# 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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