Worked examples · Sheet 2

Trigonometric and exponential forms in the various quadrants

Theory used: Trigonometric form and argument Euler's formula and exponential form Modulus and its properties

Exercise 1

Write in trigonometric and exponential form

z = 1 + i 3 .

ρ = 1 + 3 = 2

a = 1 > 0 ⟹ θ = arctan 3 = \frac{π}{3}

z = 2 (cos \frac{π}{3} + i sin \frac{π}{3}) = 2 e^{i \frac{π}{3}}

The modulus is the square root of $a^{2} + b^{2} = 1 + 3 = 4$ , so $ρ = 2$ . The point $(1, 3)$ has positive real part, therefore it lies in the first quadrant and the principal argument coincides with the arctangent of the ratio $\frac{b}{a} = 3$ , that is $\frac{π}{3}$ , without any correction.

As a check, $2 cos \frac{π}{3} = 2 \cdot \frac{1}{2} = 1$ and $2 sin \frac{π}{3} = 2 \cdot \frac{3}{2} = 3$ , which are the original real part and imaginary part.

Exercise 2

Write in trigonometric and exponential form

z = - 3 + 3 i .

ρ = 9 + 9 = 32

a = - 3 < 0, b = 3 \geq 0 ⟹ θ = arctan (- 1) + π = - \frac{π}{4} + π = \frac{3 π}{4}

z = 32 (cos \frac{3 π}{4} + i sin \frac{3 π}{4}) = 32 e^{i \frac{3 π}{4}}

The modulus is $(- 3)^{2} + 3^{2} = 18 = 32$ . The point $(- 3, 3)$ has negative real part and positive imaginary part, so it lies in the second quadrant; the arctangent of $\frac{3}{- 3} = - 1$ equals $- \frac{π}{4}$ , which is an angle in the fourth quadrant, therefore, according to the cases of the principal argument, $π$ is added and one obtains $θ = \frac{3 π}{4}$ .

As a check, $cos \frac{3 π}{4} = - \frac{2}{2}$ and $sin \frac{3 π}{4} = \frac{2}{2}$ , so $32 cos \frac{3 π}{4} = - 3$ and $32 sin \frac{3 π}{4} = 3$ , which coincide with the real part and the imaginary part of $z$ .

Exercise 3

Write in trigonometric and exponential form

z = - 1 - i 3 .

ρ = 1 + 3 = 2

a = - 1 < 0, b = - 3 < 0 ⟹ θ = arctan 3 - π = \frac{π}{3} - π = - \frac{2 π}{3}

z = 2 (cos (- \frac{2 π}{3}) + i sin (- \frac{2 π}{3})) = 2 e^{- i \frac{2 π}{3}}

z = - 1 - i 3

lies in the third quadrant: the principal argument

θ = - \frac{2 π}{3}

is measured clockwise from the positive real semi-axis

The modulus is $1 + 3 = 2$ , the same as in the first exercise, because changing the sign of the real part and the imaginary part does not change the modulus. The point $(- 1, - 3)$ has both coordinates negative, so it lies in the third quadrant; the ratio $\frac{b}{a} = \frac{- 3}{- 1} = 3$ is positive and its arctangent equals $\frac{π}{3}$ , an angle in the first quadrant. Since $a < 0$ and $b < 0$ , according to the cases of the principal argument $π$ is subtracted and one obtains $θ = - \frac{2 π}{3}$ , which belongs to the interval $(- π, π]$ .

As a check, $cos (- \frac{2 π}{3}) = - \frac{1}{2}$ and $sin (- \frac{2 π}{3}) = - \frac{3}{2}$ , so $2 cos θ = - 1$ and $2 sin θ = - 3$ , as required.