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Question: The principal argument of \(\dfrac{i-3}{i-1}\) is: \(\left( a \right)tan^{-1}\dfrac{1}{2}\) \(\l...

The principal argument of i−3i−1\dfrac{i-3}{i-1} is:
(a)tan−112\left( a \right)tan^{-1}\dfrac{1}{2}
(b)tan−132\left( b \right) tan^{-1}\dfrac{3}{2}
(c)tan−152\left( c \right) tan^{-1}\dfrac{5}{2}
(d)tan−172\left( d \right) tan^{-1}\dfrac{7}{2}

Explanation

Solution

We are asked to find the principal argument of a given complex number. But we see that it is not in the standard a+bia+bi form. So, we first convert this into the standard form and then apply the definition of principal argument to find the answer. We should be aware of some terms such as argument, principal argument which are related to the imaginary numbers in order to solve this question.

Complete step by step answer:
We have i−3i−1\dfrac{i-3}{i-1}. We need to first convert this into a standard form. For this, we multiply both numerator and denominator by i+1i+1. Doing this we get:
i−3i−1×i+1i+1=i2+i−3i−3i2−12\dfrac{i-3}{i-1}\times\dfrac{i+1}{i+1}=\dfrac{i^2+i-3i-3}{i^2-1^2}
We have used the following identity in denominator:
(a+b)(a−b)=a2−b2\left(a+b\right)\left(a-b\right)=a^2-b^2
We get:
−1−2i−3−1−1=−4−2i−2=2+i\dfrac{-1-2i-3}{-1-1}=\dfrac{-4-2i}{-2}=2+i
Hence, we have found out the complex number in standard form. Now, we know that the principal argument, Arg(z)Arg\left(z\right)of any complex number z=a+biz=a+biis found out using the following formula:
Arg(z)=tan−1(ba)Arg\left(z\right)=tan^{-1}\left(\dfrac{b}{a}\right) if a>0a>0
And
Arg(z)=tan−1(ba)+πArg\left(z\right)=tan^{-1}\left(\dfrac{b}{a}\right)+\pi if a<0a<0
We have 2+i2+i as the complex number. So a=2a=2 and b=1b=1. We conclude that a>0a>0, hence the principal argument will be:
Arg(2+i)=tan−1(12)Arg\left(2+i\right)=tan^{-1}\left(\dfrac{1}{2}\right)

So, the correct answer is “Option a”.

Note: We can also do the same using the following formula:
Arg(z1z2)=Arg(z1)−Arg(z2)Arg\left(\dfrac{z_1}{z_2}\right)=Arg\left(z_1\right)-Arg\left(z_2\right)
Here, put z1=i−3z_1=i-3 and z2=i−1z_2=i-1
Then Arg(z1)=tan−1−13Arg\left(z_1\right)=tan^{-1}\dfrac{-1}{3}
And Arg(z2)=tan−1−1Arg\left(z_2\right)=tan^{-1}-1
So, we have:
Arg(i−3i−1)=Arg(z1)−Arg(z2)Arg \left(\dfrac{i-3}{i-1}\right)= Arg\left(z_1\right)-Arg\left(z_2\right)
=tan−1−13−tan−1−1= tan^{-1}\dfrac{-1}{3}- tan^{-1}-1
Now, we use the formula below:
tan−1A−tan−1B=tan−1(A−B1+AB)tan^{-1}A- tan^{-1}B=tan^{-1}\left(\dfrac{A-B}{1+AB}\right)
Using this we obtain the following:
Arg(i−3i−1)=tan−1(−13+11+13)Arg \left(\dfrac{i-3}{i-1}\right)=tan^{-1}\left(\dfrac{-\dfrac{1}{3}+1}{1+\dfrac{1}{3}}\right)
=tan−112=tan^{-1}\dfrac{1}{2}
Hence, the answer is obtained correctly. But note that you should use this only when you remember the formula of inverse tan function correctly.