Question

Write the trigonometric form of \[2\,-\,2\,i\] ?

Answer 1

Hint: In this question we have write trigonometric form of the given complex number form of any complex number first we have to find the values of modules of \[z\] that is \[\left| z \right|\] and Also, find the value of \[\theta \]. After getting these two values we will put these two values in polar form that is \[r\left( \cos \theta +i\sin \theta \right)\]. Where \[r\] is nothing but value of \[\left| z \right|\] after simplifying it we will get our required answer.

Complete step by step solution:
In this question we have given the complex number is \[2\,-\,2\,i\]
We will complex it with a complex number \[z=a+ib\] and we get \[a=2\] & \[b=-2\] from a complex number.
In order to get trigonometric first we will find the value of \[r\].
As we know that \[r=\left| z \right|=\sqrt{{{a}^{2}}+{{b}^{2}}}\] , where \[z\,=\,a+ib\].

Which implies that \[r=\sqrt{{{(2)}^{2}}+{{(-2)}^{2}}}\]
Clearly, we can observe that \[z=2-2i\] is represented by point \[(+2,-2\] which lies on \[{{4}^{th}}\] quadrant.
As we know that \[\alpha ={{\tan }^{-1}}\left( \dfrac{Im(z)}{\operatorname{Re}(z)} \right)\]
From \[\left( z \right)\] here we have real part of \[z\], that is \[\operatorname{Re}(z)=2\] and imaginary part of \[z\], that is \[\operatorname{Im}(z)=-2\] which implies
Since we know that \[\tan \left(\dfrac{\pi }{4} \right)=1\]
Which implies \[\alpha =-\dfrac{\pi }{4}\].
Now, rename \[\alpha \] as \[\theta \].
As we know that the equation of trigonometric function (polar form) is given as \[r(\cos \theta +i\sin \theta )\] then
\[\Rightarrow \]The required trigonometric form of given complex number \[2-2i\] is \[2\sqrt{2}\left[ \cos\left(\dfrac{\pi }{4}\right) - i \sin\left(\dfrac{\pi }{4} \right) \right]\] .

Note: A complex number represented by an expression of the form \[a+ib\] where \[a\] and \[b\] are real numbers.
Complex numbers can be represented as follows.
\[z\,=\,a+ib\]; where \[a\] and \[b\]are real numbers.