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Question: How do you write the trigonometric form into a complex number in standard form \(3\left( \cos 120+i\...

How do you write the trigonometric form into a complex number in standard form 3(cos120+isin120)?3\left( \cos 120+i\sin 120 \right)?

Explanation

Solution

The trigonometric form of complex number can be written in the standard rom i.e. z=a+biz=a+bi which is rectangular form and the polar form is z=r(cosθ+isinθ)z=r\left( \cos \theta +i\sin \theta \right) where r=a+bir=\left| a+bi \right| is the modules of zz and the angle θ\theta can be calculated by formula θ=tan1(ba)\theta ={{\tan }^{-1}}\left( \dfrac{b}{a} \right)

Complete step-by-step answer:
So, here we have to write the trigonometric form into a complex number in a standard form.
In the problem given trigonometric form of complex number is 3(cos120+isin120)3\left( \cos 120+i\sin 120 \right)
Here the value of cos120\cos 120 is 12\dfrac{-1}{2}
And the value of sin120\sin 120 is 32\dfrac{\sqrt{3}}{2}
Therefore, the trigonometric form of complex number 3(cos120+isin120)3\left( \cos 120+i\sin 120 \right) can be written as
3(cos120+isin120)=3(12+i32)3\left( \cos 120+i\sin 120 \right)=3\left( \dfrac{-1}{2}+i\dfrac{\sqrt{3}}{2} \right)
Now, multiplying the 33 to the terms in the bracket. Therefore we have,
3(cos120+isin120)=32+i323\left( \cos 120+i\sin 120 \right)=\dfrac{-3}{2}+i\dfrac{\sqrt{3}}{2}

So, the trigonometric form into a complex number in standard form of 3(cos120+isin120)3\left( \cos 120+i\sin 120 \right) is (32+i332)\left( \dfrac{-3}{2}+i\dfrac{3\sqrt{3}}{2} \right)

Additional Information:
Here, Trigonometric polar form of complex number can be written as,
z=r(cosθ+isinθ)z=r\left( \cos \theta +i\sin \theta \right)
Where, r=zr=\left| z \right| and θ=\theta = angle (z)\left( z \right)
Here, z=a+bi,z=a+bi, therefore modules of zz can be calculated as,
z=a+bi\left| z \right|=\left| a+bi \right| which is equal to r'r'
Therefore,
r=a+bir=\left| a+bi \right|
r=a2+b2r=\left| \sqrt{{{a}^{2}}+{{b}^{2}}} \right|
And the angle θ\theta can be calculated as,
θ=tan1(ba)\theta ={{\tan }^{-1}}\left( \dfrac{b}{a} \right)
Then, we get the polar coordinates that is (r;θ)\left( r;\theta \right) which help in deriving the trigonometric polar form i.e.,
Trigonometric form =r(cosθ+isinθ)=r\left( \cos \theta +i\sin \theta \right)
Above is the trigonometric polar form of complex number but to convert the trigonometric polar form into a complex number of rectangular form or standard form. We have, z=r(cosθ+isinθ)z=r\left( \cos \theta +i\sin \theta \right) then convert into standard rectangular form that is z=a+biz=a+bi
Then, applying the below formula for the value of a'a' and b'b' that is.
a=rcosθa=r\cos \theta
b=rsinθb=r\sin \theta
Then put the value of aa and bb in z=a+biz=a+bi you will get the rectangular (standard) form.

Note:
Find the sine and cosine angle accurately while solving the problem. Multiply the r'r' value appropriately to the sine and cosine value. Here r'r' means the radius of the circle which corresponds to the hypotenuse of the right triangle for your angle θ'\theta '