Answer
The given point lies in quadrant II.
Work Step by Step
Here, $\left( r,\theta \right)$ is a polar coordinate defined by the formula as:
$p\left( r,\theta \right)=r\cos \theta +ir\sin \theta $ …….(1)
Define the real value $r\cos \theta$ on the x-axis and define the imaginary value $r\sin \theta $ on the y-axis.
Then, put the value of $\left( r,\theta \right)=\left( 4,135{}^\circ \right)$ in equation (1),
Therefore,
$\begin{align}
& p\left( r,\theta \right)=r\cos \theta +ir\sin \theta \\
& =\left( 4\cos 135{}^\circ \right)+i\left( 4\sin 135{}^\circ \right) \\
& =\left( 4\times \left( -0.707 \right) \right)+i\left( 4\times \left( 0.0883 \right) \right) \\
& =\left( -2.828 \right)+i\left( 0.3532 \right)
\end{align}$
Now, by comparing it with the equation in the rectangular coordinates, we get:
$p\left( x,y \right)=x+iy$
Hence, the rectangular coordinate is:
$x=-2.828\,\,\,\text{ and }\,\,y=0.3532$
So,
$\left( x,y \right)=\left( -2.828,0.3532 \right)$
Since $x=-2.828$ is negative in nature then it will lie in the second and third quadrant, while $y=0.3532$ is positive so it will lie in the first and second quadrant.
So the point lies in quadrant II.
