Question

If \[x = 1 + i\sqrt 3 \], \[y = 1 - i\sqrt 3 \], and \[z = 2\]. Then show that \[{x^P} + {y^P} = {z^P}\] for every prime \[p > 2\].

Answer 1

Hint: First, convert the given complex numbers \[x = 1 + i\sqrt 3 \], and \[y = 1 - i\sqrt 3 \] into the polar form of complex number. Then apply the De Moivre’s theorem to calculate the values of \[{x^P}\], and \[{y^P}\]. In the end, substitute the values in the given equation to get the required answer.

Formula used:
The polar form of a complex number \[z = a + ib\] is: \[z = r\left( {\cos\theta + i\sin\theta } \right)\], where \[r = \sqrt {{a^2} + {b^2}} \]

Complete step by step solution:
The given complex numbers are \[x = 1 + i\sqrt 3 \], \[y = 1 - i\sqrt 3 \], and \[z = 2\].
To Prove: \[{x^P} + {y^P} = {z^P}\] for every prime \[p > 2\].

Let’s convert the above complex numbers into polar form.
Multiply and divide \[x = 1 + i\sqrt 3 \] by 2.
We get,
\[x = 2\left( {\dfrac{1}{2} + i\dfrac{{\sqrt 3 }}{2}} \right)\]
Now convert the above complex number into the polar form.
Apply the trigonometric angles \[\cos\dfrac{\pi }{3} = \dfrac{1}{2}\] and \[\sin\dfrac{\pi }{3} = \dfrac{{\sqrt 3 }}{2}\].
\[x = 2\left( {\cos\dfrac{\pi }{3} + i\sin\dfrac{\pi }{3}} \right)\] \[.....\left( 1 \right)\]

Multiply and divide \[y = 1 - i\sqrt 3 \] by 2.
We get,
\[y = 2\left( {\dfrac{1}{2} - i\dfrac{{\sqrt 3 }}{2}} \right)\]
Now convert the above complex number into the polar form.
Apply the trigonometric angles \[\cos\left( { - \dfrac{\pi }{3}} \right) = \dfrac{1}{2}\] and \[\sin\left( { - \dfrac{\pi }{3}} \right) = \dfrac{{ - \sqrt 3 }}{2}\].
\[y = 2\left( {\cos\left( { - \dfrac{\pi }{3}} \right) + i\sin\left( { - \dfrac{\pi }{3}} \right)} \right)\]
Apply the trigonometric properties of angles \[\cos\left( { - \theta } \right) = \cos\theta \] and \[\sin\left( { - \theta } \right) = - \sin\theta \]
\[y = 2\left( {\cos\dfrac{\pi }{3} - i\sin\dfrac{\pi }{3}} \right)\] \[.....\left( 2 \right)\]

Let’s apply the De Moivre’s theorem on the equations \[\left( 1 \right)\] and \[\left( 2 \right)\].
We get,
\[{x^P} = {\left[ {2\left( {\cos\dfrac{\pi }{3} + i\sin\dfrac{\pi }{3}} \right)} \right]^P}\]
\[ \Rightarrow {x^P} = \left[ {{2^P}\left( {\cos\dfrac{{P\pi }}{3} + i\sin\dfrac{{P\pi }}{3}} \right)} \right]\] \[.....\left( 3 \right)\]
Also,
\[{y^P} = {\left[ {2\left( {\cos\dfrac{\pi }{3} - i\sin\dfrac{\pi }{3}} \right)} \right]^P}\]
\[ \Rightarrow {y^P} = \left[ {{2^P}\left( {\cos\dfrac{{P\pi }}{3} - i\sin\dfrac{{P\pi }}{3}} \right)} \right]\] \[.....\left( 4 \right)\]

Add the equations \[\left( 3 \right)\] and \[\left( 4 \right)\]. We get,
\[{x^P} + {y^P} = \left[ {{2^P}\left( {\cos\dfrac{{P\pi }}{3} + i\sin\dfrac{{P\pi }}{3}} \right)} \right] + \left[ {{2^P}\left( {\cos\dfrac{{P\pi }}{3} - i\sin\dfrac{{P\pi }}{3}} \right)} \right]\]
Simplify the above equation.
\[{x^P} + {y^P} = {2^P}\cos\dfrac{{P\pi }}{3} + {2^P}i\sin\dfrac{{P\pi }}{3} + {2^P}\cos\dfrac{{P\pi }}{3} - {2^P}i\sin\dfrac{{P\pi }}{3}\]
\[ \Rightarrow {x^P} + {y^P} = {2^P}\cos\dfrac{{P\pi }}{3} + {2^P}\cos\dfrac{{P\pi }}{3}\]
\[ \Rightarrow {x^P} + {y^P} = {2^P}\left( {2\cos\dfrac{{P\pi }}{3}} \right)\]
We know that \[\cos\left( {\dfrac{{n\pi }}{3}} \right) = \dfrac{1}{2}\].
Then,
\[{x^P} + {y^P} = {2^P}\left( {2 \times \dfrac{1}{2}} \right)\]
\[ \Rightarrow {x^P} + {y^P} = {2^P}\]
Now substitute \[z = 2\] in the above equation.
\[{x^P} + {y^P} = {z^P}\]
Hence, proved.

Note: Students often get confused about the De Moivre’s theorem.
De Moivre’s Theorem: The power of a complex number in polar form is equal to raising the modulus to the same power and multiplying the argument by the same power.