
What is the principal argument of $\left( { - 1, - i} \right)$, where $i = \sqrt { - 1} $?
$
(a){\text{ }}\dfrac{\pi }{4} \\
(b){\text{ }}\dfrac{{ - \pi }}{4} \\
(c){\text{ }}\dfrac{{ - 3\pi }}{4} \\
(d){\text{ }}\dfrac{{3\pi }}{4} \\
$
Answer
232.8k+ views
Hint: In this question we have to find the principal argument of the given coordinates. Now a complex number z can be written in the form of $x + iy$ where x is the real part and y is the imaginary part. Now the argument of any complex number z is ${\tan ^{ - 1}}\left( {\dfrac{y}{x}} \right)$. Use this concept to find the principal argument. Keep in mind that here we are asked about the principal argument and not any general argument, so solve accordingly.
Complete step-by-step answer:
We have to find out the principal argument of $\left( { - 1, - i} \right)$.
Now as we know the argument of $\left( {x + iy} \right)$ is $\theta = {\tan ^{ - 1}}\left( {\dfrac{y}{x}} \right)$
But in the given coordinate both x and y are negative and we know x and y are both negative in the third quadrant. So, the principal argument of the given coordinate lies in the third quadrant.
Now first calculate the argument of given coordinate so, on comparing,
Y = -1 and x = -1
So, the argument of $\left( { - 1, - i} \right)$ is $\theta = {\tan ^{ - 1}}\left( {\dfrac{{ - 1}}{{ - 1}}} \right) = {\tan ^{ - 1}}\left( 1 \right)$
Now as we know that the value of ${\tan ^{ - 1}}\left( 1 \right)$ is $\dfrac{\pi }{4}$ in the first quadrant.
And the value of ${\tan ^{ - 1}}\left( 1 \right)$ in third quadrant is $\left[ {\left( {\dfrac{\pi }{4} - \pi } \right) = \dfrac{{ - 3\pi }}{4}} \right]$ or $\left[ {\left( {\pi + \dfrac{\pi }{4}} \right) = \dfrac{{5\pi }}{4}} \right]$
So, the principal argument of $\left( { - 1, - i} \right)$ is $\dfrac{{ - 3\pi }}{4}$ or $\dfrac{{5\pi }}{4}$.
But in options $\dfrac{{5\pi }}{4}$ is not there.
So, the principal argument according to options is $\dfrac{{ - 3\pi }}{4}$.
Hence option (c) is correct.
Note: The note point here is we are asked to find the principal argument and not any general argument. The principal argument is a unique value of argument which always lies in the range $ - \pi < {\text{argz < }}\pi $, however a general argument is such that it repeats itself after a regular interval and is given as $2n\pi + {\text{principal argument}}$ where n is an integer.
Complete step-by-step answer:
We have to find out the principal argument of $\left( { - 1, - i} \right)$.
Now as we know the argument of $\left( {x + iy} \right)$ is $\theta = {\tan ^{ - 1}}\left( {\dfrac{y}{x}} \right)$
But in the given coordinate both x and y are negative and we know x and y are both negative in the third quadrant. So, the principal argument of the given coordinate lies in the third quadrant.
Now first calculate the argument of given coordinate so, on comparing,
Y = -1 and x = -1
So, the argument of $\left( { - 1, - i} \right)$ is $\theta = {\tan ^{ - 1}}\left( {\dfrac{{ - 1}}{{ - 1}}} \right) = {\tan ^{ - 1}}\left( 1 \right)$
Now as we know that the value of ${\tan ^{ - 1}}\left( 1 \right)$ is $\dfrac{\pi }{4}$ in the first quadrant.
And the value of ${\tan ^{ - 1}}\left( 1 \right)$ in third quadrant is $\left[ {\left( {\dfrac{\pi }{4} - \pi } \right) = \dfrac{{ - 3\pi }}{4}} \right]$ or $\left[ {\left( {\pi + \dfrac{\pi }{4}} \right) = \dfrac{{5\pi }}{4}} \right]$
So, the principal argument of $\left( { - 1, - i} \right)$ is $\dfrac{{ - 3\pi }}{4}$ or $\dfrac{{5\pi }}{4}$.
But in options $\dfrac{{5\pi }}{4}$ is not there.
So, the principal argument according to options is $\dfrac{{ - 3\pi }}{4}$.
Hence option (c) is correct.
Note: The note point here is we are asked to find the principal argument and not any general argument. The principal argument is a unique value of argument which always lies in the range $ - \pi < {\text{argz < }}\pi $, however a general argument is such that it repeats itself after a regular interval and is given as $2n\pi + {\text{principal argument}}$ where n is an integer.
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