Explanation
we have,
z is a complex number.
Then,
Let z=r\cos \theta +i\sin \theta
Now, we know that,
\overline{z}=r\left( \cos \theta -i\sin \theta \right)
So,
z\overline{z}={{r}^{2}}
\dfrac{1}{z}=\dfrac{\overline{z}}{z\overline{z}}=\dfrac{\overline{z}}{{{r}^{2}}}
Now,
According to given question,
z+\dfrac{1}{z}=\left| z+\dfrac{1}{z} \right|=1\,\,\,\left( \text{z}\,\text{and}\,\text{z+}\dfrac{\text{1}}{\text{z}}\,\text{same argument} \right)
=\left| r\left( \cos \theta +i\sin \theta \right)+\dfrac{r\left( \cos \theta -i\sin \theta \right)}{{{r}^{2}}} \right|=1
=\left| \left( r+\dfrac{1}{r} \right)\cos \theta +i\left( r-\dfrac{1}{r} \right)\sin \theta \right|=1
Now, squaring both side and we get,
\left| \left( {{r}^{2}}+\dfrac{1}{{{r}^{2}}}+2 \right){{\cos }^{2}}\theta +{{i}^{2}}\left( {{r}^{2}}+\dfrac{1}{{{r}^{2}}}-2 \right){{\sin }^{2}}\theta \right|=1
\left( {{r}^{2}}+\dfrac{1}{{{r}^{2}}}+2 \right){{\cos }^{2}}\theta =1
{{r}^{2}}+\dfrac{1}{{{r}^{2}}}+2\cos 2\theta =1
{{r}^{2}}+\dfrac{1}{{{r}^{2}}}=1-2\cos 2\theta
1-2\cos 2\theta =0
\Rightarrow \cos 2\theta =\dfrac{1}{2}
\Rightarrow \cos 2\theta =cos\dfrac{\pi }{3}
\Rightarrow 2\theta =\dfrac{\pi }{3}
\Rightarrow \theta =\dfrac{\pi }{6}
Then, Z is purely imaginary.
Hence, this is the answer.
We have,
\dfrac{2+3i\sin \theta }{1-2i\sin \theta }
On rationalize and we get,
\dfrac{2+3i\sin \theta }{1-2i\sin \theta }\times \dfrac{1+2i\sin \theta }{1+2i\sin \theta }
\Rightarrow \dfrac{2+3i\sin \theta +4i\sin \theta +6{{i}^{2}}{{\sin }^{2}}\theta }{{{1}^{2}}-4{{i}^{2}}{{\sin }^{2}}\theta }
\Rightarrow \dfrac{2-{{\sin }^{2}}\theta +7i\sin \theta }{1+4{{\sin }^{2}}\theta }
\Rightarrow \dfrac{1+1-{{\sin }^{2}}\theta +7i\sin \theta }{1+4{{\sin }^{2}}\theta }
\Rightarrow \dfrac{1+{{\cos }^{2}}\theta +7i\sin \theta }{1+4{{\sin }^{2}}\theta }
\Rightarrow \dfrac{1+{{\cos }^{2}}\theta }{1+4{{\sin }^{2}}\theta }+\dfrac{7i\sin \theta }{1+4{{\sin }^{2}}\theta }
\Rightarrow \dfrac{1+{{\cos }^{2}}\theta }{1+4{{\sin }^{2}}\theta }+i\dfrac{7\sin \theta }{1+4{{\sin }^{2}}\theta }
Then purely imaginary is
Real part of equal to zero.
\dfrac{1+{{\cos }^{2}}\theta }{1+4{{\sin }^{2}}\theta }=0
\Rightarrow 1+{{\cos }^{2}}\theta =0
\Rightarrow {{\cos }^{2}}\theta =-1
\Rightarrow \cos \theta =\sqrt{-1}
\Rightarrow \theta ={{\cos }^{-1}}\sqrt{-1}
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