How to convert \[3 - 3i\] into trigonometric form.
Answer
567k+ 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 \[ - \dfrac{\pi }{4}\].
Recently Updated Pages
Three beakers labelled as A B and C each containing 25 mL of water were taken A small amount of NaOH anhydrous CuSO4 and NaCl were added to the beakers A B and C respectively It was observed that there was an increase in the temperature of the solutions contained in beakers A and B whereas in case of beaker C the temperature of the solution falls Which one of the following statements isarecorrect i In beakers A and B exothermic process has occurred ii In beakers A and B endothermic process has occurred iii In beaker C exothermic process has occurred iv In beaker C endothermic process has occurred

Master Class 11 Social Science: Engaging Questions & Answers for Success

Master Class 11 Physics: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 Computer Science: Engaging Questions & Answers for Success

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

There are 720 permutations of the digits 1 2 3 4 5 class 11 maths CBSE

State and prove Bernoullis theorem class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

Discuss the various forms of bacteria class 11 biology CBSE

