Question

# If $P$ is the point in the argand diagram corresponding to the complex number $\sqrt{3}+i$ and if $OPQ$ is an isosceles right angled triangle, right angled at $O$, then $Q$ represents the complex number

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Solution

## Step 1: Finding the value slopesFor the graphical representation of complex numbers, an argand diagram is used, if the complex number is in the form of $x+iy$, in the complex plane. Similar to the x−axis and y−axis in the two-dimensional geometry.Given the complex number $P=\sqrt{3}+i$, and $O$is centerLet us draw a figure, according to the question for understanding The figure shows an Isosceles right-angled triangle $OPQ$If two lines are perpendicular to each other then the product of their slopes will be equal to $-1$, that is ${m}_{1}×{m}_{2}=-1$The slope of line OP is $\begin{array}{c}{m}_{1}=\frac{1-0}{\sqrt{3}-0}\\ =\frac{1}{\sqrt{3}}\end{array}$Let $Q=\left(x,y\right)$ , then the slope ${m}_{2}$ of line OQ;$\begin{array}{c}{m}_{2}=\frac{y-0}{x-0}\\ =\frac{y}{x}\end{array}$Step 2: Substitute the values of slopes into ${m}_{1}×{m}_{2}=-1$, we get $\therefore \frac{1}{\sqrt{3}}×\frac{y}{x}=-1\phantom{\rule{0ex}{0ex}}⇒y=-\sqrt{3}x......\left(i\right)$Step 3: Using the properties of isosceles triangles, Congruent sides of the isosceles triangle are equal, then $OP=OQ$,Use the distance formula between two points $d=\sqrt{{\left({x}_{2}-{x}_{1}\right)}^{2}+{\left({y}_{2}-{y}_{1}\right)}^{2}}$, then $\begin{array}{c}OP=\sqrt{{\left(\sqrt{3}-0\right)}^{2}+{\left(1-0\right)}^{2}}\\ =\sqrt{3+1}\\ =\sqrt{4}\end{array}$Similarly, $\begin{array}{c}OQ=\sqrt{{\left(x-0\right)}^{2}+{\left(y-0\right)}^{2}}\\ =\sqrt{{x}^{2}+{y}^{2}}\end{array}$Since $OP=OQ$, then $\begin{array}{c}OP=OQ\\ O{P}^{2}=O{Q}^{2}\\ 4={x}^{2}+{y}^{2}\end{array}$Put $y=-\sqrt{3}x$ from equation (iii), in the above equation, $\begin{array}{c}⇒4={x}^{2}+{\left(-\sqrt{3}x\right)}^{2}\\ ⇒4={x}^{2}+3{x}^{2}\\ ⇒4=4{x}^{2}\\ ⇒{x}^{2}=1\\ ⇒x=±1\end{array}$Step 4: Substitute $x=1\text{and}x=-1$ in the equation (iii) to find the ordered pair $Q\left(x,y\right)$When $x=1$, then $y=-\sqrt{3}$, and $x=-1$, then $y=\sqrt{3}$,Thus, there can be two possible values of $Q\left(x,y\right)$ are $\left(1,-\sqrt{3}\right)$ and $\left(-1,\sqrt{3}\right)$.Hence, the answer can be written $Q$ represents the complex number either $1-i\sqrt{3}$ or $-1+i\sqrt{3}$.

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