
How to convert \[3 - 3i\] into trigonometric form.
Answer
453.6k+ views
Hint: When such complex number as above is to be converted into it’s trigonometric form we directly compare the real and imaginary parts of given complex number to the generalized trigonometric form of complex number, which is:-
\[ \Rightarrow \]\[r\cos \theta + i\left( {r\sin \theta } \right)\], where r is a positive number, \[\theta \] is the principal argument.
Complete step by step answer:
So initially in this question we will compare the real part of the complex number and the generalized trigonometric form.
\[ \Rightarrow \]\[r\cos \theta \]=3,
Further,
\[ \Rightarrow \]\[\cos \theta \]=\[\dfrac{3}{r}\]\[\xrightarrow{{}}eqn1\]
Now we will compare the imaginary part of the complex number and the generalized trigonometric form
\[ \Rightarrow \]r\[\sin \theta \]=-3,
Further,
\[ \Rightarrow \]\[\sin \theta \]= \[\dfrac{3}{r}\] \[\xrightarrow{{}}eqn2\]
Now we will square both equation 1 and 2 and add them,
\[ \Rightarrow \]\[{\sin ^2}\theta \]+\[{\cos ^2}\theta \]=\[2\] \[{\left( {\dfrac{3}{r}} \right)^2}\],
We know that \[{\sin ^2}\theta \]+\[{\cos ^2}\theta \]=1,
\[ \Rightarrow \]1=\[\dfrac{{18}}{{{r^2}}}\],
Therefore,
\[ \Rightarrow \]\[r = \pm 3\sqrt 2 \], but since r is a positive number, therefore,
\[ \Rightarrow \]\[r = + 3\sqrt 2 \]
Now after taking \[3\sqrt 2 \]common from \[3 - 3i\], we get,
\[ \Rightarrow \]\[3\sqrt 2 \left( {\dfrac{1}{{\sqrt 2 }} + i\left( {\dfrac{{ - 1}}{{\sqrt 2 }}} \right)} \right)\]
Now we will compare above equation with \[r\left( {\cos \theta + i\sin \theta } \right)\]
After comparing we get \[\cos \theta \]=\[\dfrac{1}{{\sqrt 2 }}\] and \[\sin \theta \]=\[\left( {\dfrac{{ - 1}}{{\sqrt 2 }}} \right)\],
For \[x \in \left[ { - \pi ,\pi } \right]\], above condition is satisfied only when \[\theta \]=\[ - \dfrac{\pi }{4}\].
Therefore, \[3 - 3i\] in trigonometric form can be written as \[3\sqrt 2 \left( {\cos \left( { - \dfrac{\pi }{4}} \right) + i\sin \left( { - \dfrac{\pi }{4}} \right)} \right)\].
Note: While solving such types of questions we should always keep in mind that the principal argument lies between -pi to +pi. So while choosing the appropriate angle, we must not choose an angle which does not lie between -pi to +pi. For example in above case we could also choose \[\dfrac{{7\pi }}{4}\] as an angle, but it does not lie between the intervals. Thus, we choose \[ - \dfrac{\pi }{4}\].
\[ \Rightarrow \]\[r\cos \theta + i\left( {r\sin \theta } \right)\], where r is a positive number, \[\theta \] is the principal argument.
Complete step by step answer:
So initially in this question we will compare the real part of the complex number and the generalized trigonometric form.
\[ \Rightarrow \]\[r\cos \theta \]=3,
Further,
\[ \Rightarrow \]\[\cos \theta \]=\[\dfrac{3}{r}\]\[\xrightarrow{{}}eqn1\]
Now we will compare the imaginary part of the complex number and the generalized trigonometric form
\[ \Rightarrow \]r\[\sin \theta \]=-3,
Further,
\[ \Rightarrow \]\[\sin \theta \]= \[\dfrac{3}{r}\] \[\xrightarrow{{}}eqn2\]
Now we will square both equation 1 and 2 and add them,
\[ \Rightarrow \]\[{\sin ^2}\theta \]+\[{\cos ^2}\theta \]=\[2\] \[{\left( {\dfrac{3}{r}} \right)^2}\],
We know that \[{\sin ^2}\theta \]+\[{\cos ^2}\theta \]=1,
\[ \Rightarrow \]1=\[\dfrac{{18}}{{{r^2}}}\],
Therefore,
\[ \Rightarrow \]\[r = \pm 3\sqrt 2 \], but since r is a positive number, therefore,
\[ \Rightarrow \]\[r = + 3\sqrt 2 \]
Now after taking \[3\sqrt 2 \]common from \[3 - 3i\], we get,
\[ \Rightarrow \]\[3\sqrt 2 \left( {\dfrac{1}{{\sqrt 2 }} + i\left( {\dfrac{{ - 1}}{{\sqrt 2 }}} \right)} \right)\]
Now we will compare above equation with \[r\left( {\cos \theta + i\sin \theta } \right)\]
After comparing we get \[\cos \theta \]=\[\dfrac{1}{{\sqrt 2 }}\] and \[\sin \theta \]=\[\left( {\dfrac{{ - 1}}{{\sqrt 2 }}} \right)\],
For \[x \in \left[ { - \pi ,\pi } \right]\], above condition is satisfied only when \[\theta \]=\[ - \dfrac{\pi }{4}\].
Therefore, \[3 - 3i\] in trigonometric form can be written as \[3\sqrt 2 \left( {\cos \left( { - \dfrac{\pi }{4}} \right) + i\sin \left( { - \dfrac{\pi }{4}} \right)} \right)\].
Note: While solving such types of questions we should always keep in mind that the principal argument lies between -pi to +pi. So while choosing the appropriate angle, we must not choose an angle which does not lie between -pi to +pi. For example in above case we could also choose \[\dfrac{{7\pi }}{4}\] as an angle, but it does not lie between the intervals. Thus, we choose \[ - \dfrac{\pi }{4}\].
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