
Locate the complex numbers \[z = x + iy\] for which \[{\log _{\sqrt 3 }}\dfrac{{|z{|^2} - |z| + 1}}{{2 + |z|}} < 2\].
Answer
460.2k+ views
Hint: According to the question, complex numbers are any number that can be written in the form of \[a + bi\], where ‘i’ is the imaginary unit and ‘a’ and ‘b’ are real numbers. Here, ‘i’ is the symbol of the imaginary unit and it satisfies the equation \[{i^2} = - 1\].
Complete step-by-step solution:
A simple property of logarithm is being used here that is:
\[{\log _{{a^{}}}}b = c\] then \[{a^c} = b\]
So, here in this equation, according to the complex form, \[a\] is represented by \[\sqrt 3 \], \[b\] is represented by \[\dfrac{{\left| {{z^2}} \right| - \left| z \right| + 1}}{{2 + \left| z \right|}}\], and \[c\] is represented by \[2\].
So therefore, \[\dfrac{{\left| {{z^2}} \right| - \left| z \right| + 1}}{{2 + \left| z \right|}} < {\sqrt {{3^{}}} ^2}\]
Now, when we solve this, we get:
\[
\Rightarrow \dfrac{{\left| {{z^2}} \right| - \left| z \right| + 1}}{{2 + \left| z \right|}} < {3^{2 \times \dfrac{1}{2}}} \\
\Rightarrow \dfrac{{\left| {{z^2}} \right| - \left| z \right| + 1}}{{2 + \left| z \right|}} < 3 \\
\Rightarrow \left| {{z^2}} \right| - \left| z \right| + 1 < 3(2 + \left| z \right|) \\
\Rightarrow \left| {{z^2}} \right| - \left| z \right| + 1 < 6 + 3\left| z \right| \\
\Rightarrow \left| {{z^2}} \right| - 4\left| z \right| + ( - 5) < 0 \]
Now we will try to split the equation, and we get:
\[ \Rightarrow \left| {{z^2}} \right| + \left| z \right| - 5\left| z \right| - 5 < 0\]
Now, we will take out the common terms, and we get:
\[ \Rightarrow \left| z \right|(\left| z \right| + 1) - 5(\left| z \right| + 1) < 0 \\
\therefore (\left| z \right| - 5)(\left| z \right| + 1) < 0 \\
\therefore - 1 < \left| z \right| < 5 \]
But as we know, \[\left| z \right| > 0\]
\[ \Rightarrow 0 < \left| z \right| < 5\]
Now this represents that all points of \[z\]lie in the circle of radius \[5\] with centre \[(0,0)\]. We can see in the below diagram,
Note: We wrote \[z = x + iy\], hence it was a cartesian equation of writing a complex number. The value of \[\left| z \right|\] comes out to be between \[ - 1\] and \[5\], but as we take only positive numbers, this value of \[\left| z \right|\]is between \[0\] and \[5\]. Thus, we make a circle which comprises all the numbers between \[0\] and \[5\] as to represent them in the cartesian form.
Complete step-by-step solution:
A simple property of logarithm is being used here that is:
\[{\log _{{a^{}}}}b = c\] then \[{a^c} = b\]
So, here in this equation, according to the complex form, \[a\] is represented by \[\sqrt 3 \], \[b\] is represented by \[\dfrac{{\left| {{z^2}} \right| - \left| z \right| + 1}}{{2 + \left| z \right|}}\], and \[c\] is represented by \[2\].
So therefore, \[\dfrac{{\left| {{z^2}} \right| - \left| z \right| + 1}}{{2 + \left| z \right|}} < {\sqrt {{3^{}}} ^2}\]
Now, when we solve this, we get:
\[
\Rightarrow \dfrac{{\left| {{z^2}} \right| - \left| z \right| + 1}}{{2 + \left| z \right|}} < {3^{2 \times \dfrac{1}{2}}} \\
\Rightarrow \dfrac{{\left| {{z^2}} \right| - \left| z \right| + 1}}{{2 + \left| z \right|}} < 3 \\
\Rightarrow \left| {{z^2}} \right| - \left| z \right| + 1 < 3(2 + \left| z \right|) \\
\Rightarrow \left| {{z^2}} \right| - \left| z \right| + 1 < 6 + 3\left| z \right| \\
\Rightarrow \left| {{z^2}} \right| - 4\left| z \right| + ( - 5) < 0 \]
Now we will try to split the equation, and we get:
\[ \Rightarrow \left| {{z^2}} \right| + \left| z \right| - 5\left| z \right| - 5 < 0\]
Now, we will take out the common terms, and we get:
\[ \Rightarrow \left| z \right|(\left| z \right| + 1) - 5(\left| z \right| + 1) < 0 \\
\therefore (\left| z \right| - 5)(\left| z \right| + 1) < 0 \\
\therefore - 1 < \left| z \right| < 5 \]
But as we know, \[\left| z \right| > 0\]
\[ \Rightarrow 0 < \left| z \right| < 5\]
Now this represents that all points of \[z\]lie in the circle of radius \[5\] with centre \[(0,0)\]. We can see in the below diagram,

Note: We wrote \[z = x + iy\], hence it was a cartesian equation of writing a complex number. The value of \[\left| z \right|\] comes out to be between \[ - 1\] and \[5\], but as we take only positive numbers, this value of \[\left| z \right|\]is between \[0\] and \[5\]. Thus, we make a circle which comprises all the numbers between \[0\] and \[5\] as to represent them in the cartesian form.
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