Question
Question: Let \({{z}_{1}}\) and \({{z}_{2}}\) be two complex numbers with \(\alpha ,\beta \) as their principa...
Let z1 and z2 be two complex numbers with α,β as their principal arguments such that α+β>π, then principal arg(z1z2) is given by
A. α+β+π
B. α+β−π
C. α+β−2π
D. α+β
Solution
We explain the concept of argument for the complex number. Then we use the concept of range for argument and the theorem arg(z1z2)=arg(z1)+arg(z2) to find the arg(z1z2). We deduct 2π from the argument if it crosses π to keep it in the range.
Complete step by step answer:
We have z1 and z2 as two complex numbers with α,β as their principal arguments.
We know that −π≤α,β≤π. This range is for the argument of any complex number.
We can express any arbitrary complex number as z=eiθ. Here θ is the argument.
We denote z1=eiα and z2=eiβ. We also know that arg(z1z2)=arg(z1)+arg(z2).
Therefore, putting the values we get arg(z1z2)=arg(z1)+arg(z2)=α+β.
It is given that α+β>π which gives arg(z1z2)>π.
But as the principal argument has to be in the range of [−π,π], we deduct 2π from the argument if it crosses π to keep it in the range. We can deduct as the period of the trigonometric function is 2π.
Therefore, arg(z1z2)=arg(z1)+arg(z2)−2π if arg(z1)+arg(z2)>π.
So, principal arg(z1z2) is given by α+β−2π.
So, the correct answer is “Option C”.
Note: The same thing can be done for condition of arg(z1)+arg(z2)<−π by adding 2π to the argument if it goes less than −π to keep it in the range. The complex form can also be represented as z=eiθ=cosθ+isinθ.