Modulus of complex number $z=2sqrt3+2i$












0












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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







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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


















0












$begingroup$


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







share|cite|improve this question









$endgroup$












  • $begingroup$
    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
















0












0








0





$begingroup$


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







share|cite|improve this question









$endgroup$




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






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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




















  • $begingroup$
    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












2 Answers
2






active

oldest

votes


















1












$begingroup$

To calculate the modulus of a complex number, first, draw it on the complex plane:



enter image description here



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 $4$.






share|cite|improve this answer











$endgroup$





















    2












    $begingroup$

    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 $|a+bi|=sqrt{a^2+b^2}$






    share|cite|improve this answer









    $endgroup$













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      2 Answers
      2






      active

      oldest

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      2 Answers
      2






      active

      oldest

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      active

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      active

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      1












      $begingroup$

      To calculate the modulus of a complex number, first, draw it on the complex plane:



      enter image description here



      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 $4$.






      share|cite|improve this answer











      $endgroup$


















        1












        $begingroup$

        To calculate the modulus of a complex number, first, draw it on the complex plane:



        enter image description here



        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 $4$.






        share|cite|improve this answer











        $endgroup$
















          1












          1








          1





          $begingroup$

          To calculate the modulus of a complex number, first, draw it on the complex plane:



          enter image description here



          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 $4$.






          share|cite|improve this answer











          $endgroup$



          To calculate the modulus of a complex number, first, draw it on the complex plane:



          enter image description here



          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 $4$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 6 at 16:19

























          answered Jan 6 at 16:14









          Noble MushtakNoble Mushtak

          15.3k1835




          15.3k1835























              2












              $begingroup$

              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 $|a+bi|=sqrt{a^2+b^2}$






              share|cite|improve this answer









              $endgroup$


















                2












                $begingroup$

                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 $|a+bi|=sqrt{a^2+b^2}$






                share|cite|improve this answer









                $endgroup$
















                  2












                  2








                  2





                  $begingroup$

                  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 $|a+bi|=sqrt{a^2+b^2}$






                  share|cite|improve this answer









                  $endgroup$



                  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 $|a+bi|=sqrt{a^2+b^2}$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 6 at 16:13









                  Siong Thye GohSiong Thye Goh

                  103k1468119




                  103k1468119






























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