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: \[i^{4k+3} = i^{4k}i^3 = \sqrt{-1}^{4k}i^2i = (-1)^{2k}\sqrt{-1}^2i = 1^k*-1i = -i\] \[\therefore i^{4k+3} = -i\]

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