} = \sqrt{32}$$
$$AB = BC = CA$$
Hence, $$\triangle ABC$$ is equilateral.
SAQ-2 : Show that the four points in the Argand plane represented by the complex numbers 2+i, 4+3i, 2+5i, 3i are the vertices of a square
Given complex numbers are taken as
$$A(2,1)$$ $$B(4,3)$$ $$C(2,5)$$ $$D(0,3)$$
$$AB = \sqrt{(2-4)^2 + (1-3)^2} = \sqrt{8}$$
$$BC = \sqrt{(4-2)^2 + (3-5)^2} = \sqrt{8}$$
$$CD = \sqrt{(2-0)^2 + (5-3)^2} = \sqrt{8}$$
$$DA = \sqrt{(2-0)^2 + (1-3)^2} = \sqrt{8}$$
$$AC = \sqrt{(2-2)^2 + (1-5)^2} = \sqrt{16} = 4$$
$$BD = \sqrt{(4-0)^2 + (3-3)^2} = \sqrt{16} = 4$$
Hence, the four sides AB, BC, CD, DA are equal.
The two diagonals AC, BD are equal.
A,B,C,D form a square
SAQ-3 : Show that the points in the Argand plane represented by the complex numbers -2+7i, -3/2 + 1/2 i, 4-3i, 7/2(1+i) are the vertices of a rhombus
Given complex numbers are taken as
$$A(2,7)$$ $$B(-\frac{3}{2},\frac{1}{2})$$ $$C(4,-3)$$ $$D(\frac{7}{2},\frac{7}{2})$$
$$AB = \sqrt{(-2 + \frac{3}{2})^2 + (7 – \frac{1}{2})^2}$$
$$= \sqrt{(-\frac{1}{2})^2 + (\frac{13}{2})^2}$$
$$= \sqrt{\frac{1}{4} + \frac{169}{4}} = \sqrt{\frac{170}{4}}$$
$$BC = \sqrt{(4 + \frac{3}{2})^2 + (-3 – \frac{1}{2})^2}$$
$$= \sqrt{(\frac{11}{2})^2 + (-\frac{7}{2})^2}$$
$$= \sqrt{\frac{121}{4} + \frac{49}{4}} = \sqrt{\frac{170}{4}}$$
$$CD = \sqrt{(4 – \frac{7}{2})^2 + (-3 -\frac{7}{2})^2}$$
$$= \sqrt{(\frac{1}{2})^2 + (-\frac{13}{2})^2}$$
$$= \sqrt{\frac{1}{4} + \frac{169}{4}} = \sqrt{\frac{170}{4}}$$
$$DA = \sqrt{(\frac{7}{2} + 2)^2 + (\frac{7}{2} – 7)^2}$$
$$= \sqrt{(\frac{11}{2})^2 + (-\frac{7}{2})^2}$$
$$= \sqrt{\frac{121}{4} + \frac{49}{4}} = \sqrt{\frac{170}{4}}$$
$$AC = \sqrt{(4+2)^2 + (-3-7)^2}$$
$$= \sqrt{6^2 + (-10)^2}$$
$$= \sqrt{36+100} = \sqrt{136}$$
$$BD = \sqrt{(\frac{7}{2} + \frac{3}{2})^2 + (\frac{7}{2} – \frac{1}{2})^2}$$
$$= \sqrt{(\frac{10}{2})^2 + (\frac{6}{2})^2}$$
$$= \sqrt{\frac{100}{4} + \frac{36}{4}} = \sqrt{\frac{136}{4}}$$
Hence, the four sides AB, BC, CD, DA are equal.
The two diagonals AC, BD are unequal
A, B, C, D form a Rhombus
SAQ-4 : If (x-iy)1∕3 = a – ib then show that x/a + y/b = 4(a2 – b2)
Given that $$(x-iy)^{\frac{1}{3}} = a-ib$$ $$x-iy = (a-ib)^3$$
$$\Rightarrow x-iy = a^3 – 3a^2bi + 3a(i^2)b^2 – i^3b^3$$
$$= a^3 – 3a^2bi – 3ab^2 + ib^3$$
$$= (a^3 – 3ab^2) – i(3a^2b – b^3)$$
Equating real parts on both sides, we get $$x = a^3 – 3ab^2 = a(a^2 – 3b^2)$$
$$\Rightarrow \frac{x}{a} = a^2 – 3b^2$$
Equating imaginary parts on both sides, we get $$y = 3a^2b – b^3 = b(3a^2 – b^2)$$
$$\Rightarrow \frac{y}{b} = 3a^2 – b^2$$
$$\frac{x}{a} + \frac{y}{b} = (a^2 – 3b^2) + (3a^2 – b^2) = 4a^2 – 4b^2 = 4(a^2 – b^2)$$
SAQ-5 : Show that z1 = (2+11i)/25, z2 = (-2+i)/(1-2i)2 are conjugate to each other
Given $$z^2 = \frac{-2+i}{(1-2i)^2}$$
$$= \frac{-2+i}{1-4i+4i^2} = \frac{-(2-i)}{1-4i+4(-1)} = \frac{-(2-i)}{-(3+4i)}$$
$$= \frac{2-i}{3+4i} = \frac{(2-i)(3-4i)}{(3+4i)(3-4i)} = \frac{2\cdot3 – 2\cdot4i – i\cdot3 + i\cdot4i}{3^2+(4i)^2}$$
$$= \frac{6 – 8i – 3i + 4}{9 + 16} = \frac{10 – 11i}{25}$$
$$= \frac{2}{25} – \frac{11i}{25}$$
The conjugate of $$\frac{2}{25} – \frac{11i}{25}$$ is $$\frac{2}{25} + \frac{11i}{25} = \frac{2 + 11i}{25} = z_1$$
This demonstrates that $$\frac{2 – i}{(1 – 2i)^2}$$ and its conjugate $$\frac{2 + 11i}{25}$$ are indeed conjugates to each other
SAQ-6 : If z = 3 – 5i then show that z3 – 10z2 + 58z – 136 = 0
Given $$z = 3 – 5i$$
$$\Rightarrow z-3 = -5i$$
$$\Rightarrow (z-3)^2 = (5i)^2$$
$$\Rightarrow z^2 – 6z + 9 = -25$$
$$\Rightarrow z^2 – 6z + 34 = 0 \quad \text{(1)}$$
Now, for $$z^3 – 10z^2 + 58z – 136$$
Given the hint and the approach, we break it down as follows:
$$z^3 – 10z^2 + 58z – 136 = z(z^2 – 6z + 34) – 4(z^2 – 6z + 34)$$
$$= (z – 4)(z^2 – 6z + 34)$$
Given from (1) that $$z^2 – 6z + 34 = 0$$
$$= (z – 4)(0) = 0$$
Therefore, it is shown that if $$z = 2 – i\sqrt{7}$$ then $$3z^3 – 4z^2 + z + 88 = 0$$
SAQ-7 : If x + iy = 1/1+ cosθ + isinθ then, show that 4x2 – 1 = 0
Given $$x + iy = \frac{1}{1 + \cos\theta} + i\sin\theta = \frac{1}{2\cos^2 \frac{\theta}{2}} + i\left(2\sin\frac{\theta}{2} \cos\frac{\theta}{2}\right)$$
$$\Rightarrow x + iy = \frac{1}{2\cos \frac{\theta}{2}}\left(\cos \frac{\theta}{2} + i\sin \frac{\theta}{2}\right) = \frac{\cos \frac{\theta}{2} – i\sin \frac{\theta}{2}}{2\cos \frac{\theta}{2}}\left(\cos \frac{\theta}{2} + i\sin \frac{\theta}{2}\right)\left(\cos \frac{\theta}{2} – i\sin \frac{\theta}{2}\right)$$
$$= \frac{\cos \frac{\theta}{2} – i\sin \frac{\theta}{2}}{2\cos \frac{\theta}{2}}\left(\cos^2 \frac{\theta}{2} + \sin^2 \frac{\theta}{2}\right)$$
$$= \frac{\cos \frac{\theta}{2} – i\sin \frac{\theta}{2}}{2\cos \frac{\theta}{2}}(1)$$
$$= \frac{1}{2} – \frac{i\sin \frac{\theta}{2}}{2\cos \frac{\theta}{2}}$$
$$= \frac{1}{2} – \frac{i\tan \frac{\theta}{2}}{2}$$
Equating the real parts, we get:
$$x = \frac{1}{2}$$
$$\Rightarrow 2x = 1$$
$$\Rightarrow (2x)^2 = 1^2$$
$$\Rightarrow 4x^2 = 1$$
$$\Rightarrow 4x^2 – 1 = 0$$
Hence, the equation derived from the given transformation is $$4x^2 – 1 = 0$$
SAQ-8 : If the point P denotes the complex number z = x + iy in the Argand plane and if (z-i)/(z-1) is a purely imaginary number, find the locus of P
Given $$z = x + iy$$
$$\frac{z – i}{z – 1} = \frac{(x + iy) – i}{(x + iy) – 1} = \frac{x + i(y – 1)}{(x – 1) + iy}$$
By multiplying both the numerator and the denominator by the conjugate of the denominator:
$$= \frac{[x + i(y – 1)][(x – 1) – iy]}{[(x – 1) + iy][(x – 1) – iy]}$$
$$= \frac{x(x – 1) – ixy + i(x – 1)(y – 1) + y(y – 1)}{(x – 1)^2 + y^2} = \frac{x^2 – x + xy – xy + i(x – 1)(y – 1) – ixy + y^2 – y}{(x – 1)^2 + y^2}$$
$$= \frac{x^2 + y^2 – x – y}{(x – 1)^2 + y^2} + i\left[\frac{(x – 1)(y – 1) – xy}{(x – 1)^2 + y^2}\right]$$
For $$\frac{z – i}{z – 1}$$ to be purely imaginary, the real part must be zero:
$$\frac{x^2 + y^2 – x – y}{(x – 1)^2 + y^2} = 0$$
$$x^2 + y^2 – x – y = 0$$
SAQ-9 : If the real part of (z+1)/(z+i) is 1, find the locus of z
Given $$z = x + iy$$
$$\frac{z + 1}{z + i} = \frac{(x + iy) + 1}{(x + iy) + i}$$
$$= \frac{(x + 1) + iy}{x + i(y + 1)} \cdot \frac{x – i(y + 1)}{x – i(y + 1)}$$
$$= \frac{[(x + 1) + iy][x – i(y + 1)]}{[x + i(y + 1)][x – i(y + 1)]}$$
$$= \frac{x(x + 1) – i(x + 1)(y + 1) + ixy + y(y + 1)}{x^2 + (y + 1)^2}$$
$$= \frac{x^2 + x + xy – xy – i(x + 1)(y + 1) + ixy + y^2 + y}{x^2 + (y + 1)^2}$$
$$= \frac{x^2 + y^2 + x + y}{x^2 + (y + 1)^2} + i \frac{xy – (x + 1)(y + 1) + xy}{x^2 + (y + 1)^2}$$
Given the real part is 1,
$$\frac{x^2 + y^2 + x + y}{x^2 + (y + 1)^2} = 1$$
$$x^2 + y^2 + x + y = x^2 + y^2 + 2y + 1$$
Therefore, the locus of z is
$$x – y = 1$$
SAQ-10 : Find the real values of x and y if (x-1)/(3+i) + (y-1)/(3-i) = i
Given: $$\frac{x-1}{3+i} + \frac{y-1}{3-i} = i$$
$$\frac{(x-1)(3-i)+(y-1)(3+i)}{(3+i)(3-i)} = i$$
$$\frac{(3x-xi-3+i)+(3y+yi-3-i)}{3^2+1^2} = i$$
$$\frac{(3x-3+3y-3)+(i-xi+yi-i)}{10} = i$$
$$\frac{3x + 3y – 6}{10} + \frac{i(1 – x + y – 1)}{10} = 0 + i$$
$$3x + 3y – 6 = 0 \quad \Rightarrow \quad x + y = 2 \quad \text{…(1)}$$
$$y – x = 10 \quad \Rightarrow \quad x – y = -10 \quad \text{…(2)}$$
$$1 – x + y – 1 = 10 \quad \Rightarrow \quad -x + y = 10 \quad \text{…(2)}$$
$$2y = 12 \quad \Rightarrow \quad y = 6$$
$$x + 6 = 2 \quad \Rightarrow \quad x = -4$$
$$x = -4, \quad y = 6$$