Note that \(z = 2 + (-3)i\). To obtain its complex conjugate, we negate the imaginary part. Hence, \(\overline{z} = 2 - (-3)i = 2 + 3i\).

And the modulus of \(z\) is given by \(\sqrt{2^2 + (-3)^2} = \sqrt{4 + 9} = \sqrt{13}\).

For the multiplicative inverse of \(z\), we have \(\displaystyle z^{-1} = \frac{1}{z} = \frac{1}{z} \cdot \frac{\overline{z}}{\overline{z}} = \frac{\overline{z}}{z\overline{z}} = \frac{\overline{z}}{|z|^2} = \frac{2+3i}{13} = \frac{2}{13} + \frac{3}{13}i.\)

\begin{eqnarray*} z + 4 w^{-2} & = & z + \frac{4}{w^2} \\ & = & 1-2i + \frac{4}{(1-i)^2} \\ & = & 1-2i + \frac{4}{-2i} \\ & = & 1-2i + \frac{2}{-i} \\ & = & 1-2i + \frac{2}{-i}\cdot\frac{i}{i} \\ & = & 1-2i + \frac{2i}{-i^2} \\ & = & 1-2i + \frac{2i}{-(-1)} \\ & = & 1-2i + \frac{2i}{1} \\ & = & 1-2i + 2i \\ & = & 1 \\ \end{eqnarray*}

Recall that \(3 \operatorname{cis} \left(\frac{3}{5}\right)\) is an abbreviated way for writing \(3 \left(\cos \frac{3}{5} + i \sin \frac{3}{5}\right)\). Hence, the real part is \( 3 \cos \frac{3}{5} \approx 2.476006844729\) and the imaginary part is \( 3 \sin \frac{3}{5} \approx 1.693927420185\).

Hence, the rectangular form of \(3 \operatorname{cis} \left(\frac{3}{5}\right)\) is \(2.4760 + 1.6939 i\) with real and imaginary parts rounded to four decimal places.

Recall that the fourth roots of a complex number in polar form \(r \operatorname{cis} \theta\) are given by \(r^{\frac{1}{4}}\operatorname{cis}\left(\frac{2j \pi}{4}+ \frac{\theta}{4}\right)\) for \(j = 0, 1, 2, 3\). So let us first convert \(4+3i\) into polar form.

The modulus \(r\) is given by \(\sqrt{ 4^2 + 3^2} = \sqrt{16+9} = \sqrt{25} = 5\).

For the argument, we have \(\theta = \arctan \frac{3}{4}\).

Hence, the four fourth roots are \(5^{\frac{1}{4}}\operatorname{cis}\left(\frac{\arctan\frac{3}{4}}{4}\right)\), \(5^{\frac{1}{4}}\operatorname{cis}\left(\frac{2\pi}{4}+\frac{\arctan\frac{3}{4}}{4}\right)\), \(5^{\frac{1}{4}}\operatorname{cis}\left(\frac{4\pi}{4}+\frac{\arctan\frac{3}{4}}{4}\right)\), \(5^{\frac{1}{4}}\operatorname{cis}\left(\frac{6\pi}{4}+\frac{\arctan\frac{3}{4}}{4}\right)\).

Simplifying using a calculator, we get the following four fourth roots with real and imaginary parts accurate to four decimal places: \(1.4760 + 0.2395i\), \(-0.2395+1.4760i\), \(-1.4760-0.2395i\), \(0.2395-1.4760i\).

Let \(z = 8-5i\). Note that \(z\) has three cube roots but the question looks for one particular root with the given real part (rounded to 5 decimal places). So one strategy is to first find all the cube roots of \(z\) and look through the list to find one that matches what is given.

We first convert \(z\) to polar form. The modulus \(r\) is given by \(\sqrt{ 8^2 + (-5)^2} = \sqrt{ 64 + 25} = \sqrt{89} \approx 9.433981132\). (We carry a few more decimal places than necessary so that we can have 5-decimal place accuracy in our final answers.)

For the argument, we have \(\theta = 2\pi - \arctan \frac{5}{8}= 5.72458599\).

Hence, the cube roots of \(z\) are, approximately, \(9.433981132^{\frac{1}{3}} \operatorname{cis}\left(\frac{5.72458599}{3} \right)\), \(9.433981132^{\frac{1}{3}} \operatorname{cis}\left(\frac{2\pi}{3}+\frac{5.72458599}{3}\right)\), and \(9.433981132^{\frac{1}{3}} \operatorname{cis}\left(\frac{4\pi}{3}+\frac{5.72458599}{3}\right)\).

Converting the first one into rectangular form gives \(-0.69947 + 1.99386i\). Since the real part of this cube root matches the one given in the question, there is no need to convert the remaining roots to rectangular form.

Hence, \(b = 1.99386\).

We claim that the answer is “yes”. To see this, rewrite the equation as \((z-i)^2 = -1\). Since \(-1\) has two square roots, namely \(i\) and \(-i\), we get that \(z - i= i\) or \(z -i = -i\), implying that \(z = 2i\) or \(z = 0\). And \(z=0\) is a real solution.

Recall that for a complex number \(z = x+yi\) where \(x,y\) are real numbers, \(|z| = \sqrt{x^2+y^2}\). Hence, \(|w| = \sqrt{(-2)^2+(-1)^2} = \sqrt{5}\). Also, \(|a+u| = |a+i| = \sqrt{a^2 + 1^2}\) since \(a\) is a real number. Thus, we need \(a\) to satisfy \(\sqrt{a^2+1} = \sqrt{5}\). Squaring both sides gives \(a^2+1 = 5\). Thus, \(a = \pm 2\).

(We will solve this using the method of completing the square. The key is to transform the equation \(z^2 + pz + q=0\) to the form \((z+r)^2 = s\). Then, the solutions are \(s_1 - r\) and \(s_2 - r\) where \(s_1\) and \(s_2\) are the square roots of \(s\). The transformation can be obtained by doing the following \begin{eqnarray} & & z^2 + pz + q =0 \\ & \Leftrightarrow & z^2 + pz = -q\\ & \Leftrightarrow & z^2 + pz + (p/2)^2 = - q + (p/2)^2\\ & \Leftrightarrow & (z + p/2)^2 = (p/2)^2 - q \end{eqnarray}

The key step is, once you have rewritten the equation as \(z^2 + pz = -q\), you add the square of one-half the coefficient of \(z\) to both sides. The resulting left-hand side will be a perfect square.)

First, multiply both sides by \(i^{-1}\) (which is \(-i\)) to obtain \[z^2 + 2iz - i= 0.\]

The equation can be rewritten as \[z^2 + 2iz = i.\]

The coefficient of \(z\) is \(2i\). So we add \((2i/2)^2\) (which is \(i^2\)) to both sides to obtain \[z^2 + 2iz + i^2 = i+i^2.\]

Hence, the equation can be rewritten as \[(z+i)^2 = -1+i.\]

Thus, \(z+i\) must equal to one of the square roots of \(-1+i\).

In polar form, \(-1+i = \sqrt{2} \operatorname{cis} \frac{3\pi}{4}\). So the square roots of \(-1+i\) are \(2^{\frac{1}{4}} \operatorname{cis} \frac{3\pi}{8}\) and \(2^{\frac{1}{4}} \operatorname{cis} \frac{11\pi}{8}\), which are approximately \(0.4551 + 1.0987i\) and \(-0.4551 - 1.0987i\), respectively. Hence, the solutions, with real and imaginary parts accurate to 4 decimal places, are \(0.4551 +0.0987i\) and \(-0.4551 - 2.0987i\).

By the definition of the modulus of a complex number, we have \(|a+(2-3i)| = |(a+2)+(-3)i| = \sqrt{(a+2)^2 + (-3)^2}\). Thus, we want to find all real numbers \(a\) such that \[\sqrt{(a+2)^2+9} = 5,\] or equivalently, \[(a+2)^2+9 = 25.\] Subtracting \(9\) from both sides gives \[(a+2)^2 = 16.\] Thus, \(a+2 = 4 \text{ or } -4\), giving \(a=2\text{ or } -6\).