I want help to find tan of polar correctly

[r-soy]

Hi



I face some problem when i try to find tan in like this questions



for example : write in polar

z1 = 1 - i

the answer is -5π/4

but my answer is

tan Q = y/x = -1/1 = -1



how I get -5π/4



--------------



ex : 2



z = -root3 + i

my answer is

tanQ = -1/root3 = -0.143

but the answer is = 5π/6



How i get the answer 5π/6



----





please help me >>

 

[Susanne217]

Hi



I face some problem when i try to find tan in like this questions



for example : write in polar

z1 = 1 - i

the answer is -5π/4

but my answer is

tan Q = y/x = -1/1 = -1



how I get -5π/4



--------------



ex : 2



z = -root3 + i

my answer is

tanQ = -1/root3 = -0.143

but the answer is = 5π/6



How i get the answer 5π/6



----





please help me >>

A complex number on the form $$x+iy$$ and the point of your exercise is to find $\theta$

you know that

$$tan(\theta) = \frac{y}{x}$$ Understand?

 

[HallsofIvy]

Hi



I face some problem when i try to find tan in like this questions



for example : write in polar

z1 = 1 - i

the answer is -5π/4

but my answer is

tan Q = y/x = -1/1 = -1



how I get -5π/4

Do you even know what question you are trying to answer?
"$- 5\pi/4$" is NOT the answer to the question posed, to write 1- i in polar form but it is part of the answer. Yes, tan(Q)= -1 and a calculator will tell you that $Q= tan^{-1}(-1)= -45$ degrees or $-\pi/4$ radians. You should also be able to recognize that tangent is 1 when the "near side" and "opposite" side are the same- an isosceles right triangle with angles or 45 degrees= [ Of course, that is in the
fourth quadrant where x is positive and y is negative so that is the correct answer. You can also write that, as a positive number, $2\pi- \pi/4= 8\pi/4- \pi/4= 7\pi/4$. $-5\pi/4$ is NOT the correct argument. That is in the second quadrant and so is
the argument for -1+ i.

To find the polar form you will also need the modulus or absolute value. That is [math]\sqrt{1^2+ (-1)^2}= \sqrt{2}[/math].

--------------



ex : 2



z = -root3 + i

my answer is

tanQ = -1/root3 = -0.143

but the answer is = 5π/6



How i get the answer 5π/6

Once again, the value you are looking for is NOT "tan(Q)" but Q itself. To find that take the inverse tangent. A calculator will tell you that [itex]Q= tan^{-1}(-1/\sqrt{3})= -30 degrees= $-\pi/6$ radians. Again that is in the fourth quadrant (a calculator always gives arctan values between $-\pi/2$ and $\pi/2$). But $-\sqrt{3}+ i$ with negative x value and positive y is in the second quadrant: the correct angle is $\pi- \pi/6= 6\pi/6- \pi/6= 5\pi/6$.<br /> <br /> And, again, for polar form you will need the modulus, $\sqrt{(-\sqrt{3})^2+1^2}=\sqrt{3+ 1}= \sqrt{4}= 2$. <br /> [/quote]<br /> <br /> <br /> ----<br /> <br /> <br /> <br /> <br /> <br /> please help me >>[/QUOTE][/itex]

 

[Susanne217]

Do you even know what question you are trying to answer?
"$- 5\pi/4$" is NOT the answer to the question posed, to write 1- i in polar form but it is part of the answer. Yes, tan(Q)= -1 and a calculator will tell you that $Q= tan^{-1}(-1)= -45$ degrees or $-\pi/4$ radians. You should also be able to recognize that tangent is 1 when the "near side" and "opposite" side are the same- an isosceles right triangle with angles or 45 degrees= [ Of course, that is in the
fourth quadrant where x is positive and y is negative so that is the correct answer. You can also write that, as a positive number, $2\pi- \pi/4= 8\pi/4- \pi/4= 7\pi/4$. $-5\pi/4$ is NOT the correct argument. That is in the second quadrant and so is
the argument for -1+ i.

To find the polar form you will also need the modulus or absolute value. That is [math]\sqrt{1^2+ (-1)^2}= \sqrt{2}[/math].


Once again, the value you are looking for is NOT "tan(Q)" but Q itself. To find that take the inverse tangent. A calculator will tell you that [itex]Q= tan^{-1}(-1/\sqrt{3})= -30 degrees= $-\pi/6$ radians. Again that is in the fourth quadrant (a calculator always gives arctan values between $-\pi/2$ and $\pi/2$). But $-\sqrt{3}+ i$ with negative x value and positive y is in the second quadrant: the correct angle is $\pi- \pi/6= 6\pi/6- \pi/6= 5\pi/6$.<br /> <br /> And, again, for polar form you will need the modulus, $\sqrt{(-\sqrt{3})^2+1^2}=\sqrt{3+ 1}= \sqrt{4}= 2$. [/itex]

[itex] <br /> <br /> ----<br /> <br /> <br /> I have a different view of this <br /> <br /> let z = 1-i be a complex number first<br /> <br /> then we find the angle alpha$$\alpha = tan^{-1}(-1) = -45 = \frac{-\pi}{4}$$<br /> <br /> since z = 1-i lies within the second quadrant<br /> <br /> so $$180--45 = 225$$ degrees. Which in Radians are $$\theta = \frac{5\pi}{4<br /> }$$[/itex]

 

[r-soy]

Now i understand it

thanks


If posiible i want all angle with degrees and radians

for exaple : 45 = π/4
30 = π/6

help me

 

[Susanne217]

Now i understand it

thanks


If posiible i want all angle with degrees and radians

for exaple : 45 = π/4
30 = π/6

help me

You are welcome r-soy :)

 

[Mark44]

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I have a different view of this

let z = 1-i be a complex number first

then we find the angle alpha$$\alpha = tan^{-1}(-1) = -45 = \frac{-\pi}{4}$$

since z = 1-i lies within the second quadrant

No, z is in the fourth quadrant.

so $$180--45 = 225$$ degrees. Which in Radians are $$\theta = \frac{5\pi}{4<br /> }$$

As a positive angle it would be 7pi/4.

 

[r-soy]

If posiible i want all angle with degrees and radians

for exaple : 45 = π/4
30 = π/6

help me