Examples On Geometrical Interpretation Of Complex Equations

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Example - 19

Plot the region/locus represented by \(z\) if \(z\) satisfies:

(a)  \(\left| {z - 1} \right| = \left| {z + 1} \right|\)

(b)  \(\left| {z - 1 + i} \right| = \left| {z + 1 - i} \right|\)         

(c)  \(2 < {\mathop{\rm Re}\nolimits} (z) < 3;\,\,\,2 < {\mathop{\rm Im}\nolimits} (z) < 3\)

(d) \(\left| {z - i} \right| + \left| {z + i} \right| = 3\)

Solution: (a) As mentioned earlier, we must look for a  geometrical  interpretations of equations involving complex numbers. This equation, in particular, says that the distance of \(z\) from 1 (you must learn to view every number as a point on the complex plane; for example, 1 is a point which lies on the real axis!) must be equal to the distance of \(z\) from –1, because

\[\left| {z - 1} \right| = \left| {z - ( - 1)} \right|\]

From plane geometry, \(z\) must lie on the perpendicular bisector of 1 and –1, or equivalently \(z\) must lie anywhere on the imaginary axis.

(b)          As in part (a), this equation, which can be written as

\[\left| {z - (1 - i)} \right| = \left| {z - ( - 1 + i)} \right|\]

means that \(z\) is equidistant from \(1–i\) and \(–1+i,\) i.e., \(z\) lies on the perpendicular bisector of these two points:

 (c) Since \(2 < {\mathop{\rm Re}\nolimits} \left( z \right) < 3,\) \(z\) lies anywhere in the region between the vertical lines \(x = 2\) and \(x = 3.\)  Also, since \(2 < {\mathop{\rm Im}\nolimits} \left( z \right) < 3,\,\,\,z\) must also lie in the region between the horizontal lines \( y = 2\) and \(y = 3.\)

 (d) To solve this part, a little knowledge of co-ordinate geometry would be helpful. The given equation says that the sum of the distances of \(z\) from \( i \) and from \(–i\) must equal 3. \(z\) would therefore trace out an elliptical path in the plane with \(i\) and \(–i\) as its two foci, as shown in the figure below:

For those not conversant with co-ordinate geometry, here’s an explanation. Suppose that you fix two pegs at the points \(i\) and \(–i\) and tie a 3-unit long string between the two pegs.

Now, with a pen, pull this string to “away” from the pegs so that it becomes taut, and then, keeping the string taut, trace out a complete revolution on the plane with the tip of the pen (the taut string will automatically guide the pen)

 Example - 20

Plot the locus of \(z\) if \(z\) satisfies \(\begin{align}\left| {\arg \left( {\frac{{z - 1}}{{z + 1}}} \right)} \right| = \frac{\pi }{2}\end{align}\)

Solution: We have,

\[\begin{align}{}&\arg \left( {\frac{{z - 1}}{{z + 1}}} \right) =  \pm \frac{\pi }{2}\\ \Rightarrow \qquad &\arg \left( {z - 1} \right) - \arg \left( {z + 1} \right) =  \pm \frac{\pi }{2}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...{\rm{ }}\left( 1 \right)\end{align}\]

 \( z – 1\) is the vector drawn from the point 1 to the point \(z.\) Similarly, \(z + 1\) is the Vectron from –1 to \(z.\) The angle between these two vectors, as (1) tells us, is\(\begin{align} \pm \frac{\pi }{2}\end{align}\)   

Since the angle in a semicircle is a right angle, \(z\) can lie anywhere on a circle with 1 and –1 as the  end-points of a diameter. (1 and –1 themselves cannot lie on this circle because either \(z – 1\) or \(z + 1\) becomes a zero vector if \(z = 1\) or \(–1,\) and the argument of a zero vector cannot be uniquely defined)

TRY YOURSELF - V

Q. 1 Plot the locus / region represented by \(z\) in the following cases:

(a) \(|z|\; \in \;[1,\;2]\; \cup [3,\;4]\)             

(b) \(\begin{align}|z|\; < 3,\;\frac{\pi }{6} < \;{\rm{Arg}}(z) < \frac{\pi }{3}\end{align}\)

            
(c) \(\begin{align}1\, < \,|z|\; < \,2,\;\frac{\pi }{6}\; < \;|{\rm{Arg}}(z)|\; < \;\frac{\pi }{3}\end{align}\)             

(d) \(|z|\; < 2,\;1 < \;|{\mathop{\rm Re}\nolimits} (z)|\; < 2,\;1 < \;|{\mathop{\rm Im}\nolimits} (z)|\; < 2\)

            
(e) \(|z - 1{|^2} + |z + 1{|^2} = 4\)             

(f) \(|z - i{|^2} + |z + i{|^2} = 4\)

            
(g) \(|z - 2 + 2i|\; = \;|z + 2 + 2i|\)             

(h) \({\mathop{\rm Re}\nolimits} (z) + {\mathop{\rm Im}\nolimits} (z) = 1\)

            
(i) \(|{\mathop{\rm Re}\nolimits} (z)| + |{\mathop{\rm Im}\nolimits} (z)|\; \le 1\)             

(j) \(|z - 2|\; + \;|z + 2i|\; = 5\)

            
(k) \(\begin{align}\arg \left( {\frac{{z - 1}}{{z + 1}}} \right) = \frac{\pi }{6}\end{align}\)             

(l) \(\begin{align}|z|\; < \;3,\;|z - 4i|\; < \;2,\;\;\arg (z) < \frac{\pi }{2}\end{align}\)

            
(m) \(\sin |z|\; > 0\)             

(n) \(|z|\; < \;1,\;\;\tan (\arg (z))\; > 1\)

            
(o) \(|z{|^3} - 6|z{|^2} + 11|z| - 6 \ge 0,\) \(|z{|^2} - 5|z| + \;6 \le 0\)