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

Question

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?

Answer 1

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:

$$\text{hypotenuse}^2=2^2+(2\sqrt{3})^2=4+12=16$$ $$\text{hypotenuse}=\sqrt{16}=4$$

Therefore, the modulus of $2\sqrt{3}+2i=4$.

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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:

$$\cos(30^\circ)+i\sin(30^\circ)=\frac{\sqrt 3}{2}+i\frac 1 2$$

Now, our complex number, which is $2\sqrt{3}+2i$, is clearly four times the above complex number from the unit circle: In other words, $2\sqrt{3}+2i=4(\frac{\sqrt 3}{2}+i\frac 1 2)$. Therefore, its distance away from $0$ must be $4$ times the distance of $\frac{\sqrt 3}{2}+i\frac 1 2$ away from $0$. However, the distance of $\frac{\sqrt 3}{2}+i\frac 1 2$ away from $0$ is $1$, since it is on the unit circle of radius $1$. Therefore, the distance of $2\sqrt{3}+2i$ away from $0$ is $4$ times $1$, or $4$. This is how your teacher found the modulus of $2\sqrt{3}+2i$ to be $4$.

Answer 2

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

$$2\sqrt3+2i= 4\left(\frac{\sqrt3}2+\frac{i}{2} \right)=4\left(\cos\left(\frac{\pi}{6} \right)+i \sin \left(\frac{\pi}6 \right) \right)$$

Notice that the length of $\cos\left(\frac{\pi}{6} \right)+i \sin \left(\frac{\pi}6 \right)$ 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 $|a+bi|=\sqrt{a^2+b^2}$