Theta between them. (6 marks) c. Simplify (4+3i)^3 (3 marks) d. Differentiate between standard and general forms of a circle (4 marks) e. Express r=1-3cosTheta in rectangular coordinates (3 marks) f. Find the polar equation of the circle whose Cartesian equation is 2x^2+2y^2=8x (6 marks) g. Find the equation of hyperbola having foci at F1(-5,0) and F2(5,0) and the difference of the total radii 6. (4 marks)

c. To simplify $(4+3i)^{3}$, we can use the binomial theorem or multiply the expression step by step.<br /><br />$(4+3i)^{3} = (4+3i)(4+3i)(4+3i)$<br /><br />First, let's multiply the first two factors:<br /><br />$(4+3i)(4+3i) = 16 + 12i + 12i + 9i^{2} = 16 + 24i - 9 = 7 + 24i$<br /><br />Now, let's multiply the result with the third factor:<br /><br />$(7 + 24i)(4 + 3i) = 28 + 21i + 96i + 72i^{2} = 28 + 117i - 72 = -44 + 117i$<br /><br />Therefore, the simplified form of $(4+3i)^{3}$ is $-44 + 117i$.<br /><br />d. The standard form of a circle is $(x-h)^{2} + (y-k)^{2} = r^{2}$, where $(h,k)$ is the center of the circle and $r$ is the radius.<br /><br />The general form of a circle is $x^{2} + y^{2} + Ax + By + C = 0$, where $A$, $B$, and $C$ are constants.<br /><br />The main difference between the standard and general forms is that in the standard form, the center and radius of the circle are easily identifiable, while in the general form, additional steps are required to find the center and radius.<br /><br />e. To express $r=1-3cos\Theta$ in rectangular coordinates, we can use the relationship between polar and rectangular coordinates: $x = rcos\Theta$ and $y = rsin\Theta$.<br /><br />Given $r=1-3cos\Theta$, we can substitute $r$ with $x/cos\Theta$:<br /><br />$x/cos\Theta = 1 - 3cos\Theta$<br /><br />Multiplying both sides by $cos\Theta$, we get:<br /><br />$x = cos\Theta + 3cos^{2}\Theta$<br /><br />Using the identity $cos^{2}\Theta = (1 + cos(2\Theta))/2$, we can rewrite the equation as:<br /><br />$x = cos\Theta + 3(1 + cos(2\Theta))/2$<br /><br />Simplifying further, we have:<br /><br />$x = 3cos(2\Theta) + 2cos\Theta + 1$<br /><br />Therefore, the expression $r=1-3cos\Theta$ in rectangular coordinates is $x = 3cos(2\Theta) + 2cos\Theta + 1$.<br /><br />f. To find the polar equation of the circle whose Cartesian equation is $2x^{2}+2y^{2}=8x$, we can rewrite the equation in the standard form of a circle.<br /><br />Dividing the equation by 2, we get:<br /><br />$x^{2} + y^{2} = 4x$<br /><br />Completing the square, we have:<br /><br />$(x-2)^{2} + y^{2} = 4$<br /><br />Comparing this with the standard form $(x-h)^{2} + (y-k)^{2} = r^{2}$, we can see that the center of the circle is $(2,0)$ and the radius is 2.<br /><br />Now, we can use the relationship between polar and Cartesian coordinates to find the polar equation. In polar coordinates, $x = rcos\Theta$ and $y = rsin\Theta$.<br /><br />Substituting $x = rcos\Theta$ into the equation $(x-2)^{2} + y^{2} = 4$, we get:<br /><br />$(rcos\Theta - 2)^{2} + (rsin\Theta)^{2} = 4$<br /><br />Expanding and simplifying, we have:<br /><br />$r^{2}cos^{2}\Theta - 4rcos\Theta + 4 + r^{2}sin^{2}\Theta = 4$<br /><br />$r^{2}(cos^{2}\Theta + sin^{2}\Theta) - 4rcos\Theta = 0$<br /><br />$r^{2} - 4rcos\Theta = 0$<br /><br />$r(r - 4cos\Theta) = 0$<br /><br />Therefore, the polar equation of the circle is $r = 4cos\Theta$.<br /><br />g. To find the equation of the hyperbola with foci at $F1(-5,0)$ and $F2(5,0)$ and the difference of the total radii 6, we can use the standard form of a hyperbola equation:<br /><br />$\frac{(x-h)^{2}}{a^{2}} - \frac{(y-k)^{2}}{b^{2}} = 1$<br /><br />where $(h,k)$ is the center of the hyperbola,