
How do you find the trigonometric form of \[5-5i\]?
Answer
563.7k+ views
Hint: We have to convert the expression to polar coordinates form. For that, firstly, we will have to find the values of r using the x and y equivalent in the polar coordinates form and the value of \[\theta \] using the tangent function. The obtained values will be put in the formula of the polar coordinates which has the trigonometric form.
Complete step-by-step solution:
From the expression we have now, we will have to convert it into a form that is, \[z=r(\cos \theta +i\sin \theta )\]
But to get the given expression in the form \[z=x+iy\] into the trigonometric form we need the values of \[r\] and \[\theta \]. We know that,
\[{{r}^{2}}={{x}^{2}}+{{y}^{2}}\]
Writing in terms of r, we have
\[\Rightarrow r=\sqrt{{{x}^{2}}+{{y}^{2}}}\]
where \[x=r\cos \theta \] and \[y=r\sin \theta \]
on comparing the equation \[z=x+iy\], with the expression we have that is, \[5-5i\]
we can clearly state that,
\[x=5\] and \[y=-5\],
Putting the values of x and y in the formula of r, we have,
\[r=\sqrt{{{x}^{2}}+{{y}^{2}}}\]
\[\Rightarrow r=\sqrt{{{5}^{2}}+{{(-5)}^{2}}}\]
\[\Rightarrow r=\sqrt{25+25}\]
Adding up the terms and taking the root of the sum we have,
\[\Rightarrow r=\sqrt{50}=5\sqrt{2}\]
So, we have the value of \[r=5\sqrt{2}\]
Now, we have to find the value of \[\theta \], we know that
\[\tan \theta =\dfrac{y}{x}\]
which is according that \[\tan \theta =\dfrac{perpendicular}{base}\]
Now, substituting the values of x and y in the tangent function’s formula, we have,
\[\tan \theta =\dfrac{-5}{5}\]
\[\Rightarrow \tan \theta =-1\]
We have to find the value of angle for which \[\tan \theta =-1\]
To get the value of theta, we will take the inverse of tangent function, we now have
\[\Rightarrow \theta ={{\tan }^{-1}}(-1)\]
\[\Rightarrow \theta =-\dfrac{\pi }{4}or\dfrac{3\pi }{4}\]
As \[\tan \theta \] function is negative in the 2nd quadrant, so, we can have more than one coordinates for a single point in polar coordinates.
So, we have the values of \[r\] and \[\theta \], so we can proceed with the final expression in trigonometric form.
That is,
\[5-5i=5\sqrt{2}\left( \cos \left( -\dfrac{\pi }{4} \right)+i\sin \left( -\dfrac{\pi }{4} \right) \right)\]
Therefore, in trigonometric form we have, \[5-5i=5\sqrt{2}\left( \cos \left( -\dfrac{\pi }{4} \right)+i\sin \left( -\dfrac{\pi }{4} \right) \right)\].
Note: The trigonometric form is to be dealt with caution. The polar coordinates involved have properties that a single coordinate can be written in many forms. So, we can have many answers but its value should remain unchanged. While calculating the angle, the quadrant should be known in which a particular function is positive or negative, else it might result in the wrong answer.
Complete step-by-step solution:
From the expression we have now, we will have to convert it into a form that is, \[z=r(\cos \theta +i\sin \theta )\]
But to get the given expression in the form \[z=x+iy\] into the trigonometric form we need the values of \[r\] and \[\theta \]. We know that,
\[{{r}^{2}}={{x}^{2}}+{{y}^{2}}\]
Writing in terms of r, we have
\[\Rightarrow r=\sqrt{{{x}^{2}}+{{y}^{2}}}\]
where \[x=r\cos \theta \] and \[y=r\sin \theta \]
on comparing the equation \[z=x+iy\], with the expression we have that is, \[5-5i\]
we can clearly state that,
\[x=5\] and \[y=-5\],
Putting the values of x and y in the formula of r, we have,
\[r=\sqrt{{{x}^{2}}+{{y}^{2}}}\]
\[\Rightarrow r=\sqrt{{{5}^{2}}+{{(-5)}^{2}}}\]
\[\Rightarrow r=\sqrt{25+25}\]
Adding up the terms and taking the root of the sum we have,
\[\Rightarrow r=\sqrt{50}=5\sqrt{2}\]
So, we have the value of \[r=5\sqrt{2}\]
Now, we have to find the value of \[\theta \], we know that
\[\tan \theta =\dfrac{y}{x}\]
which is according that \[\tan \theta =\dfrac{perpendicular}{base}\]
Now, substituting the values of x and y in the tangent function’s formula, we have,
\[\tan \theta =\dfrac{-5}{5}\]
\[\Rightarrow \tan \theta =-1\]
We have to find the value of angle for which \[\tan \theta =-1\]
To get the value of theta, we will take the inverse of tangent function, we now have
\[\Rightarrow \theta ={{\tan }^{-1}}(-1)\]
\[\Rightarrow \theta =-\dfrac{\pi }{4}or\dfrac{3\pi }{4}\]
As \[\tan \theta \] function is negative in the 2nd quadrant, so, we can have more than one coordinates for a single point in polar coordinates.
So, we have the values of \[r\] and \[\theta \], so we can proceed with the final expression in trigonometric form.
That is,
\[5-5i=5\sqrt{2}\left( \cos \left( -\dfrac{\pi }{4} \right)+i\sin \left( -\dfrac{\pi }{4} \right) \right)\]
Therefore, in trigonometric form we have, \[5-5i=5\sqrt{2}\left( \cos \left( -\dfrac{\pi }{4} \right)+i\sin \left( -\dfrac{\pi }{4} \right) \right)\].
Note: The trigonometric form is to be dealt with caution. The polar coordinates involved have properties that a single coordinate can be written in many forms. So, we can have many answers but its value should remain unchanged. While calculating the angle, the quadrant should be known in which a particular function is positive or negative, else it might result in the wrong answer.
Recently Updated Pages
Basicity of sulphurous acid and sulphuric acid are

Master Class 11 Business Studies: Engaging Questions & Answers for Success

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

Master Class 11 Economics: Engaging Questions & Answers for Success

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

Master Class 11 English: Engaging Questions & Answers for Success

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

Explain zero factorial class 11 maths CBSE

State and prove Bernoullis theorem class 11 physics CBSE

What steps did the French revolutionaries take to create class 11 social science CBSE

The transition element that has lowest enthalpy of class 11 chemistry CBSE

Can anyone list 10 advantages and disadvantages of friction

