
What is the trigonometric form of \[\left( {12 - 2i} \right)\] ?
Answer
409.8k+ views
Hint: In the angle formed by the complex numbers on a polar graph with one imaginary axis. This can be found using the right angle trigonometry for the trigonometric functions. And trigonometry is that study related between the side lengths and angles of triangles.
Formula used:
There are some formula and in this problem
\[z = a + ib\]
\[ = \left| z \right|\left( {\cos \left( \theta \right) + i\sin \left( \theta \right)} \right)\]
\[ = \left| {{z^2}} \right| = {a^2} + {b^2}\]
Where,\[z = a + ib\]
\[\left| z \right| = \sqrt {{a^2} + {b^2}} \]
\[\theta = {\tan ^{ - 1}}\left( {\dfrac{y}{x}} \right)\]
Complete step-by-step solution:
We have to convert the complex number into a trigonometric form.
The given complex number is \[\left( {12 - 2i} \right)\]
This is the trigonometric form of complex number where \[\left| z \right|\] is the modulus and \[\theta \]is the angle created on the complex plane.
\[z = a + ib\]
\[ = \left| z \right|\left( {\cos \left( \theta \right) + i\sin \left( \theta \right)} \right)\]
The modulus of a complex number is the distance from the origin on the complex plane.
\[\left| z \right| = \sqrt {{a^2} + {b^2}} \] Where, \[z = a + ib\]
From this compare the equation to the given problem.
We get the values.
\[a = 12\] And \[b = - 2\]
Hence,
\[\left| z \right| = {\sqrt {{{12}^2} + \left( { - 2} \right)} ^2}\]
\[ = \sqrt {144 + 4} \]
\[ = \sqrt {148} \]
Take square root, we have,
\[
= \sqrt {4 \times 37} \\
= \sqrt {2 \times 2 \times 37} \\
= 2\sqrt {37} \\
\left| z \right| = 2\sqrt {37} \]
Or use the formula
\[\left| {{z^2}} \right| = {a^2} + {b^2}\]
Substitute the values we get,
\[\left| {{z^2}} \right| = {12^2} + \left( { - {2^2}} \right)\]
\[ = 144 + 4\]
\[\left| {{z^2}} \right| = 148\]
To find the value of \[\left| z \right|\], take square root
\[
\left| z \right| = \sqrt {148} \\
\left| z \right| = \sqrt {2 \times 2 \times 148} \\
\left| z \right| = 2\sqrt {37} \]
The angle of the point on the complex plane is the inverse tangent to the complex position over the real portion.
We have the formula,
\[\theta = {\tan ^{^{ - 1}}}\left( {\dfrac{{ - 2}}{{12}}} \right)\]
This is approximately equal,
\[ \approx - 0.165\]Radians
Therefore the trigonometric form of the given question is,
\[ = \left| z \right|\left( {\cos \left( \theta \right) + i\sin \left( \theta \right)} \right)\]
Substitute the value of
\[
\left| z \right| = 2\sqrt {37} \\
\theta = - 0.165 \\
= 2\sqrt {37} \left( {\cos \left( { - 0.165} \right) + i\sin \left( { - 0.165} \right)} \right) \]
Therefore the solution for this given problem,
\[ = 2\sqrt {37} \left( {\cos \left( { - 0.165} \right) + i\sin \left( { - 0.165} \right)} \right)\]
Note: Trigonometry is the type of mathematics. Polar was a form of complex numbers. The trigonometry is the branch of mathematics concerned with specific functions of angles and their application to calculations. There are six functions of an angle commonly used in trigonometry. Their names and abbreviations are sin, cos, tangent cotangent, secant and cosecant.
Formula used:
There are some formula and in this problem
\[z = a + ib\]
\[ = \left| z \right|\left( {\cos \left( \theta \right) + i\sin \left( \theta \right)} \right)\]
\[ = \left| {{z^2}} \right| = {a^2} + {b^2}\]
Where,\[z = a + ib\]
\[\left| z \right| = \sqrt {{a^2} + {b^2}} \]
\[\theta = {\tan ^{ - 1}}\left( {\dfrac{y}{x}} \right)\]
Complete step-by-step solution:
We have to convert the complex number into a trigonometric form.
The given complex number is \[\left( {12 - 2i} \right)\]
This is the trigonometric form of complex number where \[\left| z \right|\] is the modulus and \[\theta \]is the angle created on the complex plane.
\[z = a + ib\]
\[ = \left| z \right|\left( {\cos \left( \theta \right) + i\sin \left( \theta \right)} \right)\]
The modulus of a complex number is the distance from the origin on the complex plane.
\[\left| z \right| = \sqrt {{a^2} + {b^2}} \] Where, \[z = a + ib\]
From this compare the equation to the given problem.
We get the values.
\[a = 12\] And \[b = - 2\]
Hence,
\[\left| z \right| = {\sqrt {{{12}^2} + \left( { - 2} \right)} ^2}\]
\[ = \sqrt {144 + 4} \]
\[ = \sqrt {148} \]
Take square root, we have,
\[
= \sqrt {4 \times 37} \\
= \sqrt {2 \times 2 \times 37} \\
= 2\sqrt {37} \\
\left| z \right| = 2\sqrt {37} \]
Or use the formula
\[\left| {{z^2}} \right| = {a^2} + {b^2}\]
Substitute the values we get,
\[\left| {{z^2}} \right| = {12^2} + \left( { - {2^2}} \right)\]
\[ = 144 + 4\]
\[\left| {{z^2}} \right| = 148\]
To find the value of \[\left| z \right|\], take square root
\[
\left| z \right| = \sqrt {148} \\
\left| z \right| = \sqrt {2 \times 2 \times 148} \\
\left| z \right| = 2\sqrt {37} \]
The angle of the point on the complex plane is the inverse tangent to the complex position over the real portion.
We have the formula,
\[\theta = {\tan ^{^{ - 1}}}\left( {\dfrac{{ - 2}}{{12}}} \right)\]
This is approximately equal,
\[ \approx - 0.165\]Radians
Therefore the trigonometric form of the given question is,
\[ = \left| z \right|\left( {\cos \left( \theta \right) + i\sin \left( \theta \right)} \right)\]
Substitute the value of
\[
\left| z \right| = 2\sqrt {37} \\
\theta = - 0.165 \\
= 2\sqrt {37} \left( {\cos \left( { - 0.165} \right) + i\sin \left( { - 0.165} \right)} \right) \]
Therefore the solution for this given problem,
\[ = 2\sqrt {37} \left( {\cos \left( { - 0.165} \right) + i\sin \left( { - 0.165} \right)} \right)\]
Note: Trigonometry is the type of mathematics. Polar was a form of complex numbers. The trigonometry is the branch of mathematics concerned with specific functions of angles and their application to calculations. There are six functions of an angle commonly used in trigonometry. Their names and abbreviations are sin, cos, tangent cotangent, secant and cosecant.
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