Represent in trigonometric form the numbers: 1^(°) 1+i…
Represent in trigonometric form the numbers: $$ \begin{aligned} &1^{\circ}\; 1+i; \quad 2^{\circ}\; 1+i\sqrt{3}; \quad 3^{\circ}\; \sqrt{3}-i; \quad 4^{\circ}\; 1-\cos\alpha+i\sin\alpha,\ (0<\alpha<2\pi); \\ &5^{\circ}\; 1+\sin\alpha-i\cos\alpha; \quad 6^{\circ}\; 1+i\tan\alpha \quad \left(-\frac{\pi}{2}<\alpha<\frac{\pi}{2}\right). \end{aligned} $$
Explanation of Method
The trigonometric (polar) form of a complex number $z = x + iy$ is:
$$ z = r(\cos\theta + i\sin\theta), \qquad r = |z| = \sqrt{x^2+y^2},\quad \theta = \arg(z). $$
- Identify the real part $x$ and imaginary part $y$.
- Compute the modulus $r = \sqrt{x^2+y^2}$, simplifying with trigonometric identities where $x, y$ involve $\alpha$.
- Factor $r$ out of $x + iy$ and match the remaining expression to $\cos\theta + i\sin\theta$ to read off $\theta$.
Step 1° · $z = 1 + i$
Modulus: $r = \sqrt{1^2+1^2} = \sqrt{2}$.
Argument: point $(1,1)$ is in Quadrant I, $\tan\theta = \dfrac{1}{1} = 1 \implies \theta = \dfrac{\pi}{4}$.
$$ \boxed{z = \sqrt{2}\left(\cos\frac{\pi}{4} + i\sin\frac{\pi}{4}\right)} $$
Step 2° · $z = 1 + i\sqrt{3}$
Modulus: $r = \sqrt{1^2 + (\sqrt3)^2} = \sqrt{1+3} = 2$.
Argument: point $(1,\sqrt3)$ is in Quadrant I, $\tan\theta = \sqrt3 \implies \theta = \dfrac{\pi}{3}$.
$$ \boxed{z = 2\left(\cos\frac{\pi}{3} + i\sin\frac{\pi}{3}\right)} $$
Step 3° · $z = \sqrt{3} - i$
Modulus: $r = \sqrt{(\sqrt3)^2 + (-1)^2} = \sqrt{3+1} = 2$.
Argument: point $(\sqrt3,-1)$ is in Quadrant IV, $\tan\theta = -\dfrac{1}{\sqrt3} \implies \theta = -\dfrac{\pi}{6}$.
$$ \boxed{z = 2\left(\cos\left(-\frac{\pi}{6}\right) + i\sin\left(-\frac{\pi}{6}\right)\right)} \quad \text{or equivalently} \quad \boxed{z = 2\left(\cos\frac{11\pi}{6} + i\sin\frac{11\pi}{6}\right)} $$
Step 4° · $z = 1 - \cos\alpha + i\sin\alpha$, $\ (0 < \alpha < 2\pi)$
Here $x = 1-\cos\alpha$, $y = \sin\alpha$.
$$ r^2 = (1-\cos\alpha)^2 + \sin^2\alpha = 1 - 2\cos\alpha + \cos^2\alpha + \sin^2\alpha = 2(1-\cos\alpha) $$
Using the identity $1-\cos\alpha = 2\sin^2\dfrac{\alpha}{2}$:
$$ r^2 = 4\sin^2\frac{\alpha}{2} \implies r = 2\left|\sin\frac{\alpha}{2}\right| = 2\sin\frac{\alpha}{2} $$
(since $0 < \alpha < 2\pi \implies 0 < \frac{\alpha}{2} < \pi$, so $\sin\frac{\alpha}{2} > 0$).
Using $1-\cos\alpha = 2\sin^2\dfrac{\alpha}{2}$ and $\sin\alpha = 2\sin\dfrac{\alpha}{2}\cos\dfrac{\alpha}{2}$:
$$ x = r\sin\frac{\alpha}{2}, \qquad y = r\cos\frac{\alpha}{2} $$
so $z = r\left(\sin\dfrac{\alpha}{2} + i\cos\dfrac{\alpha}{2}\right) = r\left(\cos\left(\dfrac{\pi}{2}-\dfrac{\alpha}{2}\right) + i\sin\left(\dfrac{\pi}{2}-\dfrac{\alpha}{2}\right)\right)$.
$$ \boxed{z = 2\sin\frac{\alpha}{2}\left(\cos\left(\frac{\pi}{2} - \frac{\alpha}{2}\right) + i\sin\left(\frac{\pi}{2} - \frac{\alpha}{2}\right)\right)} $$
Step 5° · $z = 1 + \sin\alpha - i\cos\alpha$
Here $x = 1+\sin\alpha$, $y = -\cos\alpha$.
$$ r^2 = (1+\sin\alpha)^2 + \cos^2\alpha = 1 + 2\sin\alpha + \sin^2\alpha + \cos^2\alpha = 2(1+\sin\alpha) $$
Using $1+\sin\alpha = 1+\cos\left(\dfrac{\pi}{2}-\alpha\right) = 2\cos^2\left(\dfrac{\pi}{4}-\dfrac{\alpha}{2}\right)$:
$$ r^2 = 4\cos^2\left(\frac{\pi}{4}-\frac{\alpha}{2}\right) \implies r = 2\left|\cos\left(\frac{\pi}{4}-\frac{\alpha}{2}\right)\right| $$
Writing $x = 2\cos^2\left(\dfrac{\pi}{4}-\dfrac{\alpha}{2}\right)$ and, since $\cos\alpha = \sin\left(\dfrac{\pi}{2}-\alpha\right) = 2\sin\left(\dfrac{\pi}{4}-\dfrac{\alpha}{2}\right)\cos\left(\dfrac{\pi}{4}-\dfrac{\alpha}{2}\right)$, we get $y = -2\sin\left(\dfrac{\pi}{4}-\dfrac{\alpha}{2}\right)\cos\left(\dfrac{\pi}{4}-\dfrac{\alpha}{2}\right)$. Factoring out $r = 2\cos\left(\dfrac{\pi}{4}-\dfrac{\alpha}{2}\right)$:
$$ z = 2\cos\left(\frac{\pi}{4}-\frac{\alpha}{2}\right)\left[\cos\left(\frac{\pi}{4}-\frac{\alpha}{2}\right) - i\sin\left(\frac{\pi}{4}-\frac{\alpha}{2}\right)\right] $$
Since $\cos$ is even and $\sin$ is odd, $\cos\left(\dfrac{\pi}{4}-\dfrac{\alpha}{2}\right) = \cos\left(\dfrac{\alpha}{2}-\dfrac{\pi}{4}\right)$ and $-\sin\left(\dfrac{\pi}{4}-\dfrac{\alpha}{2}\right) = \sin\left(\dfrac{\alpha}{2}-\dfrac{\pi}{4}\right)$, giving the trigonometric form directly (checked at $\alpha=0$: $z=1-i$ gives $r=\sqrt2$, $\theta=-\pi/4$ ✓):
$$ \boxed{z = 2\cos\left(\frac{\pi}{4} - \frac{\alpha}{2}\right)\left(\cos\left(\frac{\alpha}{2} - \frac{\pi}{4}\right) + i\sin\left(\frac{\alpha}{2} - \frac{\pi}{4}\right)\right)} $$
Step 6° · $z = 1 + i\tan\alpha$, $\ \left(-\dfrac{\pi}{2} < \alpha < \dfrac{\pi}{2}\right)$
$$ r = \sqrt{1^2 + \tan^2\alpha} = \sqrt{\frac{\cos^2\alpha+\sin^2\alpha}{\cos^2\alpha}} = \frac{1}{|\cos\alpha|} = \frac{1}{\cos\alpha} = \sec\alpha $$
(since $-\frac{\pi}{2} < \alpha < \frac{\pi}{2} \implies \cos\alpha > 0$).
Argument: $\tan\theta = \dfrac{\tan\alpha}{1} = \tan\alpha \implies \theta = \alpha$ (point $(1,\tan\alpha)$ makes angle $\alpha$ with the positive real axis).
$$ \boxed{z = \sec\alpha\,(\cos\alpha + i\sin\alpha)} $$
Summary Table
| No. | Complex number | Trigonometric form |
|---|---|---|
| $1^{\circ}$ | $1+i$ | $\sqrt{2}\left(\cos\frac{\pi}{4}+i\sin\frac{\pi}{4}\right)$ |
| $2^{\circ}$ | $1+i\sqrt{3}$ | $2\left(\cos\frac{\pi}{3}+i\sin\frac{\pi}{3}\right)$ |
| $3^{\circ}$ | $\sqrt{3}-i$ | $2\left(\cos\left(-\frac{\pi}{6}\right)+i\sin\left(-\frac{\pi}{6}\right)\right)$ |
| $4^{\circ}$ | $1-\cos\alpha+i\sin\alpha$ | $2\sin\frac{\alpha}{2}\left(\cos\left(\frac{\pi}{2}-\frac{\alpha}{2}\right)+i\sin\left(\frac{\pi}{2}-\frac{\alpha}{2}\right)\right)$ |
| $5^{\circ}$ | $1+\sin\alpha-i\cos\alpha$ | $2\cos\left(\frac{\pi}{4}-\frac{\alpha}{2}\right)\left(\cos\left(\frac{\alpha}{2}-\frac{\pi}{4}\right)+i\sin\left(\frac{\alpha}{2}-\frac{\pi}{4}\right)\right)$ |
| $6^{\circ}$ | $1+i\tan\alpha$ | $\sec\alpha\left(\cos\alpha+i\sin\alpha\right)$ |