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Question: If \(z={{\left( \dfrac{\sqrt{3}+i}{2} \right)}^{5}}+{{\left( \dfrac{\sqrt{3}-i}{2} \right)}^{5}}\) t...

If z=(3+i2)5+(3−i2)5z={{\left( \dfrac{\sqrt{3}+i}{2} \right)}^{5}}+{{\left( \dfrac{\sqrt{3}-i}{2} \right)}^{5}} then which of the following option is correct:
(a) Re(z)=0{{R}_{e}}(z)=0
(b) Im(z)=0{{I}_{m}}(z)=0
(c) Re(z)=0{{R}_{e}}(z)=0, Im(z)>0{{I}_{m}}(z)>0
(d) Re(z)>0,Im(z)<0{{R}_{e}}(z)>0,{{I}_{m}}(z)<0

Explanation

Solution

Hint: First we solve the equation by opening brackets and then we have to rearrange some terms to make it in the form of a + ib. After doing this much then we will write the imaginary part and real part separately and find out the correct option.

Complete step-by-step answer:
First let’s try to find out the value of this one (3+i2)5{{\left( \dfrac{\sqrt{3}+i}{2} \right)}^{5}} .
The formula for (a+b)2=a2+b2+2ab{{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab , we are going to use this formula for calculating the value of z2{{z}^{2}}, where z can any complex number.
Another formula that we are going to use is i2=−1{{i}^{2}}=-1 ,
Now (3+i2)5{{\left( \dfrac{\sqrt{3}+i}{2} \right)}^{5}} will be,
=(3+i)532 =(3+i)2(3+i)2(3+i)32 \begin{aligned} & =\dfrac{{{\left( \sqrt{3}+i \right)}^{5}}}{32} \\\ & =\dfrac{{{\left( \sqrt{3}+i \right)}^{2}}{{\left( \sqrt{3}+i \right)}^{2}}\left( \sqrt{3}+i \right)}{32} \\\ \end{aligned}
Now we will use (a+b)2=a2+b2+2ab{{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab to expand,
=((3)2+i2+23i)((3)2+i2+23i)(3+i)32=\dfrac{\left( {{\left( \sqrt{3} \right)}^{2}}+{{i}^{2}}+2\sqrt{3}i \right)\left( {{\left( \sqrt{3} \right)}^{2}}+{{i}^{2}}+2\sqrt{3}i \right)\left( \sqrt{3}+i \right)}{32}
Now we know that i2=−1{{i}^{2}}=-1 , using this we get,
=(2+23i)2(3+i)32=\dfrac{{{\left( 2+2\sqrt{3}i \right)}^{2}}\left( \sqrt{3}+i \right)}{32}
Now we will use (a+b)2=a2+b2+2ab{{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab to expand,
=(22+(23i)2+83i)(3+i)32=\dfrac{\left( {{2}^{2}}+{{\left( 2\sqrt{3}i \right)}^{2}}+8\sqrt{3}i \right)\left( \sqrt{3}+i \right)}{32}
Now we know that i2=−1{{i}^{2}}=-1 , using this we get,
=(4−12+83i)(3+i)32=\dfrac{\left( 4-12+8\sqrt{3}i \right)\left( \sqrt{3}+i \right)}{32}
=(−8+83i)(3+i)32 =8(−1+3i)(3+i)32 =8(−3−i+3i+3i2)32 \begin{aligned} & =\dfrac{\left( -8+8\sqrt{3}i \right)\left( \sqrt{3}+i \right)}{32} \\\ & =\dfrac{8\left( -1+\sqrt{3}i \right)\left( \sqrt{3}+i \right)}{32} \\\ & =\dfrac{8\left( -\sqrt{3}-i+3i+\sqrt{3}{{i}^{2}} \right)}{32} \\\ \end{aligned}
Now we know that i2=−1{{i}^{2}}=-1 , using this we get,
=8(−23+2i)32 =16(−3+i)32 \begin{aligned} & =\dfrac{8\left( -2\sqrt{3}+2i \right)}{32} \\\ & =\dfrac{16\left( -\sqrt{3}+i \right)}{32} \\\ \end{aligned}
So, we have the first part now for the 2nd part we just have substitute i by –i in 16(−3+i)32\dfrac{16\left( -\sqrt{3}+i \right)}{32}
(3−i2)5{{\left( \dfrac{\sqrt{3}-i}{2} \right)}^{5}} , after substituting we get,
(3−i2)5{{\left( \dfrac{\sqrt{3}-i}{2} \right)}^{5}}= 16(−3−i)32\dfrac{16\left( -\sqrt{3}-i \right)}{32}
Now we calculated all the values that are needed to write the given equation of z in the form of a + ib.
Therefore, the value of z becomes after substituting is:
16(−3+i)32+16(−3−i)32 z=−3+i−3−i2 z=−232 z=−3 \begin{aligned} & \dfrac{16\left( -\sqrt{3}+i \right)}{32}+\dfrac{16\left( -\sqrt{3}-i \right)}{32} \\\ & z=\dfrac{-\sqrt{3}+i-\sqrt{3}-i}{2} \\\ & z=\dfrac{-2\sqrt{3}}{2} \\\ & z=-\sqrt{3} \\\ \end{aligned}
In a + ib, ‘a’ is the real part and ‘b’ is the imaginary part.
Now it is the form a + ib, now we can write the real part and the imaginary part.
Re(z){{R}_{e}}(z) = −3-\sqrt{3}
Im(z){{I}_{m}}\left( z \right) = 0
From this we can conclude that option (b) is correct.

Note: One can also solve this question by first writing all the powers of (3+i)\left( \sqrt{3}+i \right) up to 5 and then writing all the powers of (3−i)\left( \sqrt{3}-i \right) up to 5 and then substitute it in the given equation and then again we have to separate it imaginary part and real part to mark the correct option.