Mixing
Express the following complex number in polar form
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I've just began studying complex numbers for the first time and there are no solutions to the book I'm following. Are my answers correct for the following introductory questions? Thanks.
Q. Express the following in the form $re^{itheta}$.
i. $i^3$
Ans: $e^{frac{pi i}{2}} $.
ii. $1-i$
Ans: $sqrt{2} e^{frac{7pi i}{4}} $.
iii. $sqrt{2}(1+i)$
Ans: $2e^{frac{pi i}{4}}$.
complex-numbers
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add a comment |
$begingroup$
I've just began studying complex numbers for the first time and there are no solutions to the book I'm following. Are my answers correct for the following introductory questions? Thanks.
Q. Express the following in the form $re^{itheta}$.
i. $i^3$
Ans: $e^{frac{pi i}{2}} $.
ii. $1-i$
Ans: $sqrt{2} e^{frac{7pi i}{4}} $.
iii. $sqrt{2}(1+i)$
Ans: $2e^{frac{pi i}{4}}$.
complex-numbers
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How did you arrive at these answers? Show your work, please.
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– Yuval Gat
Jan 26 at 13:32
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Apologies. I'm using my phone and couldn't get some formatting correct. I took the modulus of the complex number in question to get r, and in order to obtain $theta$ I used the unit circle .
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– user606466
Jan 26 at 13:36
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While we prefer fractions to be written $frac nd$ in most places, that format doesn’t work well in an exponent. I would write $e^{pi i/4}$ rather than $e^{frac{pi i}{4}}.$
$endgroup$
– David K
Jan 26 at 14:42
add a comment |
$begingroup$
I've just began studying complex numbers for the first time and there are no solutions to the book I'm following. Are my answers correct for the following introductory questions? Thanks.
Q. Express the following in the form $re^{itheta}$.
i. $i^3$
Ans: $e^{frac{pi i}{2}} $.
ii. $1-i$
Ans: $sqrt{2} e^{frac{7pi i}{4}} $.
iii. $sqrt{2}(1+i)$
Ans: $2e^{frac{pi i}{4}}$.
complex-numbers
$endgroup$
I've just began studying complex numbers for the first time and there are no solutions to the book I'm following. Are my answers correct for the following introductory questions? Thanks.
Q. Express the following in the form $re^{itheta}$.
i. $i^3$
Ans: $e^{frac{pi i}{2}} $.
ii. $1-i$
Ans: $sqrt{2} e^{frac{7pi i}{4}} $.
iii. $sqrt{2}(1+i)$
Ans: $2e^{frac{pi i}{4}}$.
complex-numbers
complex-numbers
edited Jan 26 at 14:33
asked Jan 26 at 13:15
user606466
$begingroup$
How did you arrive at these answers? Show your work, please.
$endgroup$
– Yuval Gat
Jan 26 at 13:32
$begingroup$
Apologies. I'm using my phone and couldn't get some formatting correct. I took the modulus of the complex number in question to get r, and in order to obtain $theta$ I used the unit circle .
$endgroup$
– user606466
Jan 26 at 13:36
$begingroup$
While we prefer fractions to be written $frac nd$ in most places, that format doesn’t work well in an exponent. I would write $e^{pi i/4}$ rather than $e^{frac{pi i}{4}}.$
$endgroup$
– David K
Jan 26 at 14:42
add a comment |
$begingroup$
How did you arrive at these answers? Show your work, please.
$endgroup$
– Yuval Gat
Jan 26 at 13:32
$begingroup$
Apologies. I'm using my phone and couldn't get some formatting correct. I took the modulus of the complex number in question to get r, and in order to obtain $theta$ I used the unit circle .
$endgroup$
– user606466
Jan 26 at 13:36
$begingroup$
While we prefer fractions to be written $frac nd$ in most places, that format doesn’t work well in an exponent. I would write $e^{pi i/4}$ rather than $e^{frac{pi i}{4}}.$
$endgroup$
– David K
Jan 26 at 14:42
$begingroup$
How did you arrive at these answers? Show your work, please.
$endgroup$
– Yuval Gat
Jan 26 at 13:32
$begingroup$
How did you arrive at these answers? Show your work, please.
$endgroup$
– Yuval Gat
Jan 26 at 13:32
$begingroup$
Apologies. I'm using my phone and couldn't get some formatting correct. I took the modulus of the complex number in question to get r, and in order to obtain $theta$ I used the unit circle .
$endgroup$
– user606466
Jan 26 at 13:36
$begingroup$
Apologies. I'm using my phone and couldn't get some formatting correct. I took the modulus of the complex number in question to get r, and in order to obtain $theta$ I used the unit circle .
$endgroup$
– user606466
Jan 26 at 13:36
$begingroup$
While we prefer fractions to be written $frac nd$ in most places, that format doesn’t work well in an exponent. I would write $e^{pi i/4}$ rather than $e^{frac{pi i}{4}}.$
$endgroup$
– David K
Jan 26 at 14:42
$begingroup$
While we prefer fractions to be written $frac nd$ in most places, that format doesn’t work well in an exponent. I would write $e^{pi i/4}$ rather than $e^{frac{pi i}{4}}.$
$endgroup$
– David K
Jan 26 at 14:42
add a comment |
2 Answers
2
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oldest
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$begingroup$
Yes you are correct on the bottom two.
For the the top one, remember $i^3=-i$
Be careful in questions where they restrict the values of $theta$. For example if $-pi <theta leqpi$, your second would be $sqrt2e^{frac{-ipi}{4}}$ instead.
But yeah, just follow the steps I suspect you have been.
If your number is $x+iy$ we have:
$$r=sqrt{x^2+y^2}$$
$$theta=arctan(frac yx)$$
With the second, double check you got the right angle, and add or subtract $pi$ to get the correct angle if need be.
answered Jan 26 at 13:34
Rhys HughesRhys Hughes
7,0251630
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add a comment |
$begingroup$
For (i), recall that $i = e^{ipi/2}$. Then $i^3 = (e^{ipi/2})^3 = e^{3pi i/2} = e^{-pi i/2}$.
The other answers are correct.
answered Jan 26 at 13:36
AlexAlex
52538
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add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
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$begingroup$
Yes you are correct on the bottom two.
For the the top one, remember $i^3=-i$
Be careful in questions where they restrict the values of $theta$. For example if $-pi <theta leqpi$, your second would be $sqrt2e^{frac{-ipi}{4}}$ instead.
But yeah, just follow the steps I suspect you have been.
If your number is $x+iy$ we have:
$$r=sqrt{x^2+y^2}$$
$$theta=arctan(frac yx)$$
With the second, double check you got the right angle, and add or subtract $pi$ to get the correct angle if need be.
answered Jan 26 at 13:34
Rhys HughesRhys Hughes
7,0251630
$endgroup$
add a comment |
$begingroup$
Yes you are correct on the bottom two.
For the the top one, remember $i^3=-i$
Be careful in questions where they restrict the values of $theta$. For example if $-pi <theta leqpi$, your second would be $sqrt2e^{frac{-ipi}{4}}$ instead.
But yeah, just follow the steps I suspect you have been.
If your number is $x+iy$ we have:
$$r=sqrt{x^2+y^2}$$
$$theta=arctan(frac yx)$$
With the second, double check you got the right angle, and add or subtract $pi$ to get the correct angle if need be.
answered Jan 26 at 13:34
Rhys HughesRhys Hughes
7,0251630
$endgroup$
add a comment |
$begingroup$
Yes you are correct on the bottom two.
For the the top one, remember $i^3=-i$
Be careful in questions where they restrict the values of $theta$. For example if $-pi <theta leqpi$, your second would be $sqrt2e^{frac{-ipi}{4}}$ instead.
But yeah, just follow the steps I suspect you have been.
If your number is $x+iy$ we have:
$$r=sqrt{x^2+y^2}$$
$$theta=arctan(frac yx)$$
With the second, double check you got the right angle, and add or subtract $pi$ to get the correct angle if need be.
answered Jan 26 at 13:34
Rhys HughesRhys Hughes
7,0251630
$endgroup$
Yes you are correct on the bottom two.
For the the top one, remember $i^3=-i$
Be careful in questions where they restrict the values of $theta$. For example if $-pi <theta leqpi$, your second would be $sqrt2e^{frac{-ipi}{4}}$ instead.
But yeah, just follow the steps I suspect you have been.
If your number is $x+iy$ we have:
$$r=sqrt{x^2+y^2}$$
$$theta=arctan(frac yx)$$
With the second, double check you got the right angle, and add or subtract $pi$ to get the correct angle if need be.
answered Jan 26 at 13:34
Rhys HughesRhys Hughes
7,0251630
answered Jan 26 at 13:34
Rhys HughesRhys Hughes
7,0251630
answered Jan 26 at 13:34
Rhys HughesRhys Hughes
7,0251630
answered Jan 26 at 13:34
Rhys HughesRhys Hughes
7,0251630
7,0251630
add a comment |
add a comment |
$begingroup$
For (i), recall that $i = e^{ipi/2}$. Then $i^3 = (e^{ipi/2})^3 = e^{3pi i/2} = e^{-pi i/2}$.
The other answers are correct.
answered Jan 26 at 13:36
AlexAlex
52538
$endgroup$
add a comment |
$begingroup$
For (i), recall that $i = e^{ipi/2}$. Then $i^3 = (e^{ipi/2})^3 = e^{3pi i/2} = e^{-pi i/2}$.
The other answers are correct.
answered Jan 26 at 13:36
AlexAlex
52538
$endgroup$
add a comment |
$begingroup$
For (i), recall that $i = e^{ipi/2}$. Then $i^3 = (e^{ipi/2})^3 = e^{3pi i/2} = e^{-pi i/2}$.
The other answers are correct.
answered Jan 26 at 13:36
AlexAlex
52538
$endgroup$
For (i), recall that $i = e^{ipi/2}$. Then $i^3 = (e^{ipi/2})^3 = e^{3pi i/2} = e^{-pi i/2}$.
The other answers are correct.
answered Jan 26 at 13:36
AlexAlex
52538
answered Jan 26 at 13:36
AlexAlex
52538
answered Jan 26 at 13:36
AlexAlex
52538
answered Jan 26 at 13:36
AlexAlex
52538
52538
add a comment |
add a comment |
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$begingroup$
How did you arrive at these answers? Show your work, please.
$endgroup$
– Yuval Gat
Jan 26 at 13:32
$begingroup$
Apologies. I'm using my phone and couldn't get some formatting correct. I took the modulus of the complex number in question to get r, and in order to obtain $theta$ I used the unit circle .
$endgroup$
– user606466
Jan 26 at 13:36
$begingroup$
While we prefer fractions to be written $frac nd$ in most places, that format doesn’t work well in an exponent. I would write $e^{pi i/4}$ rather than $e^{frac{pi i}{4}}.$
$endgroup$
– David K
Jan 26 at 14:42