Theory of Intelligence
11TH•Mathematics•Complex Numbers•Question 2

(vi) $\dfrac{\sqrt{3}+i}{1+\sqrt{3}i}$

Express the following complex numbers in polar form:

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(vi) 3+i1+3i\dfrac{\sqrt{3}+i}{1+\sqrt{3}i}

3+i1+3i=(3+i)(1−3i)(1+3i)(1−3i)=23−2i4=32−12i\begin{aligned} \frac{\sqrt{3}+i}{1+\sqrt{3}i} &=\frac{(\sqrt{3}+i)(1-\sqrt{3}i)}{(1+\sqrt{3}i)(1-\sqrt{3}i)}\\ &=\frac{2\sqrt{3}-2i}{4}=\frac{\sqrt{3}}{2}-\frac{1}{2}i \end{aligned}

So z=32−12iz=\frac{\sqrt{3}}{2}-\frac{1}{2}i.

r=(32)2+(−12)2=1 r=\sqrt{\left(\frac{\sqrt{3}}{2}\right)^2+\left(-\frac{1}{2}\right)^2}=1

Since x>0x>0 and y<0y<0 (Quadrant IV):

θ=−π6\theta=-\frac{\pi}{6} 3+i1+3i=cos⁡(−π6)+isin⁡(−π6)\boxed{\frac{\sqrt{3}+i}{1+\sqrt{3}i}=\cos\left(-\frac{\pi}{6}\right)+i\sin\left(-\frac{\pi}{6}\right)}

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