Question

What is the principal argument of $\left( -1,-i \right)$ where $i= \sqrt{-1}$ ​?
A) $\dfrac{\pi }{4}~~$
B) $\dfrac{-\pi }{4}~~$
C) $\dfrac{-3\pi }{4}~~$
D) $\dfrac{3\pi }{4}~~$

Answer 1

Here the given complex number is the point with x-coordinate and y-coordinate. Here we need to find the principal argument of this complex number. Also, we will determine the quadrant in which that point lies. For that, we will calculate the ratio of the y-coordinate to the x-coordinate of that point which will give us the required argument of the given point.

Complete step by step solution:
Let $Z$ be the given complex number.
Therefore,
$\Rightarrow Z=-1-i$ …….. $\left( 1 \right)$
Let’s first write the standard form of complex numbers.
$\Rightarrow Z=r\left( \cos \theta +i\sin \theta \right)$ …….. $\left( 2 \right)$
Here $r$ is the modulus and $\theta$ is the principle argument.
Now, we will compare the real part and imaginary part of equation 1 and equation 2.
$r\cos \theta =-1$ …….. $\left( 3 \right)$
$r\sin \theta =-1$ ………. $\left( 4 \right)$
Now, we will square and then add equation 1 and equation 2.
$\Rightarrow {{r}^{2}}\cos {{\theta }^{2}}+{{r}^{2}}\sin {{\theta }^{2}}={{\left( -1 \right)}^{2}}+{{\left( -1 \right)}^{2}}$
On simplifying the terms, we get
$\Rightarrow {{r}^{2}}\left( {{\cos }^{2}}\theta +{{\sin }^{2}}\theta \right)=1+1$
We know from trigonometric identity that ${{\cos }^{2}}\theta +{{\sin }^{2}}\theta =1$
Now, we will substitute this value there.
$\Rightarrow {{r}^{2}}=2$
Taking square root on both sides, we get
$\Rightarrow \sqrt{{{r}^{2}}}=\sqrt{2} \\\ \Rightarrow r=\sqrt{2} \\\$
Thus, the modulus of complex numbers is $\sqrt{2}$.
Now, we will substitute the value of modulus obtained in equation 3, we get
$\Rightarrow \sqrt{2}\cos \theta =-1$
Dividing both sides by $\sqrt{2}$, we get
$\Rightarrow \cos \theta =\dfrac{-1}{\sqrt{2}}$
Now, we will substitute the value of modulus obtained in equation 4, we get
$\Rightarrow \sqrt{2}\sin \theta =-1$
Dividing both sides by $\sqrt{2}$, we get
$\Rightarrow \sin \theta =\dfrac{-1}{\sqrt{2}}$
Since both and both are negative, therefore, $\theta$ lies in 3rd quadrant.

**Hence,
$\theta =\dfrac{-3\pi }{4}$ **

Note:
The argument of a complex number is defined as the angle inclined from the x- axis in the direction of the complex number represented on the complex plane. It is denoted by $\theta$. It is measured in the standard unit which is known as radians. Here the x-axis is also known as the real axis and y-axis is known as imaginary axis.