That's a complex equation!

Geometry Level pending

( E ) : ( z 2 − 2 z ) cos ⁡ 2 α + 1 = 0 (E): ({ z }^{ 2 }-2z)\cos ^{ 2 }{ \alpha +1=0 }

The equation ( E ) (E) admits 2 solutions U U and V V . If U × V = cos ⁡ n α U\times V=\cos ^{ n } \alpha , find n n .

Bonus : Find the set of images of these solutions.

Clarification : z z is a complex number.

-2 -1 1 2

Mohammad Hamdar by Mohammad Hamdar , 22, Lebanon

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1 solution

Harsh Khatri
Feb 7, 2016

( E ) : ( z 2 − 2 z + 1 ) cos ⁡ 2 α + 1 − cos ⁡ 2 α = 0 (E) : (z^2 - 2z +1)\cos^2\alpha + 1 - \cos^2\alpha = 0

⇒ U = 1 + i tan ⁡ α , V = 1 − i tan ⁡ α \displaystyle \Rightarrow U = 1 + i\tan\alpha, V = 1 - i\tan\alpha

⇒ U × V = 1 + tan ⁡ 2 α = cos ⁡ − 2 α \displaystyle \Rightarrow U \times V = 1 + \tan^2\alpha = \cos^{-2}\alpha

⇒ n = − 2 \displaystyle \Rightarrow \boxed{n = - 2}

I presume you meant 'set of conjugates' when you said 'set of images'. I'll proceed with that assumption.

U ‾ = 1 − i tan ⁡ α = V \displaystyle \overline{U} = 1 - i\tan\alpha = V

V ‾ = 1 + i tan ⁡ α = U \displaystyle \overline{V} = 1 + i\tan\alpha = U

Therefore, set of conjugates of the solutions = { U ‾ , V ‾ } = { V , U } \displaystyle = \{\overline{U}, \overline{V}\} = \{V, U\}
which is the same as the set of solutions.

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When I said set of images i meant the locus of the points, may be badly said :) But let U and V be the affixes of the points M and N M ( 1 , tan ⁡ α ) N ( 1 , − tan ⁡ α ) s o t h e s e t o f p o i n t s o f t h e s o l u t i o n s i s t h e s t r a i g h t l i n e x = 1 s i n c e α ∈ ] − π 2 , π 2 [ s o , t a n α ∈ ] − ∞ , + ∞ [ M(1,\tan { \alpha )\quad } \\ N(1,-\tan { \alpha ) } \\ so\quad the\quad set\quad of\quad points\quad of\quad the\quad solutions\quad is\quad the\\ straight\quad line\quad x=1\quad since\quad \alpha \quad \in \quad ]\frac { -\pi }{ 2 } ,\frac { \pi }{ 2 } [\\ so,\quad tan\alpha \quad \in \quad ]-\infty ,+\infty [

Mohammad Hamdar - 5 years, 4 months ago

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I would like to discuss another method to find the locus.

We write U in polar form as:

U = sec ⁡ α ( cos ⁡ α + i sin ⁡ α ) \displaystyle U = \sec\alpha ( \cos\alpha + i\sin\alpha)

U = 1 + tan ⁡ 2 α ⋅ e i α \displaystyle U = \sqrt{1+\tan^2\alpha} \cdot e^{i\alpha}

⇒ ∣ U ∣ = 1 + tan ⁡ 2 α \displaystyle \Rightarrow |U| = \sqrt{1 + \tan^2\alpha}

Thus, we conclude that U , O ( 0 , 0 ) \displaystyle U, O(0,0) and P ( 1 , 0 ) \displaystyle P(1,0) always form a right-angled triangle such that ∠ U O P = α \displaystyle \angle UOP = \alpha .

Thus, P U \displaystyle PU is always perpendicular to real axis and so the locus of U \displaystyle U is the part of the line z = 1 \displaystyle z=1 above the real axis(including point P) .

Similarly, we conclude P V \displaystyle PV is always perpendicular to real axis and the locus of V \displaystyle V is the part of the line z = 1 \displaystyle z=1 below the real axis(including point P).

Thus, the combination of their loci is the line z = 1 \boxed{z=1} .

Or if plotted on the Cartesian plane, the locus is x = 1 \boxed{x=1} .

Harsh Khatri - 5 years, 4 months ago

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Yup a new method i saw there ! Thank you for sharing it .

Mohammad Hamdar - 5 years, 4 months ago

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