Question

In an Argand diagram, the loci $arg\left( z-2i \right)=\dfrac{\pi }{6}$ and $\left| z-3 \right|=\left| z-3i \right|$ intersect at point P. Express the complex number represented by P in the form $r{{e}^{i\theta }}$ giving the exact value of $\theta $ and the value of r correct to 3 significant figures.

Answer 1

Hint: Assume $z=x+iy$ and use the formula ${{\left| z \right|}^{2}}={{x}^{2}}+{{y}^{2}}$ to form a relation between x and y considering the relation $\left| z-3 \right|=\left| z-3i \right|$. Form the second relation between x and y using the formula $\arg \left( z \right)={{\tan }^{-1}}\left( \dfrac{\operatorname{Im}\left( z \right)}{\operatorname{Re}\left( z \right)} \right)$ considering the relation $arg\left( z-2i \right)=\dfrac{\pi }{6}$. Solve for the values of x and y and use the formula $r=\sqrt{{{x}^{2}}+{{y}^{2}}}$ to get the value of r. To find the value of $\theta $ using the formula $\theta ={{\tan }^{-1}}\left( \dfrac{y}{x} \right)$ and check for the quadrant of the angles from the signs of values of x and y.

Complete step by step answer:
Here we have been provided with the complex number z and two relations $arg\left( z-2i \right)=\dfrac{\pi }{6}$ and $\left| z-3 \right|=\left| z-3i \right|$. We are asked to find the point of intersection of these curves and express it in the Euler’s form $r{{e}^{i\theta }}$.
Now, let us assume the complex number as $z=x+iy$. We know that modulus of a complex number is given as ${{\left| z \right|}^{2}}={{x}^{2}}+{{y}^{2}}$, so considering the relation $\left| z-3 \right|=\left| z-3i \right|$ we get,
\[\begin{align}
  & \Rightarrow \left| x+iy-3 \right|=\left| x+iy-3i \right| \\
 & \Rightarrow \left| \left( x-3 \right)+iy \right|=\left| x+i\left( y-3 \right) \right| \\
 & \Rightarrow {{\left( x-3 \right)}^{2}}+{{y}^{2}}={{x}^{2}}+{{\left( y-3 \right)}^{2}} \\
\end{align}\]
Expanding both the sides by using the algebraic identity \[{{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab\] we get,
\[\Rightarrow {{x}^{2}}+9-6x+{{y}^{2}}={{x}^{2}}+{{y}^{2}}+9-6y\]
Cancelling the like terms from both the sides we get,
\[\Rightarrow x=y\] ………. (1)
Now, consider the relation $arg\left( z-2i \right)=\dfrac{\pi }{6}$. We know that $\arg \left( z \right)={{\tan }^{-1}}\left( \dfrac{\operatorname{Im}\left( z \right)}{\operatorname{Re}\left( z \right)} \right)$, so we have,
$\begin{align}
  & \Rightarrow arg\left( x+iy-2i \right)=\dfrac{\pi }{6} \\
 & \Rightarrow arg\left( x+i\left( y-2 \right) \right)=\dfrac{\pi }{6} \\
 & \Rightarrow {{\tan }^{-1}}\left( \dfrac{y-2}{x} \right)=\dfrac{\pi }{6} \\
\end{align}$
We know that $\tan \dfrac{\pi }{6}=\dfrac{1}{\sqrt{3}}$ so simplifying using the cross multiplication we get,
\[\begin{align}
  & \Rightarrow \left( \dfrac{y-2}{x} \right)=\tan \left( \dfrac{\pi }{6} \right) \\
 & \Rightarrow \left( \dfrac{y-2}{x} \right)=\dfrac{1}{\sqrt{3}} \\
 & \Rightarrow \sqrt{3}\left( y-2 \right)=x \\
\end{align}\]
Substituting the value of y from equation (1) in the above obtained relation we get,
\[\begin{align}
  & \Rightarrow \sqrt{3}\left( x-2 \right)=x \\
 & \Rightarrow \left( \sqrt{3}-1 \right)x=2\sqrt{3} \\
 & \Rightarrow x=\dfrac{2\sqrt{3}}{\left( \sqrt{3}-1 \right)} \\
\end{align}\]
Rationalizing the denominator by multiplying and dividing by the conjugate $\left( \sqrt{3}+1 \right)$ we get,
\[\Rightarrow x=\dfrac{2\sqrt{3}}{\left( \sqrt{3}-1 \right)}\times \dfrac{\left( \sqrt{3}+1 \right)}{\left( \sqrt{3}+1 \right)}\]
Using the algebraic identity $\left( a+b \right)\left( a-b \right)=\left( {{a}^{2}}-{{b}^{2}} \right)$ we get,
\[\begin{align}
  & \Rightarrow x=\dfrac{2\sqrt{3}\left( \sqrt{3}+1 \right)}{3-1} \\
 & \Rightarrow x=\dfrac{2\sqrt{3}\left( \sqrt{3}+1 \right)}{2} \\
 & \Rightarrow x=3+\sqrt{3} \\
\end{align}\]
Substituting the value of x in equation (1) we get,
\[\Rightarrow y=3+\sqrt{3}\]
We know that in the Euler’s form of a complex number $z=r{{e}^{i\theta }}$ we have the relations $r=\sqrt{{{x}^{2}}+{{y}^{2}}}$ and $\theta ={{\tan }^{-1}}\left( \dfrac{y}{x} \right)$, so substituting the values of x and y we get,
$\begin{align}
  & \Rightarrow r=\sqrt{{{\left( 3+\sqrt{3} \right)}^{2}}+{{\left( 3+\sqrt{3} \right)}^{2}}} \\
 & \Rightarrow r=\sqrt{2\times {{\left( 3+\sqrt{3} \right)}^{2}}} \\
 & \Rightarrow r=\sqrt{2}\times \left( 3+\sqrt{3} \right) \\
 & \Rightarrow r=3\sqrt{2}+\sqrt{6} \\
\end{align}$
Substituting the values $\sqrt{2}=1.414$ and $\sqrt{3}=1.732$ and simplifying we get,
$\Rightarrow r=6.69$
Since both x and y are positive so the angle $\theta ={{\tan }^{-1}}\left( \dfrac{y}{x} \right)$ must lie in the first quadrant, so we get,
\[\begin{align}
  & \Rightarrow \theta ={{\tan }^{-1}}\left( \dfrac{3+\sqrt{3}}{3+\sqrt{3}} \right) \\
 & \Rightarrow \theta ={{\tan }^{-1}}\left( 1 \right) \\
\end{align}\]
We know that ${{\tan }^{-1}}\left( 1 \right)=\dfrac{\pi }{4}$ so we get,
\[\Rightarrow \theta =\dfrac{\pi }{4}\]
Hence, the point P can be written in the form $r{{e}^{i\theta }}$ as $6.69{{e}^{\left( \dfrac{\pi }{4} \right)i}}$ where r is correct to three significant figures.

Note: Remember the basic terms of complex numbers like conjugate, argument, modulus etc. We do not represent complex numbers on a real plane but there is a complex plane for their representation. You must be careful while finding the value of $\theta $ because the quadrant in which the angle lies depends on the signs of values of x and y just like a point lies in different quadrants for the different signs of x and y.