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Scorpion4ik [409]
2 years ago
6

Write the complex number -2+2i in polar form

Mathematics
1 answer:
FromTheMoon [43]2 years ago
8 0
For polar form you need to find the modulus (length of the vector) and the argument (angle of the vector) and present in form rcis(Arg) or re^Argi

start with the modulus r=sqrt(a^2 +b^2)
                                     =sqrt(-2^2 +2^2)
                                     = sqrt(4+4)
                                     =sqrt(8)
                                     =2sqrt(2)

next the argument, firstly arg=tan(b/a)
                                              = tan(2/2)
                                              =tan(1)
                                             =pi/4 .     (exact values table)
Now consider the quadrant the complex number is in, as it is (-2,2) it is in the second quadrant and as such your Arg value is:
Arg=pi-arg
     = pi-pi/4
     = 3pi/4

add it all together and your complex number in polar form is:
2sqrt2cis(3pi/4)

note: cis is short hand for cos(x)+isin(x), it is possible your tutor would rather you use the complex exponential form which is simply re^Argi and your answer would look like:
2sqrt2e^(3pi/4)i

Also notice the difference between arg and Arg as this often slips students up and always present Arg in prinicple argument form ie -pi<Arg<pi

Hopefully this has been clear enough and good luck

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The absolute expression |x - y| is rewritten in the form of without using absolute value notation will be x - y.

<h3>What is an absolute function?</h3>

The absolute function is also known as the mode function. The value of the absolute function is always positive.

The absolute function is given as

f(x) = | x |

If the value inside the mode operator is greater than zero, then simply the mode operator is eliminated.

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Then the absolute function is given as,

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The absolute expression |x - y| is rewritten in the form of without using absolute value notation will be x - y.

More about the absolute function link is given below.

brainly.com/question/10664936

#SPJ1

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