Mixing
Weird stuff happening with complex numbers on a ti-84
$begingroup$
So, I'm trying to do some calculations for my Electrical engineering homework. This requires a bit of algebra with complex numbers. I have been finding that some of the calculations that my calculator has been making don't really make sense. This is especially apparent if I try to enter complex numbers into my calculator in their polar form. Correct me if I'm wrong, but it my understanding of switching back and forth between rectangular and polar forms of complex numbers is the following : A*e^(i*x) => A cos(x) + i sin(x)
However, if I input (while the calculator is in degrees mode, and it is set up to output complex quantities in rectangular form) e^(i*45) and press enter the output is : .525+.851 * i. My understanding of complex numbers leads me to believe that the output should have been: cos(45)+sin(45)*i = .707 + .707*i. Also, if I put the calculator in polar mode and input e^(i*45) the output is: 1e^(58.31). Why is it giving me a different output? What is going wrong here? Also, it is saying that e^(i*100)/e^(i*50)= e^(i*-15.211). Also an input of 1+i yields an output of 1.414*e^(i*45)...which is correct. Any idea what's going on here? Is the ti-84 just garbage when it comes to complex algebra in polar form?
complex-numbers calculator
asked Jan 31 '14 at 6:48
user3034894user3034894
1612
$endgroup$
|
show 1 more comment
$begingroup$
So, I'm trying to do some calculations for my Electrical engineering homework. This requires a bit of algebra with complex numbers. I have been finding that some of the calculations that my calculator has been making don't really make sense. This is especially apparent if I try to enter complex numbers into my calculator in their polar form. Correct me if I'm wrong, but it my understanding of switching back and forth between rectangular and polar forms of complex numbers is the following : A*e^(i*x) => A cos(x) + i sin(x)
However, if I input (while the calculator is in degrees mode, and it is set up to output complex quantities in rectangular form) e^(i*45) and press enter the output is : .525+.851 * i. My understanding of complex numbers leads me to believe that the output should have been: cos(45)+sin(45)*i = .707 + .707*i. Also, if I put the calculator in polar mode and input e^(i*45) the output is: 1e^(58.31). Why is it giving me a different output? What is going wrong here? Also, it is saying that e^(i*100)/e^(i*50)= e^(i*-15.211). Also an input of 1+i yields an output of 1.414*e^(i*45)...which is correct. Any idea what's going on here? Is the ti-84 just garbage when it comes to complex algebra in polar form?
complex-numbers calculator
asked Jan 31 '14 at 6:48
user3034894user3034894
1612
$endgroup$
2
$begingroup$
$e^{ix}$ will output $cos(x) + isin(x)$ as if $x$ was in radians, no matter what mode you're in. This is because in your calculator, $e^{ix}$ is not defined in terms of $cos$ and $sin$, but rather in terms of power series.
$endgroup$
– user98602
Jan 31 '14 at 6:50
$begingroup$
Also, something around $58$ degrees is the angle corresponding to $45$ radians.
$endgroup$
– user98602
Jan 31 '14 at 6:52
$begingroup$
So, you can only enter complex quantities in radians?
$endgroup$
– user3034894
Jan 31 '14 at 6:53
2
$begingroup$
Yes, your calculator will not play pleasantly with them in degrees. If you're working with complex numbers, it's time to forget to degrees and universally work with radians anyway :)
$endgroup$
– user98602
Jan 31 '14 at 6:55
1
$begingroup$
Hahaha, I agree wholeheartedly. It's just that my textbook poses questions in degrees, and seems to want results in degrees, so I tried to do my calculations in degrees...bad habit I guess. Thanks for the info!
$endgroup$
– user3034894
Jan 31 '14 at 7:00
|
show 1 more comment
$begingroup$
So, I'm trying to do some calculations for my Electrical engineering homework. This requires a bit of algebra with complex numbers. I have been finding that some of the calculations that my calculator has been making don't really make sense. This is especially apparent if I try to enter complex numbers into my calculator in their polar form. Correct me if I'm wrong, but it my understanding of switching back and forth between rectangular and polar forms of complex numbers is the following : A*e^(i*x) => A cos(x) + i sin(x)
However, if I input (while the calculator is in degrees mode, and it is set up to output complex quantities in rectangular form) e^(i*45) and press enter the output is : .525+.851 * i. My understanding of complex numbers leads me to believe that the output should have been: cos(45)+sin(45)*i = .707 + .707*i. Also, if I put the calculator in polar mode and input e^(i*45) the output is: 1e^(58.31). Why is it giving me a different output? What is going wrong here? Also, it is saying that e^(i*100)/e^(i*50)= e^(i*-15.211). Also an input of 1+i yields an output of 1.414*e^(i*45)...which is correct. Any idea what's going on here? Is the ti-84 just garbage when it comes to complex algebra in polar form?
complex-numbers calculator
asked Jan 31 '14 at 6:48
user3034894user3034894
1612
$endgroup$
So, I'm trying to do some calculations for my Electrical engineering homework. This requires a bit of algebra with complex numbers. I have been finding that some of the calculations that my calculator has been making don't really make sense. This is especially apparent if I try to enter complex numbers into my calculator in their polar form. Correct me if I'm wrong, but it my understanding of switching back and forth between rectangular and polar forms of complex numbers is the following : A*e^(i*x) => A cos(x) + i sin(x)
However, if I input (while the calculator is in degrees mode, and it is set up to output complex quantities in rectangular form) e^(i*45) and press enter the output is : .525+.851 * i. My understanding of complex numbers leads me to believe that the output should have been: cos(45)+sin(45)*i = .707 + .707*i. Also, if I put the calculator in polar mode and input e^(i*45) the output is: 1e^(58.31). Why is it giving me a different output? What is going wrong here? Also, it is saying that e^(i*100)/e^(i*50)= e^(i*-15.211). Also an input of 1+i yields an output of 1.414*e^(i*45)...which is correct. Any idea what's going on here? Is the ti-84 just garbage when it comes to complex algebra in polar form?
complex-numbers calculator
complex-numbers calculator
asked Jan 31 '14 at 6:48
user3034894user3034894
1612
asked Jan 31 '14 at 6:48
user3034894user3034894
1612
asked Jan 31 '14 at 6:48
user3034894user3034894
1612
asked Jan 31 '14 at 6:48
user3034894user3034894
1612
asked Jan 31 '14 at 6:48
user3034894user3034894
1612
1612
2
$begingroup$
$e^{ix}$ will output $cos(x) + isin(x)$ as if $x$ was in radians, no matter what mode you're in. This is because in your calculator, $e^{ix}$ is not defined in terms of $cos$ and $sin$, but rather in terms of power series.
$endgroup$
– user98602
Jan 31 '14 at 6:50
$begingroup$
Also, something around $58$ degrees is the angle corresponding to $45$ radians.
$endgroup$
– user98602
Jan 31 '14 at 6:52
$begingroup$
So, you can only enter complex quantities in radians?
$endgroup$
– user3034894
Jan 31 '14 at 6:53
2
$begingroup$
Yes, your calculator will not play pleasantly with them in degrees. If you're working with complex numbers, it's time to forget to degrees and universally work with radians anyway :)
$endgroup$
– user98602
Jan 31 '14 at 6:55
1
$begingroup$
Hahaha, I agree wholeheartedly. It's just that my textbook poses questions in degrees, and seems to want results in degrees, so I tried to do my calculations in degrees...bad habit I guess. Thanks for the info!
$endgroup$
– user3034894
Jan 31 '14 at 7:00
|
show 1 more comment
2
$begingroup$
$e^{ix}$ will output $cos(x) + isin(x)$ as if $x$ was in radians, no matter what mode you're in. This is because in your calculator, $e^{ix}$ is not defined in terms of $cos$ and $sin$, but rather in terms of power series.
$endgroup$
– user98602
Jan 31 '14 at 6:50
$begingroup$
Also, something around $58$ degrees is the angle corresponding to $45$ radians.
$endgroup$
– user98602
Jan 31 '14 at 6:52
$begingroup$
So, you can only enter complex quantities in radians?
$endgroup$
– user3034894
Jan 31 '14 at 6:53
2
$begingroup$
Yes, your calculator will not play pleasantly with them in degrees. If you're working with complex numbers, it's time to forget to degrees and universally work with radians anyway :)
$endgroup$
– user98602
Jan 31 '14 at 6:55
1
$begingroup$
Hahaha, I agree wholeheartedly. It's just that my textbook poses questions in degrees, and seems to want results in degrees, so I tried to do my calculations in degrees...bad habit I guess. Thanks for the info!
$endgroup$
– user3034894
Jan 31 '14 at 7:00
2
2
$begingroup$
$e^{ix}$ will output $cos(x) + isin(x)$ as if $x$ was in radians, no matter what mode you're in. This is because in your calculator, $e^{ix}$ is not defined in terms of $cos$ and $sin$, but rather in terms of power series.
$endgroup$
– user98602
Jan 31 '14 at 6:50
$begingroup$
$e^{ix}$ will output $cos(x) + isin(x)$ as if $x$ was in radians, no matter what mode you're in. This is because in your calculator, $e^{ix}$ is not defined in terms of $cos$ and $sin$, but rather in terms of power series.
$endgroup$
– user98602
Jan 31 '14 at 6:50
$begingroup$
Also, something around $58$ degrees is the angle corresponding to $45$ radians.
$endgroup$
– user98602
Jan 31 '14 at 6:52
$begingroup$
Also, something around $58$ degrees is the angle corresponding to $45$ radians.
$endgroup$
– user98602
Jan 31 '14 at 6:52
$begingroup$
So, you can only enter complex quantities in radians?
$endgroup$
– user3034894
Jan 31 '14 at 6:53
$begingroup$
So, you can only enter complex quantities in radians?
$endgroup$
– user3034894
Jan 31 '14 at 6:53
2
2
$begingroup$
Yes, your calculator will not play pleasantly with them in degrees. If you're working with complex numbers, it's time to forget to degrees and universally work with radians anyway :)
$endgroup$
– user98602
Jan 31 '14 at 6:55
$begingroup$
Yes, your calculator will not play pleasantly with them in degrees. If you're working with complex numbers, it's time to forget to degrees and universally work with radians anyway :)
$endgroup$
– user98602
Jan 31 '14 at 6:55
1
1
$begingroup$
Hahaha, I agree wholeheartedly. It's just that my textbook poses questions in degrees, and seems to want results in degrees, so I tried to do my calculations in degrees...bad habit I guess. Thanks for the info!
$endgroup$
– user3034894
Jan 31 '14 at 7:00
$begingroup$
Hahaha, I agree wholeheartedly. It's just that my textbook poses questions in degrees, and seems to want results in degrees, so I tried to do my calculations in degrees...bad habit I guess. Thanks for the info!
$endgroup$
– user3034894
Jan 31 '14 at 7:00
|
show 1 more comment
6 Answers
6
active
oldest
votes
$begingroup$
Your calculator won't work pleasantly with degrees in this case. It likely defines $e^{ix}$ as the power series $$e^{z} := sum_{n=0}^infty frac {z^n}{n!}$$ In the case of $z=ix$, this simplifies to $cos(x) + isin(x)$ by looking at the respective power series. These latter terms are, of course, in terms of radians; but changing 'radians' to 'degrees' doesn't change the above power series (or else your calculator would screw up, say, $e^2$ and instead output $e^{pi/90}$ - trying to convert the entire exponent from 'radians' to 'degrees', even though they don't apply here!) So if you're doing things in degrees, you'll either need to plug into $cos(x) + isin(x)$ directly, or first convert into radians.
$endgroup$
add a comment |
$begingroup$
I came across the same problem. You can convert the radians into degrees though. With $e^{45i}$ put in $e^{frac{45pi}{180}i}$. The angle $times frac{pi}{180}$ converts it to radians. I know it is annoying and I am still looking around for anyone who has written a program to do this. I may end up just writing one myself...
edited Aug 27 '14 at 14:12
k170
7,46931540
answered Aug 27 '14 at 13:46
JamiJami
111
$endgroup$
$begingroup$
Here's a MathJax tutorial :)
$endgroup$
– Shaun
Aug 27 '14 at 13:50
$begingroup$
Can't you make the TI-84 "remember" a value? Define $displaystyle E=e^{pi/180}approx1.01761$. Now, $E^{ix}=cos(x^circ)+isin(x^circ)$.
$endgroup$
– Akiva Weinberger
Aug 27 '14 at 14:22
add a comment |
$begingroup$
Mathematically, $e^{ix}=cos(x;mbox{radians})+isin(x;mbox{radians})$. This is independent of what setting your calculator is on. However, here's a "fix":
Can you make the TI-84 "remember" a value? Set $E=e^{pi/180}approx1.01761$ on the calculator. Now, $E^{ix}=cos(x^circ)+isin(x^circ)$.
answered Aug 27 '14 at 14:27
Akiva WeinbergerAkiva Weinberger
13.9k12168
$endgroup$
add a comment |
$begingroup$
You can store the conversion factor between degrees and radians as a variable and just insert it into the exponent in polar.
So I stored (pi/180) in the variable R. Then, for my problem, I scrolled to the answer I got from a previous calculation that I needed to use again(where the answer was in degrees as 12.12e^26.1275i, pressed enter to reuse it, and inserted my variable R into the exponent to get 12.12e^R26.1275i. Then I was able to multiply this by another number in polar, in my case 1200e^0i,to get the correct answer with the angle displayed in degrees.
So basically, you need to store the degrees to radians formula as a variable and insert it between calculations when needed.
answered Aug 30 '16 at 15:20
Erin BErin B
1
$endgroup$
add a comment |
$begingroup$
Here it is September 2017 and this problem with complex polar math still exists on the TI-84CE. I have found that the TI-89 performs complex math using my choice of polar/cartesian, radians/degrees settings. So TI knows how it is done. Should you switch to an '89? You might if you don't mind a smaller, monochrome screen.
answered Sep 3 '17 at 22:20
Larry ClarkLarry Clark
91
$endgroup$
add a comment |
$begingroup$
The gentleman that suggests going to radians always is wrong. EEs use degrees many places, not radians. For instance when dealing with phasors. TI-84 is not good at handling that.
What you have to do if you want to express $45$ degrees as $large e^{(text{degree_angle}cdot i)}$ is by entering this at your calculator: $large e^{(text{degree_angle}cdot2pi/360)}$
To check that it is true, try this on your $Ti-84$:
$e^{120i} = 0,814+0,580i$
Then to ehcek you result, use the angle from the CPX menu:
$text{angle}(0,814+0,580i) = 35,49 $---- well thats plain WRONG!
What calculator is better at handling this? I messed up on the exam due to the unreliable handling of $e^{xi} $with the $TI-84$
edited Sep 20 '14 at 13:14
fermesomme
5,18022351
answered Sep 20 '14 at 10:22
MariusMarius
1
$endgroup$
add a comment |
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6 Answers
6
active
oldest
votes
6 Answers
6
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Your calculator won't work pleasantly with degrees in this case. It likely defines $e^{ix}$ as the power series $$e^{z} := sum_{n=0}^infty frac {z^n}{n!}$$ In the case of $z=ix$, this simplifies to $cos(x) + isin(x)$ by looking at the respective power series. These latter terms are, of course, in terms of radians; but changing 'radians' to 'degrees' doesn't change the above power series (or else your calculator would screw up, say, $e^2$ and instead output $e^{pi/90}$ - trying to convert the entire exponent from 'radians' to 'degrees', even though they don't apply here!) So if you're doing things in degrees, you'll either need to plug into $cos(x) + isin(x)$ directly, or first convert into radians.
$endgroup$
add a comment |
$begingroup$
Your calculator won't work pleasantly with degrees in this case. It likely defines $e^{ix}$ as the power series $$e^{z} := sum_{n=0}^infty frac {z^n}{n!}$$ In the case of $z=ix$, this simplifies to $cos(x) + isin(x)$ by looking at the respective power series. These latter terms are, of course, in terms of radians; but changing 'radians' to 'degrees' doesn't change the above power series (or else your calculator would screw up, say, $e^2$ and instead output $e^{pi/90}$ - trying to convert the entire exponent from 'radians' to 'degrees', even though they don't apply here!) So if you're doing things in degrees, you'll either need to plug into $cos(x) + isin(x)$ directly, or first convert into radians.
$endgroup$
add a comment |
$begingroup$
Your calculator won't work pleasantly with degrees in this case. It likely defines $e^{ix}$ as the power series $$e^{z} := sum_{n=0}^infty frac {z^n}{n!}$$ In the case of $z=ix$, this simplifies to $cos(x) + isin(x)$ by looking at the respective power series. These latter terms are, of course, in terms of radians; but changing 'radians' to 'degrees' doesn't change the above power series (or else your calculator would screw up, say, $e^2$ and instead output $e^{pi/90}$ - trying to convert the entire exponent from 'radians' to 'degrees', even though they don't apply here!) So if you're doing things in degrees, you'll either need to plug into $cos(x) + isin(x)$ directly, or first convert into radians.
$endgroup$
Your calculator won't work pleasantly with degrees in this case. It likely defines $e^{ix}$ as the power series $$e^{z} := sum_{n=0}^infty frac {z^n}{n!}$$ In the case of $z=ix$, this simplifies to $cos(x) + isin(x)$ by looking at the respective power series. These latter terms are, of course, in terms of radians; but changing 'radians' to 'degrees' doesn't change the above power series (or else your calculator would screw up, say, $e^2$ and instead output $e^{pi/90}$ - trying to convert the entire exponent from 'radians' to 'degrees', even though they don't apply here!) So if you're doing things in degrees, you'll either need to plug into $cos(x) + isin(x)$ directly, or first convert into radians.
answered Jan 31 '14 at 7:06
user98602
add a comment |
add a comment |
$begingroup$
I came across the same problem. You can convert the radians into degrees though. With $e^{45i}$ put in $e^{frac{45pi}{180}i}$. The angle $times frac{pi}{180}$ converts it to radians. I know it is annoying and I am still looking around for anyone who has written a program to do this. I may end up just writing one myself...
edited Aug 27 '14 at 14:12
k170
7,46931540
answered Aug 27 '14 at 13:46
JamiJami
111
$endgroup$
$begingroup$
Here's a MathJax tutorial :)
$endgroup$
– Shaun
Aug 27 '14 at 13:50
$begingroup$
Can't you make the TI-84 "remember" a value? Define $displaystyle E=e^{pi/180}approx1.01761$. Now, $E^{ix}=cos(x^circ)+isin(x^circ)$.
$endgroup$
– Akiva Weinberger
Aug 27 '14 at 14:22
add a comment |
$begingroup$
I came across the same problem. You can convert the radians into degrees though. With $e^{45i}$ put in $e^{frac{45pi}{180}i}$. The angle $times frac{pi}{180}$ converts it to radians. I know it is annoying and I am still looking around for anyone who has written a program to do this. I may end up just writing one myself...
edited Aug 27 '14 at 14:12
k170
7,46931540
answered Aug 27 '14 at 13:46
JamiJami
111
$endgroup$
$begingroup$
Here's a MathJax tutorial :)
$endgroup$
– Shaun
Aug 27 '14 at 13:50
$begingroup$
Can't you make the TI-84 "remember" a value? Define $displaystyle E=e^{pi/180}approx1.01761$. Now, $E^{ix}=cos(x^circ)+isin(x^circ)$.
$endgroup$
– Akiva Weinberger
Aug 27 '14 at 14:22
add a comment |
$begingroup$
I came across the same problem. You can convert the radians into degrees though. With $e^{45i}$ put in $e^{frac{45pi}{180}i}$. The angle $times frac{pi}{180}$ converts it to radians. I know it is annoying and I am still looking around for anyone who has written a program to do this. I may end up just writing one myself...
edited Aug 27 '14 at 14:12
k170
7,46931540
answered Aug 27 '14 at 13:46
JamiJami
111
$endgroup$
I came across the same problem. You can convert the radians into degrees though. With $e^{45i}$ put in $e^{frac{45pi}{180}i}$. The angle $times frac{pi}{180}$ converts it to radians. I know it is annoying and I am still looking around for anyone who has written a program to do this. I may end up just writing one myself...
edited Aug 27 '14 at 14:12
k170
7,46931540
answered Aug 27 '14 at 13:46
JamiJami
111
edited Aug 27 '14 at 14:12
k170
7,46931540
edited Aug 27 '14 at 14:12
k170
7,46931540
edited Aug 27 '14 at 14:12
k170
7,46931540
7,46931540
answered Aug 27 '14 at 13:46
JamiJami
111
answered Aug 27 '14 at 13:46
JamiJami
111
answered Aug 27 '14 at 13:46
JamiJami
111
111
$begingroup$
Here's a MathJax tutorial :)
$endgroup$
– Shaun
Aug 27 '14 at 13:50
$begingroup$
Can't you make the TI-84 "remember" a value? Define $displaystyle E=e^{pi/180}approx1.01761$. Now, $E^{ix}=cos(x^circ)+isin(x^circ)$.
$endgroup$
– Akiva Weinberger
Aug 27 '14 at 14:22
add a comment |
$begingroup$
Here's a MathJax tutorial :)
$endgroup$
– Shaun
Aug 27 '14 at 13:50
$begingroup$
Can't you make the TI-84 "remember" a value? Define $displaystyle E=e^{pi/180}approx1.01761$. Now, $E^{ix}=cos(x^circ)+isin(x^circ)$.
$endgroup$
– Akiva Weinberger
Aug 27 '14 at 14:22
$begingroup$
Here's a MathJax tutorial :)
$endgroup$
– Shaun
Aug 27 '14 at 13:50
$begingroup$
Here's a MathJax tutorial :)
$endgroup$
– Shaun
Aug 27 '14 at 13:50
$begingroup$
Can't you make the TI-84 "remember" a value? Define $displaystyle E=e^{pi/180}approx1.01761$. Now, $E^{ix}=cos(x^circ)+isin(x^circ)$.
$endgroup$
– Akiva Weinberger
Aug 27 '14 at 14:22
$begingroup$
Can't you make the TI-84 "remember" a value? Define $displaystyle E=e^{pi/180}approx1.01761$. Now, $E^{ix}=cos(x^circ)+isin(x^circ)$.
$endgroup$
– Akiva Weinberger
Aug 27 '14 at 14:22
add a comment |
$begingroup$
Mathematically, $e^{ix}=cos(x;mbox{radians})+isin(x;mbox{radians})$. This is independent of what setting your calculator is on. However, here's a "fix":
Can you make the TI-84 "remember" a value? Set $E=e^{pi/180}approx1.01761$ on the calculator. Now, $E^{ix}=cos(x^circ)+isin(x^circ)$.
answered Aug 27 '14 at 14:27
Akiva WeinbergerAkiva Weinberger
13.9k12168
$endgroup$
add a comment |
$begingroup$
Mathematically, $e^{ix}=cos(x;mbox{radians})+isin(x;mbox{radians})$. This is independent of what setting your calculator is on. However, here's a "fix":
Can you make the TI-84 "remember" a value? Set $E=e^{pi/180}approx1.01761$ on the calculator. Now, $E^{ix}=cos(x^circ)+isin(x^circ)$.
answered Aug 27 '14 at 14:27
Akiva WeinbergerAkiva Weinberger
13.9k12168
$endgroup$
add a comment |
$begingroup$
Mathematically, $e^{ix}=cos(x;mbox{radians})+isin(x;mbox{radians})$. This is independent of what setting your calculator is on. However, here's a "fix":
Can you make the TI-84 "remember" a value? Set $E=e^{pi/180}approx1.01761$ on the calculator. Now, $E^{ix}=cos(x^circ)+isin(x^circ)$.
answered Aug 27 '14 at 14:27
Akiva WeinbergerAkiva Weinberger
13.9k12168
$endgroup$
Mathematically, $e^{ix}=cos(x;mbox{radians})+isin(x;mbox{radians})$. This is independent of what setting your calculator is on. However, here's a "fix":
Can you make the TI-84 "remember" a value? Set $E=e^{pi/180}approx1.01761$ on the calculator. Now, $E^{ix}=cos(x^circ)+isin(x^circ)$.
answered Aug 27 '14 at 14:27
Akiva WeinbergerAkiva Weinberger
13.9k12168
answered Aug 27 '14 at 14:27
Akiva WeinbergerAkiva Weinberger
13.9k12168
answered Aug 27 '14 at 14:27
Akiva WeinbergerAkiva Weinberger
13.9k12168
answered Aug 27 '14 at 14:27
Akiva WeinbergerAkiva Weinberger
13.9k12168
13.9k12168
add a comment |
add a comment |
$begingroup$
You can store the conversion factor between degrees and radians as a variable and just insert it into the exponent in polar.
So I stored (pi/180) in the variable R. Then, for my problem, I scrolled to the answer I got from a previous calculation that I needed to use again(where the answer was in degrees as 12.12e^26.1275i, pressed enter to reuse it, and inserted my variable R into the exponent to get 12.12e^R26.1275i. Then I was able to multiply this by another number in polar, in my case 1200e^0i,to get the correct answer with the angle displayed in degrees.
So basically, you need to store the degrees to radians formula as a variable and insert it between calculations when needed.
answered Aug 30 '16 at 15:20
Erin BErin B
1
$endgroup$
add a comment |
$begingroup$
You can store the conversion factor between degrees and radians as a variable and just insert it into the exponent in polar.
So I stored (pi/180) in the variable R. Then, for my problem, I scrolled to the answer I got from a previous calculation that I needed to use again(where the answer was in degrees as 12.12e^26.1275i, pressed enter to reuse it, and inserted my variable R into the exponent to get 12.12e^R26.1275i. Then I was able to multiply this by another number in polar, in my case 1200e^0i,to get the correct answer with the angle displayed in degrees.
So basically, you need to store the degrees to radians formula as a variable and insert it between calculations when needed.
answered Aug 30 '16 at 15:20
Erin BErin B
1
$endgroup$
add a comment |
$begingroup$
You can store the conversion factor between degrees and radians as a variable and just insert it into the exponent in polar.
So I stored (pi/180) in the variable R. Then, for my problem, I scrolled to the answer I got from a previous calculation that I needed to use again(where the answer was in degrees as 12.12e^26.1275i, pressed enter to reuse it, and inserted my variable R into the exponent to get 12.12e^R26.1275i. Then I was able to multiply this by another number in polar, in my case 1200e^0i,to get the correct answer with the angle displayed in degrees.
So basically, you need to store the degrees to radians formula as a variable and insert it between calculations when needed.
answered Aug 30 '16 at 15:20
Erin BErin B
1
$endgroup$
You can store the conversion factor between degrees and radians as a variable and just insert it into the exponent in polar.
So I stored (pi/180) in the variable R. Then, for my problem, I scrolled to the answer I got from a previous calculation that I needed to use again(where the answer was in degrees as 12.12e^26.1275i, pressed enter to reuse it, and inserted my variable R into the exponent to get 12.12e^R26.1275i. Then I was able to multiply this by another number in polar, in my case 1200e^0i,to get the correct answer with the angle displayed in degrees.
So basically, you need to store the degrees to radians formula as a variable and insert it between calculations when needed.
answered Aug 30 '16 at 15:20
Erin BErin B
1
answered Aug 30 '16 at 15:20
Erin BErin B
1
answered Aug 30 '16 at 15:20
Erin BErin B
1
answered Aug 30 '16 at 15:20
Erin BErin B
1
1
add a comment |
add a comment |
$begingroup$
Here it is September 2017 and this problem with complex polar math still exists on the TI-84CE. I have found that the TI-89 performs complex math using my choice of polar/cartesian, radians/degrees settings. So TI knows how it is done. Should you switch to an '89? You might if you don't mind a smaller, monochrome screen.
answered Sep 3 '17 at 22:20
Larry ClarkLarry Clark
91
$endgroup$
add a comment |
$begingroup$
Here it is September 2017 and this problem with complex polar math still exists on the TI-84CE. I have found that the TI-89 performs complex math using my choice of polar/cartesian, radians/degrees settings. So TI knows how it is done. Should you switch to an '89? You might if you don't mind a smaller, monochrome screen.
answered Sep 3 '17 at 22:20
Larry ClarkLarry Clark
91
$endgroup$
add a comment |
$begingroup$
Here it is September 2017 and this problem with complex polar math still exists on the TI-84CE. I have found that the TI-89 performs complex math using my choice of polar/cartesian, radians/degrees settings. So TI knows how it is done. Should you switch to an '89? You might if you don't mind a smaller, monochrome screen.
answered Sep 3 '17 at 22:20
Larry ClarkLarry Clark
91
$endgroup$
Here it is September 2017 and this problem with complex polar math still exists on the TI-84CE. I have found that the TI-89 performs complex math using my choice of polar/cartesian, radians/degrees settings. So TI knows how it is done. Should you switch to an '89? You might if you don't mind a smaller, monochrome screen.
answered Sep 3 '17 at 22:20
Larry ClarkLarry Clark
91
answered Sep 3 '17 at 22:20
Larry ClarkLarry Clark
91
answered Sep 3 '17 at 22:20
Larry ClarkLarry Clark
91
answered Sep 3 '17 at 22:20
Larry ClarkLarry Clark
91
91
add a comment |
add a comment |
$begingroup$
The gentleman that suggests going to radians always is wrong. EEs use degrees many places, not radians. For instance when dealing with phasors. TI-84 is not good at handling that.
What you have to do if you want to express $45$ degrees as $large e^{(text{degree_angle}cdot i)}$ is by entering this at your calculator: $large e^{(text{degree_angle}cdot2pi/360)}$
To check that it is true, try this on your $Ti-84$:
$e^{120i} = 0,814+0,580i$
Then to ehcek you result, use the angle from the CPX menu:
$text{angle}(0,814+0,580i) = 35,49 $---- well thats plain WRONG!
What calculator is better at handling this? I messed up on the exam due to the unreliable handling of $e^{xi} $with the $TI-84$
edited Sep 20 '14 at 13:14
fermesomme
5,18022351
answered Sep 20 '14 at 10:22
MariusMarius
1
$endgroup$
add a comment |
$begingroup$
The gentleman that suggests going to radians always is wrong. EEs use degrees many places, not radians. For instance when dealing with phasors. TI-84 is not good at handling that.
What you have to do if you want to express $45$ degrees as $large e^{(text{degree_angle}cdot i)}$ is by entering this at your calculator: $large e^{(text{degree_angle}cdot2pi/360)}$
To check that it is true, try this on your $Ti-84$:
$e^{120i} = 0,814+0,580i$
Then to ehcek you result, use the angle from the CPX menu:
$text{angle}(0,814+0,580i) = 35,49 $---- well thats plain WRONG!
What calculator is better at handling this? I messed up on the exam due to the unreliable handling of $e^{xi} $with the $TI-84$
edited Sep 20 '14 at 13:14
fermesomme
5,18022351
answered Sep 20 '14 at 10:22
MariusMarius
1
$endgroup$
add a comment |
$begingroup$
The gentleman that suggests going to radians always is wrong. EEs use degrees many places, not radians. For instance when dealing with phasors. TI-84 is not good at handling that.
What you have to do if you want to express $45$ degrees as $large e^{(text{degree_angle}cdot i)}$ is by entering this at your calculator: $large e^{(text{degree_angle}cdot2pi/360)}$
To check that it is true, try this on your $Ti-84$:
$e^{120i} = 0,814+0,580i$
Then to ehcek you result, use the angle from the CPX menu:
$text{angle}(0,814+0,580i) = 35,49 $---- well thats plain WRONG!
What calculator is better at handling this? I messed up on the exam due to the unreliable handling of $e^{xi} $with the $TI-84$
edited Sep 20 '14 at 13:14
fermesomme
5,18022351
answered Sep 20 '14 at 10:22
MariusMarius
1
$endgroup$
The gentleman that suggests going to radians always is wrong. EEs use degrees many places, not radians. For instance when dealing with phasors. TI-84 is not good at handling that.
What you have to do if you want to express $45$ degrees as $large e^{(text{degree_angle}cdot i)}$ is by entering this at your calculator: $large e^{(text{degree_angle}cdot2pi/360)}$
To check that it is true, try this on your $Ti-84$:
$e^{120i} = 0,814+0,580i$
Then to ehcek you result, use the angle from the CPX menu:
$text{angle}(0,814+0,580i) = 35,49 $---- well thats plain WRONG!
What calculator is better at handling this? I messed up on the exam due to the unreliable handling of $e^{xi} $with the $TI-84$
edited Sep 20 '14 at 13:14
fermesomme
5,18022351
answered Sep 20 '14 at 10:22
MariusMarius
1
edited Sep 20 '14 at 13:14
fermesomme
5,18022351
edited Sep 20 '14 at 13:14
fermesomme
5,18022351
edited Sep 20 '14 at 13:14
fermesomme
5,18022351
5,18022351
answered Sep 20 '14 at 10:22
MariusMarius
1
answered Sep 20 '14 at 10:22
MariusMarius
1
answered Sep 20 '14 at 10:22
MariusMarius
1
1
add a comment |
add a comment |
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2
$begingroup$
$e^{ix}$ will output $cos(x) + isin(x)$ as if $x$ was in radians, no matter what mode you're in. This is because in your calculator, $e^{ix}$ is not defined in terms of $cos$ and $sin$, but rather in terms of power series.
$endgroup$
– user98602
Jan 31 '14 at 6:50
$begingroup$
Also, something around $58$ degrees is the angle corresponding to $45$ radians.
$endgroup$
– user98602
Jan 31 '14 at 6:52
$begingroup$
So, you can only enter complex quantities in radians?
$endgroup$
– user3034894
Jan 31 '14 at 6:53
2
$begingroup$
Yes, your calculator will not play pleasantly with them in degrees. If you're working with complex numbers, it's time to forget to degrees and universally work with radians anyway :)
$endgroup$
– user98602
Jan 31 '14 at 6:55
1
$begingroup$
Hahaha, I agree wholeheartedly. It's just that my textbook poses questions in degrees, and seems to want results in degrees, so I tried to do my calculations in degrees...bad habit I guess. Thanks for the info!
$endgroup$
– user3034894
Jan 31 '14 at 7:00