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Question: Express the following complex number in polar form and exponential form: \[ - 1 - \sqrt 3 i\]....

Express the following complex number in polar form and exponential form: −1−3i - 1 - \sqrt 3 i.

Explanation

Solution

Complex numbers are the numbers that are expressed in the form of a+iba + ib where, a, b are real numbers and ‘i’ is an imaginary number. Here in the given question, we have the terms as; −1−3i - 1 - \sqrt 3 i, in which -1 is a real number and -3i is a complex number, hence here to Express the following complex number in polar form and exponential form we need to apply the formula and express the given complex number.

Formula used:
To express the given number in polar form we have:z=r(cos⁡θ+isin⁡θ)z = r\left( {\cos \theta + i\sin \theta } \right)
r=a2+b2r = \sqrt {{a^2} + {b^2}} ; r,θr,\theta are the polar coordinates and a, b are the given complex numbers.
To express the given number in exponential form we have:z=reiθz = r{e^{i\theta }}

Complete step by step solution:
Given,
−1−3i- 1 - \sqrt 3 i
Which is of the complex form a+iba + ib, i.e.,
z=−1−i3z = - 1 - i\sqrt 3
Hence, let us find the complex number in polar form i.e., z=r(cos⁡θ+isin⁡θ)z = r\left( {\cos \theta + i\sin \theta } \right), hence
Here, a=−1a = - 1 and b=−3b = - \sqrt 3 , we need to find the value of r and θ\theta , so
r=a2+b2r = \sqrt {{a^2} + {b^2}}
Substitute the value of a and b as:
=(−1)2+(−3)2= \sqrt {{{\left( { - 1} \right)}^2} + {{\left( { - \sqrt 3 } \right)}^2}}
=1+3= \sqrt {1 + 3}
Evaluating the terms, we get:
r=4r = \sqrt 4
⇒r=2\Rightarrow r = 2
Now, we need to find the angle θ\theta , hence we have:
tan⁡θ=ba\tan \theta = \dfrac{b}{a}
As, both the values of a and b are negative, hence they lie in third quadrant, hence we need to add 180 degrees to the angle as:
tan⁡θ=ba+180∘;a<0\tan \theta = \dfrac{b}{a} + 180^\circ ;a < 0 (since, it lies in third quadrant)
Substitute the value of a and b as:
tan⁡θ=−3−1+180\tan \theta = \dfrac{{ - \sqrt 3 }}{{ - 1}} + 180
Since, we need to find the angle θ\theta , hence we need to take tan inverse of that number as:
θ=tan⁡−1(3)+180\theta = {\tan ^{ - 1}}\left( {\sqrt 3 } \right) + 180
We know that, tan⁡−1(3)=π3{\tan ^{ - 1}}\left( {\sqrt 3 } \right) = \dfrac{\pi }{3}or π3=60\dfrac{\pi }{3} = 60 hence we get:
θ=π3+180\theta = \dfrac{\pi }{3} + 180
or θ=60∘+180∘\theta = 60^\circ + 180^\circ
⇒θ=240∘\Rightarrow \theta = 240^\circ
We know that 240 degrees in radians we have:240∘=3π2240^\circ = \dfrac{{3\pi }}{2}
Now, substitute the values of r and θ\theta in the formula we get:
z=r(cos⁡θ+isin⁡θ)z = r\left( {\cos \theta + i\sin \theta } \right)
⇒z=2(cos⁡3π2+isin⁡3π2)\Rightarrow z = 2\left( {\cos \dfrac{{3\pi }}{2} + i\sin \dfrac{{3\pi }}{2}} \right)
As, z we have z=−1−i3z = - 1 - i\sqrt 3 , hence the polar form is:
−1−i3=2(cos⁡3π2+isin⁡3π2)- 1 - i\sqrt 3 = 2\left( {\cos \dfrac{{3\pi }}{2} + i\sin \dfrac{{3\pi }}{2}} \right)
Now, the given complex number in exponential form we have:
z=reiθz = r{e^{i\theta }}
⇒z=2ei3π2\Rightarrow z = 2{e^{i\dfrac{{3\pi }}{2}}}.
As, z we have z=−1−i3z = - 1 - i\sqrt 3 , hence the exponential form is:
−1−i3=2ei3π2- 1 - i\sqrt 3 = 2{e^{i\dfrac{{3\pi }}{2}}}.

Note: The key point to solve the given question, is that we must note the formulas to express the given numbers in polar and exponential form, such that while finding the angles we must know all the trigonometric identities ratios. And we must note the imaginary number is usually represented by ‘i’ or ‘j’, which is equal to −1\sqrt { - 1} . As we know, 0 is a real number and real numbers are part of complex numbers, therefore, 0 is also a complex number and is represented as 0+0i0 + 0i.