Math · Precalculus · Worked example

Write a complex number in polar form

Write −2 + 2i in polar form.

Find the modulus

The modulus is the distance from 0 to the point (−2, 2).

r = \sqrt{{(- 2)}^{2} + 2^{2}} = \sqrt{8} = 2 \sqrt{2}

Find the argument

tan θ = 2/(−2) = −1, and the point (−2, 2) is in quadrant II, so θ = 3π/4, not the −π/4 that arctan(−1) gives.

θ = \frac{3 π}{4}

Write the polar form

- 2 + 2 i = 2 \sqrt{2} (cos \frac{3 π}{4} + i sin \frac{3 π}{4})

\begin{array}{c} - 2 + 2 i \\ = 2 \sqrt{2} (cos \frac{3 π}{4} + i sin \frac{3 π}{4}) \end{array}

Check the coordinates

r cos θ and r sin θ must give back −2 and 2.

\begin{array}{l} 2 \sqrt{2} cos \frac{3 π}{4} = - 2 \\ 2 \sqrt{2} sin \frac{3 π}{4} = 2 \end{array}

Result

−2 + 2i = 2√2(cos 3π/4 + i sin 3π/4): modulus 2√2 ≈ 2.83, argument 3π/4 = 135°.

Your turn

Write 1 − √3 i in polar form.

Show the answer and explanation

2(cos(−π/3) + i sin(−π/3)).

r = √(1 + 3) = 2. The point (1, −√3) is in quadrant IV with tan θ = −√3, so θ = −π/3, or equivalently 5π/3.

\begin{array}{l} \sqrt{1^{2} + {(\sqrt{3})}^{2}} = 2 \\ tan (- \frac{π}{3}) = - \sqrt{3} \end{array}

Keep exploring

In Complex plane & roots, inspect −2 + 2i: the modulus reads 2.828 and the principal argument 135°. Change it to −2 − 2i and the argument becomes −135°.

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