Write the given numbers in the form a+ib and find the polar coordinate form of the complex numbers.

$$-3i - 1 - 7i + 10 - 9$$ $$(-1+10-9) + (-3-7)i$$ $$0 - 10i$$

The final answer is $\boxed{0 - 10i}$.

b) $2i^6 + \left(\frac{2}{-i}\right)^3 + 5i^5 - 12i$

$$\frac{2}{-i} = \frac{2i}{-i^2} = \frac{2i}{1} = 2i$$ $$(2i)^3 = 8i^3 = 8(-i) = -8i$$

$$2(-1) + (-8i) + 5(i) - 12i$$ $$-2 - 8i + 5i - 12i$$

$$-2 + (-8+5-12)i$$ $$-2 - 15i$$

The final answer is $\boxed{-2 - 15i}$.

: Find the polar coordinate form of the complex numbers.

a) $z=i$

$x=0, y=1$.

$$r = \sqrt{0^2+1^2} = 1$$

Since $z=i$ is on the positive imaginary axis, $\theta = \frac{\pi}{2}$.

$$z = 1\left(\cos\left(\frac{\pi}{2}\right) + i\sin\left(\frac{\pi}{2}\right)\right)$$

The final answer is $\boxed{\cos\left(\frac{\pi}{2}\right) + i\sin\left(\frac{\pi}{2}\right)}$.

b) $z=3-4i$

$x=3, y=-4$.

$$r = \sqrt{3^2+(-4)^2} = \sqrt{9+16} = \sqrt{25} = 5$$

The number is in the fourth quadrant.

$$\theta = \arctan\left(\frac{-4}{3}\right) = -\arctan\left(\frac{4}{3}\right)$$

$$z = 5\left(\cos\left(-\arctan\left(\frac{4}{3}\right)\right) + i\sin\left(-\arctan\left(\frac{4}{3}\right)\right)\right)$$

The final answer is $\boxed{5\left(\cos\left(-\arctan\left(\frac{4}{3}\right)\right) + i\sin\left(-\arctan\left(\frac{4}{3}\right)\right)\right)}$.

: Find the values of $x$ and $y$ that satisfy the equation $(x+y)+i(x-y)=14.8+6.2i$.

$$x+y = 14.8 \quad (1)$$ $$x-y = 6.2 \quad (2)$$

Add (1) and (2):

$$(x+y)+(x-y) = 14.8+6.2$$ $$2x = 21 \implies x = 10.5$$

Substitute $x=10.5$ into (1):

$$10.5+y = 14.8 \implies y = 14.8-10.5 \implies y = 4.3$$

The final answer is $\boxed{x=10.5, y=4.3}$.

: Express $(-1+i)$ in the form $re^{i\theta}$, where $r$ is positive and $-\pi < \theta < \pi$.

$x=-1, y=1$.

$$r = \sqrt{(-1)^2+1^2} = \sqrt{1+1} = \sqrt{2}$$

The number is in the second quadrant. The reference angle is $\arctan\left(\frac{1}{1}\right) = \frac{\pi}{4}$.

$$\theta = \pi - \frac{\pi}{4} = \frac{3\pi}{4}$$

This value is within $(-\pi, \pi)$.

$$z = \sqrt{2}e^{i\frac{3\pi}{4}}$$

The final answer is $\boxed{\sqrt{2}e^{i\frac{3\pi}{4}}}$.

: If $(a+b)+i(a-b)=(2+5i)^2+i(2-3i)$, find the values of $a$ and $b$.

$$(2+5i)^2 = 4 + 20i + 25i^2 = 4 + 20i - 25 = -21 + 20i$$ $$i(2-3i) = 2i - 3i^2 = 2i - 3(-1) = 3 + 2i$$

Add the simplified terms:

$$(-21 + 20i) + (3 + 2i) = (-21+3) + (20+2)i = -18 + 22i$$

$$a+b = -18 \quad (1)$$ $$a-b = 22 \quad (2)$$

Add (1) and (2):

$$(a+b)+(a-b) = -18+22$$ $$2a = 4 \implies a = 2$$

Substitute $a=2$ into (1):

$$2+b = -18 \implies b = -18-2 \implies b = -20$$

The final answer is $\boxed{a=2, b=-20}$.