How do you write \[1-i\] in trigonometric form?

Hint: To solve the given question, first we need to write the trigonometric form of a complex number. The trigonometric form of a complex number \[z=a+bi\], is
\[z=r\left( \cos \theta +i\sin \theta \right)\], where \[r=\left| a+bi \right|\] is the modulus of \[z\], and\[\tan \theta =\dfrac{b}{a}\]. \[\theta \] is the argument of \[z\]. Normally, we will require \[0\le \theta \le 2\pi \].

Formula used:
The trigonometric form of a complex number \[z=a+bi\], is
\[z=r\left( \cos \theta +i\sin \theta \right)\], where
\[r=\left| a+bi \right|\] is the modulus of \[z\], and\[\tan \theta =\dfrac{b}{a}\].
\[\theta \] is the argument of \[z\].
Normally, we will require \[0\le \theta \le 2\pi \].

Complete step by step solution:
To express a complex number \[\left( a+ib \right)\]in trigonometric form,
i.e. \[r\left( \cos \theta +i\sin \theta \right)\], first we need to polar form of the given complex number, that will be written as,
\[\Rightarrow r{{e}^{i\theta }}\], where \['r'\] is the magnitude of the complex number.
\[r=\sqrt{{{a}^{2}}+{{b}^{2}}}\], and
\[\theta \]is the argument of the complex number which is equal to \[{{\tan }^{-1}}\dfrac{b}{a}\]. We have given the following complex number,
\[1-i\], where we can get the value of \[a\]and \[b\], i.e.
\[a=1,b=-1\]
Now, we find the magnitude of the complex number , we obtain
\[r=\sqrt{{{a}^{2}}+{{b}^{2}}}\]
\[\Rightarrow r=\sqrt{{{1}^{2}}+{{\left( -1 \right)}^{2}}}\]
\[\Rightarrow r=\sqrt{1+1}=\sqrt{2}\]
\[\Rightarrow r=\sqrt{2}\]
Now, we will find the argument of the complex number, we obtain
\[\theta ={{\tan }^{-1}}\left( \dfrac{b}{a} \right)\]
\[\Rightarrow\tan \theta =\dfrac{b}{a}\]
\[\Rightarrow\tan \theta =\dfrac{-1}{1}=-1\]
\[\tan \theta =1\], as magnitude accepts only positive values.
By using trigonometric ratios table,
\[\tan \theta =\tan \dfrac{\pi }{4}\]
\[\theta =\dfrac{\pi }{4}\] (it is the reference angle)
Now, if we consider that the complex number \[1-i\] is graphed in Quadrant IV of the complex plane.
In this case,
\[\theta =2\pi -\dfrac{\pi }{4}=\dfrac{7\pi }{4}\]
Thus, the complex number \[1-i\] in trigonometric form will be written as,
\[r\left( \cos \theta +i\sin \theta \right)\], where
\[r=\sqrt{2}\]and \[\theta =\dfrac{7\pi }{4}\]
After substituting the value ,we get
\[\sqrt{2}\left( \cos \left( \dfrac{7\pi }{4} \right)+i\sin \left( \dfrac{7\pi }{4} \right) \right)\]
Thus, it is the trigonometric form.

Note: In this type of question students need to have basic knowledge of trigonometry and carefully watch the corresponding quadrant. Check the quadrant after finding the argument because in trigonometry in a complete period of \[2\pi \] there exists two equal values for any argument of trigonometric functions, so checking the quadrant will solve this problem.