Complex Numbers and Quadratic Equations Test -9

Let z = sin x + i cos 2x

\(\bar z\)= sin x – i cos 2x

But we are given that \(\bar z\) =cos x – i sin 2x

∴ sin x – i cos 2x = cos x – i sin 2x

Comparing the real and imaginary part, we get sin x = cos x and cos 2x = sin 2x

⇒ tan x = 1 and tan 2x = 1

\(tan x = tan \frac{\pi}{4} \) and \(tan 2x = tan \frac{\pi}{4}\)

\(\therefore x = n\pi + \frac{\pi}{4}, n \in I \) and \(2x = n\pi + \frac{\pi}{4}\)

⇒ x = 2x ⇒ 2x – x = 0 ⇒ x = 0

Let \(z = \frac{1 - isin \alpha}{1 + 2i sin \alpha} = \frac{(1 - i \,sin \alpha)(1 - 2 i sin\alpha)}{(1 + 2i sin \alpha)(1 - 2i sin \alpha)}\)

\(= \frac{1 - 2i sin \alpha - i sin \alpha + 2i^2 sin^2 \alpha}{(1)^2 - (2 i sin \alpha)^2}\)

\(= \frac{1 - 3i sin \alpha - 2 sin^2 \alpha}{1 - 4i^2sin^2 \alpha}\)

\(= \frac{(1 - 2 sin^2 \alpha) - 3i sin \alpha}{1 + 4 sin^2 \alpha}\)

\(= \frac{1 - 2 sin^2 \alpha}{1 + 4 sin^2\alpha} - \frac{3 sin \alpha}{1 + 4 sin^2 \alpha}i\)

So, z is purely real, then

\(\frac{-3 sin \alpha}{1 + 4sin^2 \alpha} = 0\)

⇒ sin α = 0

So, α = nπ, n \(\in\) N

Let z = x + yi

|z| = |x + yi| and \(|z|^2 = |x + yi| ^2\)

⇒ \(|z| ^2 = x^ 2 + y ^2\) ….(i)

\(Now \, z ^2 = x ^2 + y ^2 i ^2 + 2xyi\)

\(z ^2 = x ^2 – y ^2 + 2xyi\)

\(|z^2| = \sqrt{(x^2 - y^2)^2 + (2xy)^2}\)

\(= \sqrt{x^4 + y^4 - 2x^2y^2 + 4x^2y^2}\)

\(= \sqrt{x^4 + y^4 + 2x^2y^2} \)

\(=\sqrt{(x^2 + y^2)^2}\)

\(So \,|z| ^2 = x ^2 + y ^2 = |z| ^2\)

so, \(|z| ^2 = |z ^2 |\)

Given that: z = x + iy

If z lies in third quadrant.

So x < 0 and y < 0.

\(\bar z\) = x - iy

\(\frac{\bar z}{z} = \frac{x - iy}{x + iy} = \frac{x - iy}{x + iy} \times \frac{x - iy}{x - iy} \)

\( = \frac{x^2 + i^2y^2 - 2xyi}{x^2 - i^2y^2} = \frac{x^2 - y^2-2xyi}{x^2 + y^2}\)

= \(\frac{x^2 - y^2}{x^2 + y^2} - \frac{2xy}{x^2 + y^2} i\)

When z lies in third quadrant then \(\frac{\bar z}{z}\) will also be lie in third quadrant

\(\therefore \frac{x^2 - y^2}{x^2 + y^2} < 0 \) and \(\frac{-2xy}{x^2 + y^2}\) < 0

⇒ \(x^2 – y^2\) < 0 and 2xy > 0

⇒ \(x^2 – y^2\) <0 and xy > 0

So x < y < 0.

Given that: (z + 3) (\(\bar z\) + 3)

Let z = x + yi

So (z + 3) (\(\bar z\) + 3) = (x + yi + 3)(x – yi + 3)

= [(x + 3) + yi][(x + 3) – yi]

\(= (x + 3)^2 – y^ 2 i^ 2 = (x + 3) ^2 + y^ 2\)

\(= |x + 3 + iy| ^2 = |z + 3| ^2\)

Given that: \((\frac{1 + i}{1 - i})^x = 1\)

⇒ \((\frac{(1 + i)(1 + i)}{(1 - i)(1 + i)})^x = 1\)

⇒ \((\frac{1 + i^2 + 2i}{1 + i^2})^x = 1\)

⇒ \((\frac{1 - 1 + 2i}{1 + 1})^x = 1\)

⇒ \((\frac{2i}{2})^x = 1\)

⇒ \((i)^x =1 \implies x=4n, n \in N\)

Given that: \((\frac{3 - 4ix}{3 + 4ix})\)= \(\alpha - i \beta\)

\( \implies (\frac{3 - 4ix}{3 + 4ix} \times \frac{3 - 4ix}{3 - 4ix}) = \alpha - i\beta \)

\(\implies \frac{9 - 12ix - 12ix + 16i^2x^2}{9 - 16i^2x^2} = \alpha - i\beta \)

\(\implies \frac{9 - 24ix - 16x^2}{9 + 16x^2} = \alpha - i \beta\)

\(\implies \frac{9 - 16x^2}{9 + 16x^2} - \frac{24x}{9 + 16x^2} i = \alpha - i\beta \dots{(i)} \)

\(\implies \frac{9 - 16x^2}{9 + 16x^2} + \frac{24x}{9 + 16x^2} i = \alpha + i\beta \dots{(ii)}\)

Multiplying eqn. (i) and (ii) we get

\((\frac{9 - 16x^2}{9 + 16x^2})^2 + (\frac{24x}{9 + 16x^2})^2 = \alpha^2 + \beta^2 \)

\(\implies (\frac{(9 - 16x^2)^2 + (24x)^2}{(9 + 16x^2)^2} = \alpha^2 + \beta^2\)

\(\implies \frac{81 + 256 x^4 - 288 x^2 + 576x^2}{(9 + 16x^2)^2} = \alpha^2 + \beta^2 \)

\(\implies \frac{81 + 256x^4 + 288x^2}{(9 + 16x^2)^2} = \alpha^2 + \beta^2\)

\(\implies \frac{(9 + 16x^2)^2}{(9 + 16x^2)^2} = \alpha^2 + \beta^2\)

\(so, \alpha^2 + \beta^2 = 1\)

Let \(z_1 = r_1 (cos \theta_1 + i sin \theta_2 )\)

\(\therefore |z_1| = r_1\)

\(and \,z_2 = r_2 (cos \theta_1 + I sin \theta_2)\)

\(\therefore |z_2| = r_2\)

\(z_1z_2 = r_1 (cos \theta_1 + i sin \theta_1 ). r_2 (cos \theta_2 + i sin \theta_2 ) \)

\(= r_1 r_2 (cos \theta_1 + i sin \theta_1 ). (cos \theta_2 + i sin \theta_2 ) \)

\(= r_1 r_2 (cos \theta_1 cos \theta_2 + i sin \theta_2 cos \theta_1\) \(+ i sin \theta_1 cos \theta_2 + i^2 sin \theta_1 sin \theta_2 )\)

\(= r_1 r_2 [(cos \theta_1 cos \theta_2 – sin \theta_1 sin \theta_2 ) \) \(+ i(sin \theta_1 cos \theta_2 + cos \theta_1 sin \theta_2 )]\)

\(=r_1 r_2 [cos (\theta_1 + \theta_2 ) + i sin (\theta_1 + \theta_2 )]\)

\(\therefore |z_1z_2| = |z_1||z_2|\)

Given that: z = 2 – i

If z rotated through an angle of \(\frac{\pi}{2}\) about the origin in clockwise direction.

Then the new position = z.\(e^{– (\pi/2)}\)

= (2 – i) \(e^{– (\pi/2)}\)

\(= (2 - i)[cos(\frac{-\pi}{2}) + i sin (\frac{-\pi}{2})]\)

= (2 – i) (0 – i) = - 1 – 2i

x + yi is a non-real complex number if y ≠ 0. If x, y ∈ R.

Given that: a + ib = c + id

⇒ |a + ib| = |c + id|

⇒ \(\sqrt{a^2 + b^2} = \sqrt{c^2 + d^2}\)

Squaring both sides, we get

\(a ^2 + b^ 2 = c ^2 + d^ 2\)

Given that: \(|\frac{i + z}{i - z}| = 1\)

Let z = x + yi

\(\therefore |\frac{i + x + yi}{i - x - yi}| = 1 \\ \implies |\frac{x + (y + 1)i}{-x - (y - 1)i}| = 1\)

⇒ |x + (y + 1)i| = | - x – (y – 1)i|

⇒ \(\sqrt{x^2 + (y + 1)^2} = \sqrt{x^2 +(y -1)^2}\)

\(x ^2 + (y + 1) ^2 = x ^2 + (y – 1) ^2 \)

\(\implies(y + 1) ^2 = (y – 1) ^2\)

\(\implies y ^2 + 2y + 1 = y ^2 – 2y + 1 \)

\(\implies 2y = - 2y\)

⇒ 4y = 0 ⇒ y = 0 ⇒ x-axis.

\(Let \,z_1 = r_1 (cos \theta_1 + i sin \theta_1 ) \,and \) \(\,z_2 = r_2 (cos \theta_2 + i sin \theta_2 )\)

\(Since\, |z_1 + z_2| = |z_1| + |z_2|\)

\(|z_1 + z_2| = r_1 cos \theta_1+ i r_1 sin \theta_1 + r_2 cos\theta_2 + i r_2 sin \theta_2\)

\(|z_1 + z_2| \) \(= \sqrt{r_1^2 cos^2 \theta_1 + r_2^2 cos^2 \theta_2 + 2r_1r_2cos\theta_1 \\ cos\theta_2 + r_1^2sin^2 \theta_1 + r_2^2 sin^2 \theta_2 + 2r_1r_2 \\ sin \theta_1 sin \theta_2}\)

\(= \sqrt{r_1^2 + r_2^2 + 2r_1r_2 \,cos(\theta_1 - \theta_2)}\)

But \( |z_1 + z_2| = |z_1| + |z_2|\)

So, \(\sqrt{r_1^2 + r_2^2 + 2r_1r_2 \,cos(\theta_1 - \theta_2)} \)

\(= r_1 + r_2\)

Squaring both sides, we get

\(r_1^2+r_2^2 + 2r_1r_2 \,cos(\theta_1 - \theta_2) = r_1^2 + r_2^2 + 2r_1r_2\)

\(\implies 2r_1 r_2 – 2r_1 r_2 \,cos (\theta_1 – \theta_2 ) = 0\)

\(\implies 1 – cos (\theta_1 – \theta_2 ) = 0 \)

\(\implies cos (\theta_1 – \theta_2 ) = 1\)

\(\implies \theta_1 – \theta_2 = 0 \implies \theta_1 = \theta_2\)

So, arg \((z _1)\) = arg \((z _2)\)

Let \(z = \frac{1 + i cos \theta}{1 - 2i cos \theta} = \frac{1 + i cos \theta}{1 - 2 i cos \theta}\times \frac{1 + 2i cos \theta}{1 + 2 i cos \theta}\)

\( = \frac{1 + 2 i cos \theta + i cos \theta + 2i^2 cos^2 \theta}{1 - 4i^2 cos^2 \theta}\)

\(= \frac{1 + 3 i cos \theta - 2 cos^2 \theta}{1 + 4 cos^2 \theta}\)

\(= \frac{1 - 2 cos^2 \theta}{1 + 4 cos^2 \theta} + \frac{3 cos \theta}{1 + 4 cos^2 \theta} i\)

If z is real number, then

\(\frac{3 cos \theta}{1 + 4 cos^2 \theta} = 0\)

⇒ 3 cos θ = 0 ⇒ cos θ = 0

\(\therefore \theta = (2n + 1)\frac{\pi}{2}, n \in N\)

Let z = - x + 0i and x < 0

\(\therefore |z| = \sqrt{(-1)^2 + (0)^2} = 1, x < 0\)

Since, the point (- x, 0) lies on the negative side of the real axis (\(\therefore\) x < 0).

\(\therefore\) principal argument (z) = π