Modulus of complex number $z=2sqrt3+2i$
$begingroup$
So I know that there is a way to calculate it but our teacher said that we can skip that if we recognize any known coordinates which are on the unit circle. The answer to this one is supposed to be $z = 4*(frac{sqrt 3}{2} + frac{1}{2}j)$ Which means that the modulus is 4. But he didn't explain exactly what we have to do to reach this answer. Could someone help?
complex-numbers
asked Jan 6 at 16:04
ythhtrgythhtrg
333
$endgroup$
add a comment |
$begingroup$
So I know that there is a way to calculate it but our teacher said that we can skip that if we recognize any known coordinates which are on the unit circle. The answer to this one is supposed to be $z = 4*(frac{sqrt 3}{2} + frac{1}{2}j)$ Which means that the modulus is 4. But he didn't explain exactly what we have to do to reach this answer. Could someone help?
complex-numbers
asked Jan 6 at 16:04
ythhtrgythhtrg
333
$endgroup$
$begingroup$
The modulus is multiplicative; since $|frac{sqrt{3}}{2}+frac{1}{2}i|=1$ and $|4|=4$, you have $|z|=4$.
$endgroup$
– egreg
Jan 6 at 16:07
add a comment |
$begingroup$
So I know that there is a way to calculate it but our teacher said that we can skip that if we recognize any known coordinates which are on the unit circle. The answer to this one is supposed to be $z = 4*(frac{sqrt 3}{2} + frac{1}{2}j)$ Which means that the modulus is 4. But he didn't explain exactly what we have to do to reach this answer. Could someone help?
complex-numbers
asked Jan 6 at 16:04
ythhtrgythhtrg
333
$endgroup$
So I know that there is a way to calculate it but our teacher said that we can skip that if we recognize any known coordinates which are on the unit circle. The answer to this one is supposed to be $z = 4*(frac{sqrt 3}{2} + frac{1}{2}j)$ Which means that the modulus is 4. But he didn't explain exactly what we have to do to reach this answer. Could someone help?
complex-numbers
complex-numbers
asked Jan 6 at 16:04
ythhtrgythhtrg
333
asked Jan 6 at 16:04
ythhtrgythhtrg
333
asked Jan 6 at 16:04
ythhtrgythhtrg
333
asked Jan 6 at 16:04
ythhtrgythhtrg
333
asked Jan 6 at 16:04
ythhtrgythhtrg
333
333
$begingroup$
The modulus is multiplicative; since $|frac{sqrt{3}}{2}+frac{1}{2}i|=1$ and $|4|=4$, you have $|z|=4$.
$endgroup$
– egreg
Jan 6 at 16:07
add a comment |
$begingroup$
The modulus is multiplicative; since $|frac{sqrt{3}}{2}+frac{1}{2}i|=1$ and $|4|=4$, you have $|z|=4$.
$endgroup$
– egreg
Jan 6 at 16:07
$begingroup$
The modulus is multiplicative; since $|frac{sqrt{3}}{2}+frac{1}{2}i|=1$ and $|4|=4$, you have $|z|=4$.
$endgroup$
– egreg
Jan 6 at 16:07
$begingroup$
The modulus is multiplicative; since $|frac{sqrt{3}}{2}+frac{1}{2}i|=1$ and $|4|=4$, you have $|z|=4$.
$endgroup$
– egreg
Jan 6 at 16:07
add a comment |
2 Answers
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oldest
votes
$begingroup$
To calculate the modulus of a complex number, first, draw it on the complex plane:
Now, the modulus is the distance of a complex number and $0$, which means we need to solve for the hypotenuse of this right triangle. One way to do this is with the Pythagorean Theorem:
$$text{hypotenuse}^2=2^2+(2sqrt{3})^2=4+12=16$$
$$text{hypotenuse}=sqrt{16}=4$$
Therefore, the modulus of $2sqrt{3}+2i=4$.
However, your teacher used an alternative method. What he did is that he realized the above triangle was a 30-60-90 triangle because one of the legs is $sqrt 3$ times the other leg. Thus, the angle of the right triangle which is adjacent to the real number line is $30^circ$ because it is opposite the smaller side. On the unit circle, $30^circ$ corresponds to the following complex number:
$$cos(30^circ)+isin(30^circ)=frac{sqrt 3}{2}+ifrac 1 2$$
Now, our complex number, which is $2sqrt{3}+2i$, is clearly four times the above complex number from the unit circle: In other words, $2sqrt{3}+2i=4(frac{sqrt 3}{2}+ifrac 1 2)$. Therefore, its distance away from $0$ must be $4$ times the distance of $frac{sqrt 3}{2}+ifrac 1 2$ away from $0$. However, the distance of $frac{sqrt 3}{2}+ifrac 1 2$ away from $0$ is $1$, since it is on the unit circle of radius $1$. Therefore, the distance of $2sqrt{3}+2i$ away from $0$ is $4$ times $1$, or $4$. This is how your teacher found the modulus of $2sqrt{3}+2i$ to be $4$.
answered Jan 6 at 16:14
Noble MushtakNoble Mushtak
15.3k1835
$endgroup$
add a comment |
$begingroup$
If you are familiar with some trigonometry values, it's possible to recognize from $cos left(frac{pi}6 right)=frac{sqrt3}{2}$ and $sin left(frac{pi}6 right)=frac{1}{2}$ that
$$2sqrt3+2i= 4left(frac{sqrt3}2+frac{i}{2} right)=4left(cosleft(frac{pi}{6} right)+i sin left(frac{pi}6 right) right)$$
Notice that the length of $cosleft(frac{pi}{6} right)+i sin left(frac{pi}6 right)$ is $1$ and multiply it by $4$, hence the length is $4$.
Alternatively, just use the formula that if $a,b in mathbb{R}$, then by Pythagoras theorem $|a+bi|=sqrt{a^2+b^2}$
answered Jan 6 at 16:13
Siong Thye GohSiong Thye Goh
103k1468119
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
To calculate the modulus of a complex number, first, draw it on the complex plane:
Now, the modulus is the distance of a complex number and $0$, which means we need to solve for the hypotenuse of this right triangle. One way to do this is with the Pythagorean Theorem:
$$text{hypotenuse}^2=2^2+(2sqrt{3})^2=4+12=16$$
$$text{hypotenuse}=sqrt{16}=4$$
Therefore, the modulus of $2sqrt{3}+2i=4$.
However, your teacher used an alternative method. What he did is that he realized the above triangle was a 30-60-90 triangle because one of the legs is $sqrt 3$ times the other leg. Thus, the angle of the right triangle which is adjacent to the real number line is $30^circ$ because it is opposite the smaller side. On the unit circle, $30^circ$ corresponds to the following complex number:
$$cos(30^circ)+isin(30^circ)=frac{sqrt 3}{2}+ifrac 1 2$$
Now, our complex number, which is $2sqrt{3}+2i$, is clearly four times the above complex number from the unit circle: In other words, $2sqrt{3}+2i=4(frac{sqrt 3}{2}+ifrac 1 2)$. Therefore, its distance away from $0$ must be $4$ times the distance of $frac{sqrt 3}{2}+ifrac 1 2$ away from $0$. However, the distance of $frac{sqrt 3}{2}+ifrac 1 2$ away from $0$ is $1$, since it is on the unit circle of radius $1$. Therefore, the distance of $2sqrt{3}+2i$ away from $0$ is $4$ times $1$, or $4$. This is how your teacher found the modulus of $2sqrt{3}+2i$ to be $4$.
answered Jan 6 at 16:14
Noble MushtakNoble Mushtak
15.3k1835
$endgroup$
add a comment |
$begingroup$
To calculate the modulus of a complex number, first, draw it on the complex plane:
Now, the modulus is the distance of a complex number and $0$, which means we need to solve for the hypotenuse of this right triangle. One way to do this is with the Pythagorean Theorem:
$$text{hypotenuse}^2=2^2+(2sqrt{3})^2=4+12=16$$
$$text{hypotenuse}=sqrt{16}=4$$
Therefore, the modulus of $2sqrt{3}+2i=4$.
However, your teacher used an alternative method. What he did is that he realized the above triangle was a 30-60-90 triangle because one of the legs is $sqrt 3$ times the other leg. Thus, the angle of the right triangle which is adjacent to the real number line is $30^circ$ because it is opposite the smaller side. On the unit circle, $30^circ$ corresponds to the following complex number:
$$cos(30^circ)+isin(30^circ)=frac{sqrt 3}{2}+ifrac 1 2$$
Now, our complex number, which is $2sqrt{3}+2i$, is clearly four times the above complex number from the unit circle: In other words, $2sqrt{3}+2i=4(frac{sqrt 3}{2}+ifrac 1 2)$. Therefore, its distance away from $0$ must be $4$ times the distance of $frac{sqrt 3}{2}+ifrac 1 2$ away from $0$. However, the distance of $frac{sqrt 3}{2}+ifrac 1 2$ away from $0$ is $1$, since it is on the unit circle of radius $1$. Therefore, the distance of $2sqrt{3}+2i$ away from $0$ is $4$ times $1$, or $4$. This is how your teacher found the modulus of $2sqrt{3}+2i$ to be $4$.
answered Jan 6 at 16:14
Noble MushtakNoble Mushtak
15.3k1835
$endgroup$
add a comment |
$begingroup$
To calculate the modulus of a complex number, first, draw it on the complex plane:
Now, the modulus is the distance of a complex number and $0$, which means we need to solve for the hypotenuse of this right triangle. One way to do this is with the Pythagorean Theorem:
$$text{hypotenuse}^2=2^2+(2sqrt{3})^2=4+12=16$$
$$text{hypotenuse}=sqrt{16}=4$$
Therefore, the modulus of $2sqrt{3}+2i=4$.
However, your teacher used an alternative method. What he did is that he realized the above triangle was a 30-60-90 triangle because one of the legs is $sqrt 3$ times the other leg. Thus, the angle of the right triangle which is adjacent to the real number line is $30^circ$ because it is opposite the smaller side. On the unit circle, $30^circ$ corresponds to the following complex number:
$$cos(30^circ)+isin(30^circ)=frac{sqrt 3}{2}+ifrac 1 2$$
Now, our complex number, which is $2sqrt{3}+2i$, is clearly four times the above complex number from the unit circle: In other words, $2sqrt{3}+2i=4(frac{sqrt 3}{2}+ifrac 1 2)$. Therefore, its distance away from $0$ must be $4$ times the distance of $frac{sqrt 3}{2}+ifrac 1 2$ away from $0$. However, the distance of $frac{sqrt 3}{2}+ifrac 1 2$ away from $0$ is $1$, since it is on the unit circle of radius $1$. Therefore, the distance of $2sqrt{3}+2i$ away from $0$ is $4$ times $1$, or $4$. This is how your teacher found the modulus of $2sqrt{3}+2i$ to be $4$.
answered Jan 6 at 16:14
Noble MushtakNoble Mushtak
15.3k1835
$endgroup$
To calculate the modulus of a complex number, first, draw it on the complex plane:
Now, the modulus is the distance of a complex number and $0$, which means we need to solve for the hypotenuse of this right triangle. One way to do this is with the Pythagorean Theorem:
$$text{hypotenuse}^2=2^2+(2sqrt{3})^2=4+12=16$$
$$text{hypotenuse}=sqrt{16}=4$$
Therefore, the modulus of $2sqrt{3}+2i=4$.
However, your teacher used an alternative method. What he did is that he realized the above triangle was a 30-60-90 triangle because one of the legs is $sqrt 3$ times the other leg. Thus, the angle of the right triangle which is adjacent to the real number line is $30^circ$ because it is opposite the smaller side. On the unit circle, $30^circ$ corresponds to the following complex number:
$$cos(30^circ)+isin(30^circ)=frac{sqrt 3}{2}+ifrac 1 2$$
Now, our complex number, which is $2sqrt{3}+2i$, is clearly four times the above complex number from the unit circle: In other words, $2sqrt{3}+2i=4(frac{sqrt 3}{2}+ifrac 1 2)$. Therefore, its distance away from $0$ must be $4$ times the distance of $frac{sqrt 3}{2}+ifrac 1 2$ away from $0$. However, the distance of $frac{sqrt 3}{2}+ifrac 1 2$ away from $0$ is $1$, since it is on the unit circle of radius $1$. Therefore, the distance of $2sqrt{3}+2i$ away from $0$ is $4$ times $1$, or $4$. This is how your teacher found the modulus of $2sqrt{3}+2i$ to be $4$.
answered Jan 6 at 16:14
Noble MushtakNoble Mushtak
15.3k1835
edited Jan 6 at 16:19
answered Jan 6 at 16:14
Noble MushtakNoble Mushtak
15.3k1835
answered Jan 6 at 16:14
Noble MushtakNoble Mushtak
15.3k1835
answered Jan 6 at 16:14
Noble MushtakNoble Mushtak
15.3k1835
15.3k1835
add a comment |
add a comment |
$begingroup$
If you are familiar with some trigonometry values, it's possible to recognize from $cos left(frac{pi}6 right)=frac{sqrt3}{2}$ and $sin left(frac{pi}6 right)=frac{1}{2}$ that
$$2sqrt3+2i= 4left(frac{sqrt3}2+frac{i}{2} right)=4left(cosleft(frac{pi}{6} right)+i sin left(frac{pi}6 right) right)$$
Notice that the length of $cosleft(frac{pi}{6} right)+i sin left(frac{pi}6 right)$ is $1$ and multiply it by $4$, hence the length is $4$.
Alternatively, just use the formula that if $a,b in mathbb{R}$, then by Pythagoras theorem $|a+bi|=sqrt{a^2+b^2}$
answered Jan 6 at 16:13
Siong Thye GohSiong Thye Goh
103k1468119
$endgroup$
add a comment |
$begingroup$
If you are familiar with some trigonometry values, it's possible to recognize from $cos left(frac{pi}6 right)=frac{sqrt3}{2}$ and $sin left(frac{pi}6 right)=frac{1}{2}$ that
$$2sqrt3+2i= 4left(frac{sqrt3}2+frac{i}{2} right)=4left(cosleft(frac{pi}{6} right)+i sin left(frac{pi}6 right) right)$$
Notice that the length of $cosleft(frac{pi}{6} right)+i sin left(frac{pi}6 right)$ is $1$ and multiply it by $4$, hence the length is $4$.
Alternatively, just use the formula that if $a,b in mathbb{R}$, then by Pythagoras theorem $|a+bi|=sqrt{a^2+b^2}$
answered Jan 6 at 16:13
Siong Thye GohSiong Thye Goh
103k1468119
$endgroup$
add a comment |
$begingroup$
If you are familiar with some trigonometry values, it's possible to recognize from $cos left(frac{pi}6 right)=frac{sqrt3}{2}$ and $sin left(frac{pi}6 right)=frac{1}{2}$ that
$$2sqrt3+2i= 4left(frac{sqrt3}2+frac{i}{2} right)=4left(cosleft(frac{pi}{6} right)+i sin left(frac{pi}6 right) right)$$
Notice that the length of $cosleft(frac{pi}{6} right)+i sin left(frac{pi}6 right)$ is $1$ and multiply it by $4$, hence the length is $4$.
Alternatively, just use the formula that if $a,b in mathbb{R}$, then by Pythagoras theorem $|a+bi|=sqrt{a^2+b^2}$
answered Jan 6 at 16:13
Siong Thye GohSiong Thye Goh
103k1468119
$endgroup$
If you are familiar with some trigonometry values, it's possible to recognize from $cos left(frac{pi}6 right)=frac{sqrt3}{2}$ and $sin left(frac{pi}6 right)=frac{1}{2}$ that
$$2sqrt3+2i= 4left(frac{sqrt3}2+frac{i}{2} right)=4left(cosleft(frac{pi}{6} right)+i sin left(frac{pi}6 right) right)$$
Notice that the length of $cosleft(frac{pi}{6} right)+i sin left(frac{pi}6 right)$ is $1$ and multiply it by $4$, hence the length is $4$.
Alternatively, just use the formula that if $a,b in mathbb{R}$, then by Pythagoras theorem $|a+bi|=sqrt{a^2+b^2}$
answered Jan 6 at 16:13
Siong Thye GohSiong Thye Goh
103k1468119
answered Jan 6 at 16:13
Siong Thye GohSiong Thye Goh
103k1468119
answered Jan 6 at 16:13
Siong Thye GohSiong Thye Goh
103k1468119
answered Jan 6 at 16:13
Siong Thye GohSiong Thye Goh
103k1468119
103k1468119
add a comment |
add a comment |
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$begingroup$
The modulus is multiplicative; since $|frac{sqrt{3}}{2}+frac{1}{2}i|=1$ and $|4|=4$, you have $|z|=4$.
$endgroup$
– egreg
Jan 6 at 16:07