Math, asked by Anonymous, 1 year ago

how to write polar form of a complex numbers?

Answers

Answered by rajat21
0
if n=x+iy
r = root x^2+y^2
theta =tan inverse y/x
Answered by Tanya2405
1
The polar form of a complex number is another way to represent a complex number. The form z=a+biz=a+bi is called the rectangular coordinate form of a complex number.

The horizontal axis is the real axis and the vertical axis is the imaginary axis. We find the real and complex components in terms of rr and θθ where rr is the length of the vector and θθ is the angle made with the real axis.

From Pythagorean Theorem :

r2=a2+b2r2=a2+b2

By using the basic trigonometric ratios :

cosθ=arcosθ=ar and sinθ=brsinθ=br .

Multiplying each side by rr :

rcosθ=a  and  rsinθ=brcosθ=a  and  rsinθ=b

The rectangular form of a complex number is given by

z=a+biz=a+bi .

Substitute the values of aa and bb .

z=a+bi    =rcosθ+(rsinθ)i    =r(cosθ+isinθ)z=a+bi    =rcosθ+(rsinθ)i    =r(cosθ+isinθ)

In the case of a complex number, rrrepresents the absolute value or modulus and the angle θθ is called the argument of the complex number.

This can be summarized as follows:

The polar form of a complex numberz=a+biz=a+bi is z=r(cosθ+isinθ)z=r(cosθ+isinθ) , wherer=|z|=a2+b2−−−−−−√r=|z|=a2+b2 ,a=rcosθ  and  b=rsinθa=rcosθ  and  b=rsinθ , andθ=tan−1(ba)θ=tan−1(ba) for a>0a>0 andθ=tan−1(ba)+πθ=tan−1(ba)+π orθ=tan−1(ba)+180°θ=tan−1(ba)+180° for a<0a<0 .

Example:

Express the complex number in polar form.

5+2i5+2i

The polar form of a complex numberz=a+biz=a+bi is z=r(cosθ+isinθ)z=r(cosθ+isinθ) .

So, first find the absolute value of rr .

r=|z|=a2+b2−−−−−−√    =52+22−−−−−−√    =25+4−−−−−√    =29−−√     ≈5.39r=|z|=a2+b2    =52+22    =25+4    =29     ≈5.39

Now find the argument θθ .

Since a>0a>0 , use the formulaθ=tan−1(ba)θ=tan−1(ba) .

θ=tan−1(25)    ≈0.38θ=tan−1(25)    ≈0.38

Note that here θθ is measured in radians.

Therefore, the polar form of 5+2i5+2i is about 5.39(cos(0.38)+isin(0.38))5.39(cos(0.38)+isin(0.38)) .

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Anonymous: so so much thnx dear
Tanya2405: welcome
rajat21: plz check my ans
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