Mixing
How can I draw the complex number $z_1 = 2e^{{piover6}i -1}$
$begingroup$
I'm not sure how to draw the following number in the complex plane:
$$z_1 = 2e^{({piover6}i -1)}$$
My guess would be that you seperate the exponents:
$$z_1 = 2e^{({piover6}i)} e^{-1}$$
To get:
$$z_1 = frac{sqrt 3}{e} + frac{1}{e}i$$
So $ operatorname{Re}(z_1) =frac{sqrt 3}{e}$ and $ operatorname{Im}(z_1) =frac{1}{e}$.
Is this correct? This is part of a previous exam question, but my teacher doesn't provide any answers.
calculus complex-numbers
edited Jan 27 at 12:47
Larry
2,53531131
asked Jan 27 at 12:40
d salined saline
162
$endgroup$
add a comment |
$begingroup$
I'm not sure how to draw the following number in the complex plane:
$$z_1 = 2e^{({piover6}i -1)}$$
My guess would be that you seperate the exponents:
$$z_1 = 2e^{({piover6}i)} e^{-1}$$
To get:
$$z_1 = frac{sqrt 3}{e} + frac{1}{e}i$$
So $ operatorname{Re}(z_1) =frac{sqrt 3}{e}$ and $ operatorname{Im}(z_1) =frac{1}{e}$.
Is this correct? This is part of a previous exam question, but my teacher doesn't provide any answers.
calculus complex-numbers
edited Jan 27 at 12:47
Larry
2,53531131
asked Jan 27 at 12:40
d salined saline
162
$endgroup$
$begingroup$
Quite correct.${}$
$endgroup$
– Bernard
Jan 27 at 12:44
add a comment |
$begingroup$
I'm not sure how to draw the following number in the complex plane:
$$z_1 = 2e^{({piover6}i -1)}$$
My guess would be that you seperate the exponents:
$$z_1 = 2e^{({piover6}i)} e^{-1}$$
To get:
$$z_1 = frac{sqrt 3}{e} + frac{1}{e}i$$
So $ operatorname{Re}(z_1) =frac{sqrt 3}{e}$ and $ operatorname{Im}(z_1) =frac{1}{e}$.
Is this correct? This is part of a previous exam question, but my teacher doesn't provide any answers.
calculus complex-numbers
edited Jan 27 at 12:47
Larry
2,53531131
asked Jan 27 at 12:40
d salined saline
162
$endgroup$
I'm not sure how to draw the following number in the complex plane:
$$z_1 = 2e^{({piover6}i -1)}$$
My guess would be that you seperate the exponents:
$$z_1 = 2e^{({piover6}i)} e^{-1}$$
To get:
$$z_1 = frac{sqrt 3}{e} + frac{1}{e}i$$
So $ operatorname{Re}(z_1) =frac{sqrt 3}{e}$ and $ operatorname{Im}(z_1) =frac{1}{e}$.
Is this correct? This is part of a previous exam question, but my teacher doesn't provide any answers.
calculus complex-numbers
calculus complex-numbers
edited Jan 27 at 12:47
Larry
2,53531131
asked Jan 27 at 12:40
d salined saline
162
edited Jan 27 at 12:47
Larry
2,53531131
asked Jan 27 at 12:40
d salined saline
162
edited Jan 27 at 12:47
Larry
2,53531131
edited Jan 27 at 12:47
Larry
2,53531131
edited Jan 27 at 12:47
Larry
2,53531131
2,53531131
asked Jan 27 at 12:40
d salined saline
162
asked Jan 27 at 12:40
d salined saline
162
asked Jan 27 at 12:40
d salined saline
162
162
$begingroup$
Quite correct.${}$
$endgroup$
– Bernard
Jan 27 at 12:44
add a comment |
$begingroup$
Quite correct.${}$
$endgroup$
– Bernard
Jan 27 at 12:44
$begingroup$
Quite correct.${}$
$endgroup$
– Bernard
Jan 27 at 12:44
$begingroup$
Quite correct.${}$
$endgroup$
– Bernard
Jan 27 at 12:44
add a comment |
1 Answer
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$begingroup$
This is correct. You get the complex number $z$ with modulus $frac2e$ and argument $frac{pi}6$.
answered Jan 27 at 12:46
Chris CusterChris Custer
14.2k3827
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add a comment |
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1 Answer
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1 Answer
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$begingroup$
This is correct. You get the complex number $z$ with modulus $frac2e$ and argument $frac{pi}6$.
answered Jan 27 at 12:46
Chris CusterChris Custer
14.2k3827
$endgroup$
add a comment |
$begingroup$
This is correct. You get the complex number $z$ with modulus $frac2e$ and argument $frac{pi}6$.
answered Jan 27 at 12:46
Chris CusterChris Custer
14.2k3827
$endgroup$
add a comment |
$begingroup$
This is correct. You get the complex number $z$ with modulus $frac2e$ and argument $frac{pi}6$.
answered Jan 27 at 12:46
Chris CusterChris Custer
14.2k3827
$endgroup$
This is correct. You get the complex number $z$ with modulus $frac2e$ and argument $frac{pi}6$.
answered Jan 27 at 12:46
Chris CusterChris Custer
14.2k3827
answered Jan 27 at 12:46
Chris CusterChris Custer
14.2k3827
answered Jan 27 at 12:46
Chris CusterChris Custer
14.2k3827
answered Jan 27 at 12:46
Chris CusterChris Custer
14.2k3827
14.2k3827
add a comment |
add a comment |
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$begingroup$
Quite correct.${}$
$endgroup$
– Bernard
Jan 27 at 12:44