
If $\left| {{\text{z - 1}}} \right|{\text{ = 1}}$ and $\arg ({\text{z) = }}\theta $, where ${\text{z }} \ne {\text{ 0}}$ and $\theta $ is acute, then $1{\text{ }} - {\text{ }}\dfrac{2}{{\text{z}}}$ is equals to
A. $\tan \theta $
B. ${\text{i}}\tan \theta $
C. $ - \tan \dfrac{\theta }{2}$
D. $\tan \dfrac{\theta }{2}$
Answer
612.9k+ views
Hint: To solve this problem, we will let z = x + iy and apply this value of z in the given condition $\left| {{\text{z - 1}}} \right|{\text{ = 1}}$. Then, we will use the expression formed to find the value of $1{\text{ }} - {\text{ }}\dfrac{2}{{\text{z}}}$.
Complete step-by-step solution -
Now, we let z = x + iy. Putting this value of z in $\left| {{\text{z - 1}}} \right|{\text{ = 1}}$, we get
$\left| {{\text{x + iy - 1}}} \right|{\text{ = 1}}$
Now, taking real parts together and imaginary part together and solving the modulus, we get
$\left| {{\text{(x - 1) + iy}}} \right|{\text{ = 1}}$
$\sqrt {{{({\text{x - 1)}}}^2}{\text{ + (iy}}{{\text{)}}^2}} {\text{ = }} \pm {\text{1}}$
Squaring both sides in the above equation, we get
${{\text{(x - 1)}}^2}{\text{ + }}{\text{ (iy}}{{\text{)}}^2}{\text{ = 1}}$
Here, we will use the property ${{\text{(x - y)}}^2}{\text{ = }}{{\text{x}}^2}{\text{ - 2xy + }}{{\text{y}}^2}$ to solve the above equation,
Therefore, ${{\text{x}}^2}{\text{ - 2x + 1 + }}{{\text{y}}^2}{\text{ = 1}}$
${{\text{x}}^2}{\text{ - 2x + }}{{\text{y}}^2}{\text{ = 0}}$ … (1)
Now, $\arg ({\text{z) = }}\theta $ which means the angle of complex number z is $\theta $. Now, z = x + iy. Therefore, $\tan \theta {\text{ = }}\dfrac{{\text{y}}}{{\text{x}}}$, where $\theta $ is acute.
Now, we have to find the value of $1{\text{ }} - {\text{ }}\dfrac{2}{{\text{z}}}$
Therefore, $1{\text{ }} - {\text{ }}\dfrac{2}{{\text{z}}}$ = $\dfrac{{{\text{z - 2}}}}{{\text{z}}}$
Putting z = x + iy, we get
$1{\text{ }} - {\text{ }}\dfrac{2}{{\text{z}}}$ = $\dfrac{{{\text{x + }}{\text{iy - 2}}}}{{{\text{x + iy}}}}$. Taking real part together and complex part together, we get
$1{\text{ }} - {\text{ }}\dfrac{2}{{\text{z}}}$ = $\dfrac{{{\text{(x - 2) + }}{\text{iy }}}}{{{\text{x + iy}}}}$
Now, any complex number should not have a term in the denominator which contains an imaginary part. So, to make the complex number in the standard form, we will multiply and divide the complex number with the conjugate of the denominator.
As, $1{\text{ }} - {\text{ }}\dfrac{2}{{\text{z}}}$ = $\dfrac{{{\text{(x - 2) + }}{\text{iy }}}}{{{\text{x + iy}}}}$. So, multiplying both the numerator and denominator term with the conjugate of denominator. Now, the conjugate of a complex number z = $\overline {\text{z}} $ and is calculated by reversing the sign of the imaginary part. So, the conjugate of denominator = ${\text{x - iy}}$.
Now, multiplying both numerator and denominator by ${\text{x - iy}}$.
$1{\text{ }} - {\text{ }}\dfrac{2}{{\text{z}}}$ = $\dfrac{{{\text{(x - 2) + }}{\text{iy }}}}{{{\text{x + iy}}}}{\text{ }} \times {\text{ }}\dfrac{{{\text{x - iy}}}}{{{\text{x - iy}}}}$
$1{\text{ }} - {\text{ }}\dfrac{2}{{\text{z}}}$ = $\dfrac{{{\text{(x - 2)(x - iy) + }}{\text{iy(x - iy) }}}}{{{\text{(x + iy)(x - iy)}}}}$
$1{\text{ }} - {\text{ }}\dfrac{2}{{\text{z}}}$ = $\dfrac{{{\text{x(x - iy) - 2(x - iy) + iy(x - iy)}}}}{{({\text{x + iy)(x - iy)}}}}$
Here, we will the property $({\text{x + y)(x - y) = }}{{\text{x}}^2}{\text{ - }}{{\text{y}}^2}$ in the denominator.
$1{\text{ }} - {\text{ }}\dfrac{2}{{\text{z}}}$ = $\dfrac{{{{\text{x}}^2}{\text{ - 2x + }}{{\text{y}}^2} + {\text{ iy(x - x + 2)}}}}{{{{\text{x}}^2}{\text{ + }}{{\text{y}}^2}}}$
Now, from equation (1), we know that ${{\text{x}}^2}{\text{ - 2x + }}{{\text{y}}^2}{\text{ = 0}}$.
Therefore, $1{\text{ }} - {\text{ }}\dfrac{2}{{\text{z}}}$ = $\dfrac{{2{\text{iy}}}}{{{{\text{x}}^2}{\text{ + }}{\text{ }}{{\text{y}}^2}}}$
Also, from equation (1), ${{\text{x}}^2}{\text{ - 2x + }}{{\text{y}}^2}{\text{ = 0}}$ $ \Rightarrow $ ${{\text{x}}^2}{\text{ + }}{{\text{y}}^2}{\text{ = 2x}}$
Therefore, $1{\text{ }} - {\text{ }}\dfrac{2}{{\text{z}}}$ = $\dfrac{{2{\text{iy}}}}{{2{\text{x}}}}{\text{ = i}}\dfrac{{\text{y}}}{{\text{x}}}$
So, $1{\text{ }} - {\text{ }}\dfrac{2}{{\text{z}}}$ = ${\text{itan}}\theta $
So, option (B) is correct.
Note: When we come up with such types of questions, we will start the question by taking z = x + iy. After it, we will put the value of z in the condition given in the question and make an expression. After it, we will find the value asked in the question by putting the value of z and by using the expression formed. We should be careful when we write any complex number. The complex number should be in standard form with no imaginary term in the denominator. If there is an imaginary term in the denominator, then we will multiply both the numerator and denominator by the conjugate of the denominator to make the complex number in the standard form.
Complete step-by-step solution -
Now, we let z = x + iy. Putting this value of z in $\left| {{\text{z - 1}}} \right|{\text{ = 1}}$, we get
$\left| {{\text{x + iy - 1}}} \right|{\text{ = 1}}$
Now, taking real parts together and imaginary part together and solving the modulus, we get
$\left| {{\text{(x - 1) + iy}}} \right|{\text{ = 1}}$
$\sqrt {{{({\text{x - 1)}}}^2}{\text{ + (iy}}{{\text{)}}^2}} {\text{ = }} \pm {\text{1}}$
Squaring both sides in the above equation, we get
${{\text{(x - 1)}}^2}{\text{ + }}{\text{ (iy}}{{\text{)}}^2}{\text{ = 1}}$
Here, we will use the property ${{\text{(x - y)}}^2}{\text{ = }}{{\text{x}}^2}{\text{ - 2xy + }}{{\text{y}}^2}$ to solve the above equation,
Therefore, ${{\text{x}}^2}{\text{ - 2x + 1 + }}{{\text{y}}^2}{\text{ = 1}}$
${{\text{x}}^2}{\text{ - 2x + }}{{\text{y}}^2}{\text{ = 0}}$ … (1)
Now, $\arg ({\text{z) = }}\theta $ which means the angle of complex number z is $\theta $. Now, z = x + iy. Therefore, $\tan \theta {\text{ = }}\dfrac{{\text{y}}}{{\text{x}}}$, where $\theta $ is acute.
Now, we have to find the value of $1{\text{ }} - {\text{ }}\dfrac{2}{{\text{z}}}$
Therefore, $1{\text{ }} - {\text{ }}\dfrac{2}{{\text{z}}}$ = $\dfrac{{{\text{z - 2}}}}{{\text{z}}}$
Putting z = x + iy, we get
$1{\text{ }} - {\text{ }}\dfrac{2}{{\text{z}}}$ = $\dfrac{{{\text{x + }}{\text{iy - 2}}}}{{{\text{x + iy}}}}$. Taking real part together and complex part together, we get
$1{\text{ }} - {\text{ }}\dfrac{2}{{\text{z}}}$ = $\dfrac{{{\text{(x - 2) + }}{\text{iy }}}}{{{\text{x + iy}}}}$
Now, any complex number should not have a term in the denominator which contains an imaginary part. So, to make the complex number in the standard form, we will multiply and divide the complex number with the conjugate of the denominator.
As, $1{\text{ }} - {\text{ }}\dfrac{2}{{\text{z}}}$ = $\dfrac{{{\text{(x - 2) + }}{\text{iy }}}}{{{\text{x + iy}}}}$. So, multiplying both the numerator and denominator term with the conjugate of denominator. Now, the conjugate of a complex number z = $\overline {\text{z}} $ and is calculated by reversing the sign of the imaginary part. So, the conjugate of denominator = ${\text{x - iy}}$.
Now, multiplying both numerator and denominator by ${\text{x - iy}}$.
$1{\text{ }} - {\text{ }}\dfrac{2}{{\text{z}}}$ = $\dfrac{{{\text{(x - 2) + }}{\text{iy }}}}{{{\text{x + iy}}}}{\text{ }} \times {\text{ }}\dfrac{{{\text{x - iy}}}}{{{\text{x - iy}}}}$
$1{\text{ }} - {\text{ }}\dfrac{2}{{\text{z}}}$ = $\dfrac{{{\text{(x - 2)(x - iy) + }}{\text{iy(x - iy) }}}}{{{\text{(x + iy)(x - iy)}}}}$
$1{\text{ }} - {\text{ }}\dfrac{2}{{\text{z}}}$ = $\dfrac{{{\text{x(x - iy) - 2(x - iy) + iy(x - iy)}}}}{{({\text{x + iy)(x - iy)}}}}$
Here, we will the property $({\text{x + y)(x - y) = }}{{\text{x}}^2}{\text{ - }}{{\text{y}}^2}$ in the denominator.
$1{\text{ }} - {\text{ }}\dfrac{2}{{\text{z}}}$ = $\dfrac{{{{\text{x}}^2}{\text{ - 2x + }}{{\text{y}}^2} + {\text{ iy(x - x + 2)}}}}{{{{\text{x}}^2}{\text{ + }}{{\text{y}}^2}}}$
Now, from equation (1), we know that ${{\text{x}}^2}{\text{ - 2x + }}{{\text{y}}^2}{\text{ = 0}}$.
Therefore, $1{\text{ }} - {\text{ }}\dfrac{2}{{\text{z}}}$ = $\dfrac{{2{\text{iy}}}}{{{{\text{x}}^2}{\text{ + }}{\text{ }}{{\text{y}}^2}}}$
Also, from equation (1), ${{\text{x}}^2}{\text{ - 2x + }}{{\text{y}}^2}{\text{ = 0}}$ $ \Rightarrow $ ${{\text{x}}^2}{\text{ + }}{{\text{y}}^2}{\text{ = 2x}}$
Therefore, $1{\text{ }} - {\text{ }}\dfrac{2}{{\text{z}}}$ = $\dfrac{{2{\text{iy}}}}{{2{\text{x}}}}{\text{ = i}}\dfrac{{\text{y}}}{{\text{x}}}$
So, $1{\text{ }} - {\text{ }}\dfrac{2}{{\text{z}}}$ = ${\text{itan}}\theta $
So, option (B) is correct.
Note: When we come up with such types of questions, we will start the question by taking z = x + iy. After it, we will put the value of z in the condition given in the question and make an expression. After it, we will find the value asked in the question by putting the value of z and by using the expression formed. We should be careful when we write any complex number. The complex number should be in standard form with no imaginary term in the denominator. If there is an imaginary term in the denominator, then we will multiply both the numerator and denominator by the conjugate of the denominator to make the complex number in the standard form.
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