How do you convert −2−2√3i to polar form?

1 Answer
KiAndres
Jul 19, 2016

You can simplify this as −2(1+√3i). This is not needed, but I like to work with easier terms. Next, you remember that z=r(cos(θ)+isin(θ)). r is called the modulus of z and it is defined as √x2+y2=|z|. Yes, that's the same symbol for the absolute value, but in complex numbers, it defines the modulus. (This has to do with metric spaces)
Hence, r=2

Now, to find θ which is also called the argument of z, you use tan−1(√31)=π3
The proof of this lies in basic trigonometry, but it is more easily seen from Euler's Formula and unit vectors.
Now, not forgetting the −2 at the beginning, we get:
−2(1+√3i)=−4(cos(π3)+isin(π3))

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