Question

Let z and $w$ be complex numbers such that $\overline z + i\overline w = 0$ and $\arg zw = \pi $, then $\arg z = $

Answer 1

Hint: First, we will see what are complex numbers, the numbers that are starting with natural numbers, with the whole numbers that contain respectively rational and irrational followed by real numbers.
Hence in that real numbers if there exists an imaginary axis and that was known as the complex numbers.
Examples; $x,y$ are the real numbers such that $i$ is the imaginary number if there exists $x + iy$ and that is known as the complex numbers (real and imaginary combined).
Formula used: $\arg ( - i) = - \dfrac{\pi }{2}$ which is a formula of the argument of the imaginary axis.

Complete step-by-step solution:
Since from the given question we have $\overline z + i\overline w = 0$, we can able to rewrite the expression as follows;
$\overline z + i\overline w = 0 \Rightarrow \overline z = - i\overline w $ (in the left side of the equation only the real values and the right side only imaginary values).
Now we are going to cancel the bar concerning the common real terms, since $z,w$ are the real values where $i$ is the imaginary.
Thus, we get$ \Rightarrow z = iw$ (on canceling the bar(conjugate) the imaginary signs change like if the imaginary is negative, it will be changed to a positive value).
Let us place the value $w$ as the real, then we get $ \Rightarrow w = - iz$(inverse of i).
Since from the given $\arg zw = \pi $ (product of the two real values), now apply the value$ \Rightarrow w = - iz$ into this, we get $\arg ( - i{z^2}) = \pi $.
Let us define the formula of the product of the argument $\arg zw = \arg z + \arg \pi $.
Thus, we get for $\arg ( - i{z^2}) = \pi \Rightarrow \arg ( - i) + \arg ({z^2}) = \pi $, since the argument of the power will become down values into the multiplication as same the logarithm, thus we get $\arg ( - i) + 2\arg (z) = \pi $.
From the formula part we have $\arg ( - i) = - \dfrac{\pi }{2}$ applied this into the above equation we get $ - \dfrac{\pi }{2} + 2\arg (z) = \pi $ replacing the terms on the right-hand side we get $2\arg (z) = \pi + \dfrac{\pi }{2} \Rightarrow \dfrac{{3\pi }}{2}$
Finlay divides the terms into two; thus, we get $\arg (z) = \dfrac{{3\pi }}{4}$.
Hence, we get the required argument $\arg (z) = \dfrac{{3\pi }}{4}$.

Note: Since $\overline z $ which is the conjugation of the value $z$; that means the positive or negative terms get changes like if $z = x + iy$ then the conjugation of the $\overline z = x - iy$.
As we see the imaginary axis only changes, thus we used the above solution.
Thus, there are many results according to the conjugation like $z + \overline z = 2\operatorname{Re} (x)$