Question

Let \[z\] and \[w\] be two complex numbers such that \[w = z\bar z - 2z + 2\], \[\left| {\dfrac{{z + i}}{{z - 3i}}} \right| = 1\] and \[{\mathop{\rm Re}\nolimits} \left( w \right)\] has a minimum value. Then what is the minimum value of \[n \in \mathbb{N}\], for which \[{w^n}\] is real?

Answer 1

Hint: Let \[z = x + iy\], where \[x \in \mathbb{R}\] and \[y \in \mathbb{R}\]. Substituting \[z = x + iy\] in the given expression \[\left| {\dfrac{{z + i}}{{z - 3i}}} \right| = 1\] and simplifying it the value of \[y\] will be found. Put the value of \[y\] in \[z = x + iy\]. Find \[\bar z\] and substituting in \[w = z\bar z - 2z + 2\], find \[w\] and hence find its real part i.e. \[{\mathop{\rm Re}\nolimits} \left( w \right)\]. Find the value of \[x\] for which \[{\mathop{\rm Re}\nolimits} \left( w \right)\] has a minimum value. Then put the value of \[x\] in \[w\] and find its polar form. Find \[{w^n}\] and put \[n = 4\].

Formula Used:
For two complex numbers \[{z_1}\] and \[{z_2}\], \[\left| {\dfrac{{{z_1}}}{{{z_2}}}} \right| = \dfrac{{\left| {{z_1}} \right|}}{{\left| {{z_2}} \right|}}\], provided \[{z_2} \ne 0\]
The modulus of \[z = x + iy\] is \[\left| z \right| = \sqrt {{x^2} + {y^2}} \]
The conjugate of \[z = x + iy\] is \[\bar z = x - iy\]
Polar form of \[z = x + iy\] is \[z = r\left( {\cos \theta + i\sin \theta } \right)\], where \[r = \left| z \right|\] and \[\theta = {\tan ^{ - 1}}\left( {\dfrac{y}{x}} \right)\]
\[\cos \left( { - \theta } \right) = \cos \theta \] and \[\sin \left( { - \theta } \right) = - \sin \theta \]
\[{e^{iy}} = \cos y + i\sin y\]

Complete step-by-step solution:
Let \[z = x + iy\], where \[x \in \mathbb{R}\] and \[y \in \mathbb{R}\]
Given that \[\left| {\dfrac{{z + i}}{{z - 3i}}} \right| = 1\]
Use the property \[\left| {\dfrac{{{z_1}}}{{{z_2}}}} \right| = \dfrac{{\left| {{z_1}} \right|}}{{\left| {{z_2}} \right|}}\]
\[ \Rightarrow \dfrac{{\left| {z + i} \right|}}{{\left| {z - 3i} \right|}} = 1\]
Cross multiply.
\[ \Rightarrow \left| {z + i} \right| = \left| {z - 3i} \right|\]
We assumed \[z = x + iy\]
\[ \Rightarrow \left| {x + iy + i} \right| = \left| {x + iy - 3i} \right|\]
Separate the real part and the imaginary part.
\[ \Rightarrow \left| {x + i\left( {y + 1} \right)} \right| = \left| {x + i\left( {y - 3} \right)} \right|\]
Use the definition \[\left| {x + iy} \right| = \sqrt {{x^2} + {y^2}} \]
\[ \Rightarrow \sqrt {{x^2} + {{\left( {y + 1} \right)}^2}} = \sqrt {{x^2} + {{\left( {y - 3} \right)}^2}} \]
Take square on both sides.
\[ \Rightarrow {x^2} + {\left( {y + 1} \right)^2} = {x^2} + {\left( {y - 3} \right)^2}\]
Cancel the term \[{x^2}\] from both sides.
\[ \Rightarrow {\left( {y + 1} \right)^2} = {\left( {y - 3} \right)^2}\]
Use the identities \[{\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}\] and \[{\left( {a - b} \right)^2} = {a^2} - 2ab + {b^2}\]
\[ \Rightarrow {y^2} + 2y + 1 = {y^2} - 6y + 9\]
Cancel the term \[{y^2}\] from both sides
\[ \Rightarrow 2y + 1 = - 6y + 9\]
\[ \Rightarrow 2y + 6y = 9 - 1\]
\[ \Rightarrow 8y = 8\]
\[ \Rightarrow y = 1\]
\[\therefore z = x + i\]
\[ \Rightarrow \bar z = x - i\]
Given that \[w = z\bar z - 2z + 2\]
Substitute \[z = x + i\] and \[\bar z = x - i\]
\[ \Rightarrow w = \left( {x + i} \right)\left( {x - i} \right) - 2\left( {x + i} \right) + 2\]
Use the identity \[\left( {a + b} \right)\left( {a - b} \right) = {a^2} - {b^2}\]
\[ \Rightarrow w = {x^2} - {i^2} - 2x - 2i + 2\]
Put \[{i^2} = - 1\]
\[ \Rightarrow w = {x^2} + 1 - 2x - 2i + 2\]
\[ \Rightarrow w = {x^2} - 2x + 3 - 2i\]
\[\therefore {\mathop{\rm Re}\nolimits} \left( w \right) = {x^2} - 2x + 3\]
It can be expressed as \[{\mathop{\rm Re}\nolimits} \left( w \right) = {\left( {x - 1} \right)^2} + 2\]
Since a square expression is always non-negative, so the minimum value of \[{\left( {x - 1} \right)^2}\] is \[0\] and hence the minimum value of \[{\mathop{\rm Re}\nolimits} \left( w \right)\] is \[2\], which occurs at \[x = 1\].
At \[x = 1\], \[ \Rightarrow w = 1 - 2 + 3 - 2i = 2 - 2i = 2\left( {1 - i} \right)\]
Let us find the polar form of \[w\].
Multiply and divide by \[\sqrt 2 \]
\[ \Rightarrow w = 2\sqrt 2 \left( {\dfrac{1}{{\sqrt 2 }} - i\dfrac{1}{{\sqrt 2 }}} \right)\]
We know that \[\sin \dfrac{\pi }{4} = \cos \dfrac{\pi }{4} = \dfrac{1}{{\sqrt 2 }}\]
\[ \Rightarrow w = 2\sqrt 2 \left( {\cos \dfrac{\pi }{4} - i\sin \dfrac{\pi }{4}} \right)\]
\[\cos \left( { - \theta } \right) = \cos \theta \] and \[\sin \left( { - \theta } \right) = - \sin \theta \]
\[ \Rightarrow w = 2\sqrt 2 \left\{ {\cos \left( { - \dfrac{\pi }{4}} \right) + i\sin \left( { - \dfrac{\pi }{4}} \right)} \right\}\]
Use the identity \[{e^{iy}} = \cos y + i\sin y\]
\[ \Rightarrow w = 2\sqrt 2 {e^{ - i\dfrac{\pi }{4}}}\]
\[\therefore {w^n} = {\left( {2\sqrt 2 {e^{ - i\dfrac{\pi }{4}}}} \right)^n} = 2\sqrt 2 {e^{ - i\dfrac{\pi }{4}n}}\]
If we take \[n = 4\], then
\[{w^4} = 2\sqrt 2 {e^{ - i\dfrac{\pi }{4} \times 4}} = 2\sqrt 2 {e^{ - i\pi }} = 2\sqrt 2 \left\{ {\cos \left( { - \pi } \right) + i\sin \left( { - \pi } \right)} \right\} = 2\sqrt 2 \left\{ {\cos \left( \pi \right) + i\sin \left( \pi \right)} \right\} = 2\sqrt 2 \left\{ {\left( { - 1} \right) + i\left( 0 \right)} \right\} = - 2\sqrt 2 \], which is real.
Hence the minimum value of \[n\] is \[4\].

Note: Any square expression is always non-negative and hence its minimum value is equal to zero. A complex number will be real whenever it’s imaginary part will be equal to zero. To find conjugate of a complex number, you just need to change the sign of the imaginary part.