Let ${{z}_{1}}$ and ${{z}_{2}}$ be two complex numbers with $\alpha ,\beta $ as their principal arguments such that $\alpha +\beta >\pi $, then principal $\arg \left( {{z}_{1}}{{z}_{2}} \right)$ is given by
A. $\alpha +\beta +\pi $
B. $\alpha +\beta -\pi $
C. $\alpha +\beta -2\pi $
D. $\alpha +\beta $
Answer
575.1k+ views
Hint: We explain the concept of argument for the complex number. Then we use the concept of range for argument and the theorem $\arg \left( {{z}_{1}}{{z}_{2}} \right)=\arg \left( {{z}_{1}} \right)+\arg \left( {{z}_{2}} \right)$ to find the $\arg \left( {{z}_{1}}{{z}_{2}} \right)$. We deduct $2\pi $ from the argument if it crosses $\pi $ to keep it in the range.
Complete step by step answer:
We have ${{z}_{1}}$ and ${{z}_{2}}$ as two complex numbers with $\alpha ,\beta $ as their principal arguments.
We know that $-\pi \le \alpha ,\beta \le \pi $. This range is for the argument of any complex number.
We can express any arbitrary complex number as $z={{e}^{i\theta }}$. Here $\theta $ is the argument.
We denote ${{z}_{1}}={{e}^{i\alpha }}$ and ${{z}_{2}}={{e}^{i\beta }}$. We also know that $\arg \left( {{z}_{1}}{{z}_{2}} \right)=\arg \left( {{z}_{1}} \right)+\arg \left( {{z}_{2}} \right)$.
Therefore, putting the values we get $\arg \left( {{z}_{1}}{{z}_{2}} \right)=\arg \left( {{z}_{1}} \right)+\arg \left( {{z}_{2}} \right)=\alpha +\beta $.
It is given that $\alpha +\beta >\pi $ which gives $\arg \left( {{z}_{1}}{{z}_{2}} \right)>\pi $.
But as the principal argument has to be in the range of $\left[ -\pi ,\pi \right]$, we deduct $2\pi $ from the argument if it crosses $\pi $ to keep it in the range. We can deduct as the period of the trigonometric function is $2\pi $.
Therefore, $\arg \left( {{z}_{1}}{{z}_{2}} \right)=\arg \left( {{z}_{1}} \right)+\arg \left( {{z}_{2}} \right)-2\pi $ if $\arg \left( {{z}_{1}} \right)+\arg \left( {{z}_{2}} \right)>\pi $.
So, principal $\arg \left( {{z}_{1}}{{z}_{2}} \right)$ is given by $\alpha +\beta -2\pi $.
So, the correct answer is “Option C”.
Note: The same thing can be done for condition of $\arg \left( {{z}_{1}} \right)+\arg \left( {{z}_{2}} \right)<-\pi $ by adding $2\pi $ to the argument if it goes less than $-\pi $ to keep it in the range. The complex form can also be represented as $z={{e}^{i\theta }}=\cos \theta +i\sin \theta $.
Complete step by step answer:
We have ${{z}_{1}}$ and ${{z}_{2}}$ as two complex numbers with $\alpha ,\beta $ as their principal arguments.
We know that $-\pi \le \alpha ,\beta \le \pi $. This range is for the argument of any complex number.
We can express any arbitrary complex number as $z={{e}^{i\theta }}$. Here $\theta $ is the argument.
We denote ${{z}_{1}}={{e}^{i\alpha }}$ and ${{z}_{2}}={{e}^{i\beta }}$. We also know that $\arg \left( {{z}_{1}}{{z}_{2}} \right)=\arg \left( {{z}_{1}} \right)+\arg \left( {{z}_{2}} \right)$.
Therefore, putting the values we get $\arg \left( {{z}_{1}}{{z}_{2}} \right)=\arg \left( {{z}_{1}} \right)+\arg \left( {{z}_{2}} \right)=\alpha +\beta $.
It is given that $\alpha +\beta >\pi $ which gives $\arg \left( {{z}_{1}}{{z}_{2}} \right)>\pi $.
But as the principal argument has to be in the range of $\left[ -\pi ,\pi \right]$, we deduct $2\pi $ from the argument if it crosses $\pi $ to keep it in the range. We can deduct as the period of the trigonometric function is $2\pi $.
Therefore, $\arg \left( {{z}_{1}}{{z}_{2}} \right)=\arg \left( {{z}_{1}} \right)+\arg \left( {{z}_{2}} \right)-2\pi $ if $\arg \left( {{z}_{1}} \right)+\arg \left( {{z}_{2}} \right)>\pi $.
So, principal $\arg \left( {{z}_{1}}{{z}_{2}} \right)$ is given by $\alpha +\beta -2\pi $.
So, the correct answer is “Option C”.
Note: The same thing can be done for condition of $\arg \left( {{z}_{1}} \right)+\arg \left( {{z}_{2}} \right)<-\pi $ by adding $2\pi $ to the argument if it goes less than $-\pi $ to keep it in the range. The complex form can also be represented as $z={{e}^{i\theta }}=\cos \theta +i\sin \theta $.
Recently Updated Pages
If x a + bt + ct2 where x is in meters and t is in class 11 physics CBSE

A car covers the first half distance between two places class 11 physics CBSE

The resultant of two vectors overrightarrow P and overrightarrow class 11 physics CBSE

Find the value of cos 135 class 11 maths CBSE

A mass M is held in place by an applied force F and class 11 physics CBSE

A solution of glucose in water is labelled as 10 dfracwv class 11 chemistry CBSE

Trending doubts
Find the value of the expression given below sin 30circ class 11 maths CBSE

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Draw a diagram of nephron and explain its structur class 11 biology CBSE

10 examples of friction in our daily life

Proton was discovered by A Thomson B Rutherford C Chadwick class 11 chemistry CBSE

Bond order ofO2 O2+ O2 and O22 is in order A O2 langle class 11 chemistry CBSE

