Question

How do you write the complex number \[-2-3i\] in polar form?

Answer 1

Hint: We are given an expression of complex numbers which is in standard form. We are asked to write it in polar form. Polar form is represented as \[r(\cos \theta +i\sin \theta )\], where r is the modulus and \[\theta \] is called the argument. We can assume, \[-2-3i=x+iy\]. We will then find the ‘r’ using the formula \[r=\sqrt{{{x}^{2}}+{{y}^{2}}}\]. Then, we will find the argument, \[\theta \] using the formula \[\theta ={{\tan }^{-1}}\left( \dfrac{y}{x} \right)\]. After finding the values, we will substitute these in the equation of the polar form.

Complete step by step answer:
According to the given question, we are given an expression on complex numbers which is in the standard form. We are now asked to write the given expression in the polar form.
Complex numbers can be expressed using polar form. The general representation of a polar equation is,
\[r(\cos \theta +i\sin \theta )\]-----(1)
 where r is the modulus and \[\theta \] is called the argument.
We will first assume that,
\[-2-3i=x+iy\]-----(2)
So, \[x=-2\] and \[y=-3\] is what we got.
Now, we will first find the ‘r’, which is the modulus or the absolute value, we have,
\[r=\sqrt{{{x}^{2}}+{{y}^{2}}}\]
Substituting the values of \[x=-2\] and , we get,
\[\Rightarrow r=\sqrt{{{(-2)}^{2}}+{{(-3)}^{2}}}\]
Solving further, we get the value of the expression as,
\[\Rightarrow r=\sqrt{4+9}\]
\[\Rightarrow r=\sqrt{13}\]
\[\Rightarrow r=3.6\]----(3)
Now, we will find the argument, \[\theta \], we have,
\[\theta ={{\tan }^{-1}}\left( \dfrac{y}{x} \right)\]
Substituting the values of x and y, we get,
\[\Rightarrow \theta ={{\tan }^{-1}}\left( \dfrac{-3}{-2} \right)\]
The negative sign gets cancelled, we get,
\[\Rightarrow \theta ={{\tan }^{-1}}\left( \dfrac{3}{2} \right)\]
\[\Rightarrow \theta ={{\tan }^{-1}}\left( 1.5 \right)\]
\[\Rightarrow \theta ={{56.18}^{\circ }}\]-----(4)
Now, we will substitute the equations (3) and (4) in equation (1) and we get,
\[3.6\left( \cos ({{56.18}^{\circ }})+i\sin ({{56.18}^{\circ }}) \right)\]

Therefore, the polar form of the given expression is \[3.6\left( \cos ({{56.18}^{\circ }})+i\sin ({{56.18}^{\circ }}) \right)\].

Note: The polar form must be neatly calculated. Also, care should be taken while deriving the value of the ‘r’ and \[\theta \]. The later substitution and further solving should be done in a progressive manner. If you don’t know the values of inverse functions, you can skip that part and straightaway write it as,
\[3.6\left( \cos \left( {{\tan }^{-1}}\left( 1.5 \right) \right)+i\sin \left( {{\tan }^{-1}}\left( 1.5 \right) \right) \right)\].