Complex Analysis Homework #1

To Write \(\left( \frac{2+i}{6i-(1-2i)} \right) ^2\), we must first expand it by multiplying it by itself, as so: \[\frac{2+i}{6i-(1-2i)}*\frac{2+i}{6i-(1-2i)} = \frac{4+4i-1}{-64-16i+1} = \frac{3+4i}{-(63+16i)}\]

and then multiplying by its complex conjugate to obtain a separable real and imaginary part: \[\frac{3+4i}{-(63+16i)}*\frac{-(63-16i)}{-(63-16i)} = \frac{189-48i+252i+64}{-(63^2+16^2)}\] which simplifies to \[- \left( \frac{253}{4225}+\frac{204}{4225}i \right)\]

We must show \(Re(iz) = -im(z), \forall z \in \mathbb {C}\)

To show this, we will examine both sides of this equation in the context that every complex number z can be expressed as \(z=a+ib\). Looking first at the left hand side: \[Re(iz) = Re(i(a+ib)) = Re(ia-b) = -b\]

Looking at the right hand side: \[-im(z) = -im (a+ib) =-b\] So one can see that \[Re(iz) = -b = -im(z)\] implying that \[Re(iz) = -im(z)\]

We wish to show: \(i^{4k}=1, k \in \mathbb {z} \)

This can be easily seen if we use the definition of i: \(i=\sqrt{-i}\) which implies that \( i^{4k} = (-1)^{2k} = 1^k = 1 \)

We wish to show: \(i^{4k+1} = i, k \in \mathbb {z}\)

We also want to use the definion of i here, but first we want to manipulate the exponent as \(i^{4k+1} = i^{4k}i^1\) and then using the definition: \[i^{4k}i^1 = \sqrt{-1}^{4k}i = (-1)^{2k}i = 1^{k}i = i\] \[\therefore i^{4k+1} = i\]

We wish to show: \(i^{4k+2} = -1, k \in \mathbb {z}\)

We will use a similar method as in b: \[i^{4k+2} = i^{4k}i^2 = \sqrt{-1}^{4k}\sqrt{-1}^{2} = (-1)^{2k} * (-1) = 1^k * -1 = -1\] \[\therefore i^{4k+2} = -1\]

We wish to show: \(i^{4k+3} = -i, k \in \mathbb {z}\)

We will use a similar method as in b and c: \[i^{4k+3} = i^{4k}i^3 = \sqrt{-1}^{4k}i^2i = (-1)^{2k}\sqrt{-1}^2i = 1^k*-1i = -i\] \[\therefore i^{4k+3} = -i\]

We wish to use the results of the previous problem to solve \(3i^{11} + 6i^3 + \frac{8}{i^{20}} + i^{-1}\).

We will approach each term individually. For the first \[i^{11} = i^{4k+3} \text{ where k=2, so: }i^{11} = -i\]

Continuing in this vein, we see: \[i^{3} = i^{4k+3} \text{ where k=0, so: }i^{3} = -i\] \[i^{20} = i^{4k} \text{ where k=5, so: }i^{20} = 1\] \[i^{-1} = i^{4k+3} \text{ where k=-1, so: }i^{-1} = -i\]

Putting it all together: \[3i^{11} +6i^3 + \frac{8}{i^{20}} + i^{-1} = -3i-6i +8 -i = 8-10i\]

We wish to solve for Z in the expression \(iz=4-iz\). This can be done using simple algebra: \[iz=4-iz \Rightarrow 2iz = 4 \Rightarrow z=\frac{4}{2i} = -2i\] \[\therefore z=-2i\]

We wish to solve for Z in the expression \(\frac{z}{1-z} = 1-5i \). We will continue as previously:\[z=(1-5i)(1-z) = 1-z-5i+5iz\] \[\Rightarrow 2z-5iz=1-5i=z(2-5i) \Rightarrow z=\frac{1-5i}{2-5i}\]

We can simplify this further into a real and imaginary part by multiplying both the numerator and denominator by the complex conjugate of the denominator: \[z=\frac{1-5i}{2-5i} = \frac{1-5i}{2-5i} * \frac{2+5i}{2+5i} = \frac{2+5i-10i +25}{4+25} = \frac{27-5i}{29}\] \[\therefore z= \frac{27}{29} -i \frac{5}{29}\]

We wish to solve for Z in the expression \((2-i)z+8z^2 = 0 \). Using algebra: \[(2-i)z+8z^2 = 0 \Rightarrow z(8z+(2-i) =0\]

This implies two solutions, as we might expect from a quadratic. Since \[8z+(2-i) =0 \Rightarrow 8z =i-2 \Rightarrow z=-1/4 +i/8\]

The two solutions can be seen to be \[z=0 \text { or } z=-1/4 +i/8\]

We wish to solve for Z in the expressi