Solveeit Logo
Login

Question

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

How do you write 1−i1-i in trigonometric form?

Explanation

Solution

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+biz=a+bi, is
z=r(cos⁡θ+isin⁡θ)z=r\left( \cos \theta +i\sin \theta \right), where r=∣a+bi∣r=\left| a+bi \right| is the modulus of zz, andtan⁡θ=ba\tan \theta =\dfrac{b}{a}. θ\theta is the argument of zz. Normally, we will require 0≤θ≤2π0\le \theta \le 2\pi .

Formula used:
The trigonometric form of a complex number z=a+biz=a+bi, is
z=r(cos⁡θ+isin⁡θ)z=r\left( \cos \theta +i\sin \theta \right), where
r=∣a+bi∣r=\left| a+bi \right| is the modulus of zz, andtan⁡θ=ba\tan \theta =\dfrac{b}{a}.
θ\theta is the argument of zz.
Normally, we will require 0≤θ≤2π0\le \theta \le 2\pi .

Complete step by step solution:
To express a complex number (a+ib)\left( a+ib \right)in trigonometric form,
i.e. r(cos⁡θ+isin⁡θ)r\left( \cos \theta +i\sin \theta \right), first we need to polar form of the given complex number, that will be written as,
⇒reiθ\Rightarrow r{{e}^{i\theta }}, where ′r′'r' is the magnitude of the complex number.
r=a2+b2r=\sqrt{{{a}^{2}}+{{b}^{2}}}, and
θ\theta is the argument of the complex number which is equal to tan⁡−1ba{{\tan }^{-1}}\dfrac{b}{a}. We have given the following complex number,
1−i1-i, where we can get the value of aaand bb, i.e.
a=1,b=−1a=1,b=-1
Now, we find the magnitude of the complex number , we obtain
r=a2+b2r=\sqrt{{{a}^{2}}+{{b}^{2}}}
⇒r=12+(−1)2\Rightarrow r=\sqrt{{{1}^{2}}+{{\left( -1 \right)}^{2}}}
⇒r=1+1=2\Rightarrow r=\sqrt{1+1}=\sqrt{2}
⇒r=2\Rightarrow r=\sqrt{2}
Now, we will find the argument of the complex number, we obtain
θ=tan⁡−1(ba)\theta ={{\tan }^{-1}}\left( \dfrac{b}{a} \right)
⇒tan⁡θ=ba\Rightarrow\tan \theta =\dfrac{b}{a}
⇒tan⁡θ=−11=−1\Rightarrow\tan \theta =\dfrac{-1}{1}=-1
tan⁡θ=1\tan \theta =1, as magnitude accepts only positive values.
By using trigonometric ratios table,
tan⁡θ=tan⁡π4\tan \theta =\tan \dfrac{\pi }{4}
θ=π4\theta =\dfrac{\pi }{4} (it is the reference angle)
Now, if we consider that the complex number 1−i1-i is graphed in Quadrant IV of the complex plane.
In this case,
θ=2π−π4=7π4\theta =2\pi -\dfrac{\pi }{4}=\dfrac{7\pi }{4}
Thus, the complex number 1−i1-i in trigonometric form will be written as,
r(cos⁡θ+isin⁡θ)r\left( \cos \theta +i\sin \theta \right), where
r=2r=\sqrt{2}and θ=7π4\theta =\dfrac{7\pi }{4}
After substituting the value ,we get
2(cos⁡(7π4)+isin⁡(7π4))\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 2π2\pi there exists two equal values for any argument of trigonometric functions, so checking the quadrant will solve this problem.