Name: Reverse Triangle Inequality for Complex Numbers
Uploaded: 2019-07-19
Duration: 5 min 1 s
Description: In this interesting video, let us study and prove the triangle inequality associated with the difference of two complex numbers. Happy learning!

Question

Interpret the following locus in

z \in C

.
(i)

1 < |z - 2 i| < 3

(ii)

lm (z) \geq 1

(iii)

Arg (z - 3 - 4 i) = \frac{π}{3}

(iv)

Re (\frac{z + 2 i}{i z + 2}) \leq 4 (z \neq 2 i)

Accepted Answer

$(i)$ Given, $1 < |z - 2 i| < 3$

We know, $|z - a| = r$ represents the circle with radius $Arg (z - 3 - 4 i) = \frac{π}{3}$ .

The region between the concentric circles with centre at $(0, 2)$ & radii $1$ and $3$ units.

$(ii)$ Given, $Im (z) \geq 1$

We know, $Im (z) = 1$ represent the line $y = 1$ .

So, it represents the part of the complex plane, on or above the line $y = 1$ .

$(iii)$ Given, $Arg (z - 3 - 4 i) = \frac{π}{3}$

Assume, $z = x + i y$

So,

$Arg (x + i y - 3 - 4 i) = \frac{π}{3}$

$\Rightarrow \frac{y - 4}{x - 3} = \tan \frac{π}{3}$

$\Rightarrow \frac{y - 4}{x - 3} = \sqrt{3}$

$\Rightarrow y - 4 = x \sqrt{3} - 3 \sqrt{3}$

So, it represents a ray emanating from the point $(3 + 4 i)$ directed away from the origin & having equation, $\sqrt{3} x - y + 4 - 3 \sqrt{3} = 0, x > 3$ .

$(iv)$ Let $z = x + i y$

Given, $Re (\frac{z + 2 i}{i z + 2}) \leq 4$

$\Rightarrow Re (\frac{x + i y + 2 i}{i x - y + 2}) \leq 4$

$\Rightarrow Re (\frac{x + (y + 2) i}{(2 - y) + i x} \times \frac{2 - y - i x}{2 - y - i x})$

$\Rightarrow Re (\frac{(x + (y + 2) i) (2 - y - i x)}{{(2 - y)}^{2} + x^{2}})$

$\Rightarrow Re (\frac{2 x - x y - i x^{2} + 2 i y - y^{2} i + x y + 4 i - 2 i y + 2 x}{{(2 - y)}^{2} + x^{2}})$

$\Rightarrow \frac{y x}{x^{2} + {(2 - y)}^{2}} \leq 4$

$\Rightarrow 4 x \leq 4 x^{2} + 4 (4 + y^{2} - 4 y)$

$\Rightarrow x \leq x^{2} + 4 + y^{2} - 4 y$

$\Rightarrow x^{2} + y^{2} - x - 4 y + 4 \geq 0$

So, it represents Region outside or on the circle with centre $\frac{1}{2} + 2 i$ and radius $\frac{1}{2}$ .

Interpret the following locus in $z \in C$ .
(i) $1 < |z - 2 i| < 3$
(ii) $lm (z) \geq 1$
(iii) $Arg (z - 3 - 4 i) = \frac{π}{3}$
(iv) $Re (\frac{z + 2 i}{i z + 2}) \leq 4 (z \neq 2 i)$

Important Questions on Complex Numbers

Learn and Prepare for any exam you want !

Interpret the following locus in z∈C. (i) 1<z-2i<3 (ii) lmz≥1 (iii) Argz-3-4i=π3 (iv) Rez+2iiz+2≤4z≠2i

Important Questions on Complex Numbers

Learn and Prepare for any exam you want !

Interpret the following locus in $z \in C$ .
(i) $1 < |z - 2 i| < 3$
(ii) $lm (z) \geq 1$
(iii) $Arg (z - 3 - 4 i) = \frac{π}{3}$
(iv) $Re (\frac{z + 2 i}{i z + 2}) \leq 4 (z \neq 2 i)$