Complex Numbers Set 2

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Example- 5

Prove that if \(p \in \mathbb{Z}\) the sum of the pth powers of the nth roots of unity is 0, unless p is a multiple of n. What is the sum in that case?

Solution:   Let the nth roots of unity be

    \(1,\alpha ,{\alpha ^2}..........{\alpha ^{n - 1}}\) where \(\alpha  = {e^{i2\pi {\rm{/}}n}}\)

The sum of the pth powers of these roots is

\[\begin{array}{l}{S_p} = {(1)^p} + {(\alpha )^p} + {({\alpha ^2})^p} + ....... + {({\alpha ^{n - 1}})^p}\\\,\,\,\,\,\,\, = 1 + {\alpha ^p} + {({\alpha ^p})^2} + ....... + {({\alpha ^p})^{n - 1}}     \ldots (1)\end{array}\]

This is a G.P. with common ratio   \({\alpha ^p}\)

\[ \Rightarrow {S_p} = \frac{{{{({\alpha ^p})}^n} - 1}}{{{\alpha ^p} - 1}} \;\;\;\;\;    \ldots (2)\]

If p is not a multiple of n, \({\alpha ^p} \ne 1\) so that the expression for\({S_p}\)is defined in (2). The numerator \({S_p}\) of \({({\alpha ^p})^n} - 1 = {({\alpha ^n})^p} - 1\,\,\, = \,\,\,\,\,{1^p} - 1\,\,\, = \,\,\,0.\) is  Thus, \({S_p} = 0.\)

Suppose now that p is a multiple of n. In that case, \({\alpha ^p} = 1,\) so that\({S_p}\) is directly obtainable from (1).

\[\begin{array}{l}{S_p} = 1 + 1 + 1 + ....... + 1\,\,\,(n\,\,{\rm{times}})\\\,\,\,\,\,\,\, = n\end{array}\]

Thus,

\[{S_p} = \left\{ \begin{array}{l}0,\;\;\; {\rm{if\; }}p\;{\rm{ is\; not\; a \;multiple\; of }}n\\n{\rm{,\;\;\;\;}} {\rm{if }}p\;{\rm{ is\; a \;multiple \;of }}n\end{array} \right\}\]

Example- 6

If \(\begin{align}\arg ({z^{1{\rm{/3}}}}) = \frac{1}{2}\arg ({z^2} + \bar z{z^{{\rm{1/3}}}}),\end{align}\) find \(\left| z \right|\). (z is a non real complex number).

Solution: Since the given relation contains only arguments, we can use the properties that arguments satisfy, to simplify this relation:

\[\begin{align}& 2\arg ({z^{1{\rm{/3}}}}) = \arg ({z^2} + \bar z{z^{1{\rm{/3}}}})\\\\ &\Rightarrow \arg ({z^{2{\rm{/3}}}}) = \arg ({z^2} + \bar z{z^{1{\rm{/3}}}})\\\\ &\Rightarrow \arg ({z^2} + \bar z{z^{1{\rm{/3}}}}) - \arg ({z^{2{\rm{/3}}}}) = 0\\\\ &\Rightarrow \arg \left( {\frac{{{z^2} + \bar z{z^{1{\rm{/3}}}}}}{{{z^{2{\rm{/3}}}}}}} \right) = 0\\\\ &\Rightarrow \arg \left( {{z^{4{\rm{/3}}}} + \frac{{\bar z}}{{{z^{1{\rm{/3}}}}}}} \right) = 0\\\\ &\Rightarrow {z^{{\rm{4/3}}}} + \frac{{\bar z}}{{{z^{{\rm{1/3}}}}}} = {{\bar z}^{{\rm{4/3}}}} + \frac{z}{{{{\bar z}^{{\rm{1/3}}}}}}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\left( \begin{array}{l}{\rm{because\ if\; arg (}}z{\rm{) = 0,}}\\z{\rm{\; is \;purely\; real }} \Rightarrow {\rm{ }}z\,{\rm{ = }}\bar z\end{array} \right)\end{align}\]

\[\begin{align}& \Rightarrow  {{\bar z}^{{\rm{1/3}}}}({z^{{\rm{5/3}}}} + \bar z) = {z^{{\rm{1/3}}}}({{\bar z}^{{\rm{5/3}}}} + z)\\\\ &\Rightarrow  {{\bar z}^{{\rm{1/3}}}}{z^{{\rm{1/3}}}}{z^{{\rm{4/3}}}} + {{\bar z}^{{\rm{4/3}}}} = {z^{{\rm{1/3}}}}{{\bar z}^{{\rm{1/3}}}}{{\bar z}^{{\rm{4/3}}}} + {z^{{\rm{4/3}}}}\\\\  &\Rightarrow  |z{|^{{\rm{2/3}}}}{z^{{\rm{4/3}}}} + {{\bar z}^{{\rm{4/3}}}} = |z{|^{{\rm{2/3}}}}{{\bar z}^{{\rm{4/3}}}} + {z^{{\rm{4/3}}}}\\\\  &\Rightarrow  {z^{{\rm{4/3}}}}(1 - |z{|^{{\rm{2/3}}}}) - {{\bar z}^{{\rm{4/3}}}}(1 - |z{|^{{\rm{2/3}}}}) = 0\\\\  &\Rightarrow  ({z^{{\rm{4/3}}}} - {{\bar z}^{{\rm{4/3}}}})(1 - |z{|^{{\rm{2/3}}}}) = 0\end{align}\]

Since z is a non-real complex number, \(z \ne \bar z\)

                                          \[ \Rightarrow  {z^{{\rm{4/3}}}} \ne {\bar z^{{\rm{4/3}}}}.\]

Therefore,

\[\begin{align}{}  &{\left| z \right|^{{\rm{2/3}}}} = 1\\ \Rightarrow \quad  &|z|\, = 1.\end{align}\]

Example- 7

If \(|z - 25i|\, \le 15,\) find \(\max (\arg (z))\) and \(\min (\arg (z)).\)

Solution:  From the given relation, it is clear that z must lie inside (or on) a circle of radius 15 centred at 25i. To obtain \(\max (\arg (z))\) and \(\min (\arg ((z)),\) what we can do is draw two tangents to the circle from the origin:

Now, \(OP = 25, AP = 15\)

\[ \Rightarrow  \angle POA = {\sin ^{ - 1}}\left( {\frac{{15}}{{25}}} \right) = {\sin ^{ - 1}}\left( {\frac{3}{5}} \right) = \angle POB\]

Therefore,  \(\begin{align}\arg ({z_1}) = \frac{\pi }{2} - {\sin ^{ - 1}}\frac{3}{5}\end{align}\)

                    and     \(\begin{align}\arg ({z_2}) = \frac{\pi }{2} + {\sin ^{ - 1}}\frac{3}{5}\end{align}\)

These are the minimum and maximum values respectively.

Example- 8

If \(\sin \alpha  + 2\sin \beta  + 3\sin \gamma  = 0\) and \(\cos \gamma + 2\cos \beta + 3\cos \gamma = 0,\) find:

(a)  \(\cos 3\alpha  + 8\cos 3\beta  + 27\cos 3\gamma \)

(b) \(\sin (\beta  + \gamma ) + 2\sin (\alpha  + \gamma ) + 3\sin (\alpha  + \beta )\)

Solution:   Construct three complex number \({z_1},{z_2}\) and \({z_3},\) such that

\({z_1} = \cos \alpha  + i\sin \alpha ;\,\,\,\,\,{z_2} = \cos \beta  + i\sin \beta ;\,\,\,\,\,\,{z_3} = \cos \gamma  + i\sin \gamma \)

Note that \(|{z_1}|\, = \,|{z_2}|\, = \,|{z_3}|\, = 1\)

Also, from the given relations,

     \[{z_1} + 2{z_2} + 3{z_3} = 0   \ldots (1)\]

          and                                              \(\begin{align}\frac{1}{{{z_1}}} + \frac{2}{{{z_2}}} + \frac{3}{{{z_3}}} = 0   \ldots (2) \left( {{\rm{because}}\,\,{z_i} = \frac{1}{{{z_i}}};i = 1,\,\,2,\,\,3} \right)\end{align}\)  

 (a)    From (1), since \({z_1} + 2{z_2} + 3{z_3} = 0,\,\,{\rm{we have}}\)

        \[\begin{array}{l}  z_1^3 + 8z_2^3 + 27z_3^3 = 3 \cdot {z_1} \cdot 2{z_2} \cdot 3{z_3}\\ \Rightarrow  {e^{i3\alpha }} + 8{e^{i3\beta }} + 27{e^{i3\gamma }} = 18{e^{i(\alpha  + \beta  + \gamma )}}\\ \Rightarrow  \cos 3\alpha  + 8\cos 3\beta  + 27\cos 3\gamma  = 18\cos (\alpha  + \beta  + \gamma )\end{array}\]

 (By comparing the real parts)

 (b)   From (2),

\[\begin{array}{l}  {z_2}{z_3} + 2{z_1}{z_3} + 3{z_1}{z_2} = 0\\ \Rightarrow  {e^{i(\beta  + \gamma )}} + 2{e^{i(\alpha  + \gamma )}} + 3{e^{i(\alpha  + \beta )}} = 0\end{array}\]

Comparing the imaginary parts on both sides,

\[\sin (\beta  + \gamma ) + 2\sin (\alpha  + \gamma ) + 3\sin (\alpha  + \beta ) = 0\]

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