Math 300

1. Homework 1 solutions

1.1: 6(a)

$(-1+i)^2 = (-1)^2 -2i + i^2 = 1 - 2i -1=-2i$


1.1: 8

$\dfrac{(8+2i) - (1-i)}{(2+i)^2} = \left(\dfrac{7+3i}{3+4i}\right)\left(\dfrac{3-4i}{3-4i}\right) = \dfrac{33}{25} - i\dfrac{19}{25}$


1.1: 10

$\left[\dfrac{2+i}{6i - (1-2i)}\right]^2 = \dfrac{3+4i}{(-1+8i)^2} = \dfrac{-7+24i}{-63-16i}= -\dfrac{253}{4225} - i\dfrac{204}{4225}$


1.1: 20(a)

$iz=4-iz \iff 2iz=4 \iff z = 4/2i = -2i$


1.1: 20(b)

$\dfrac{z}{1-z}=1-5i \iff z = (1-5i)(1-z) = (1-5i) - (1-5i)z\iff (2-5i)z=(1-5i)\iff z = (1-5i)/(2-5i) = \dfrac{27}{29} - i\dfrac{5}{29}$


1.1: 20(c)

$(2-i)z+8z^2=0 \iff z(2-i+8z)=0 \iff z=0 \hbox{\ or\ } z=(-2+i)/8 = -1/4 + (1/8)i$


1.1: 30

Suppose that a set $\cal P$ having the properties (i) (ii) and (ii) exists.

Suppose that $i\in{\cal P}$. Then by property (iii) with $\alpha=\beta = i$ we have $i^2=-1\in{\cal P}$. Using property (iii) again with $\alpha=\beta = -1$ yields $(-1)(-1)=1\in{\cal P}$. But this is impossible because property (i) says that $1$ and $-1$ can't both be in $\cal P$. So it must be that $i\not\in{\cal P}$.

On the other hand suppose that $-i\in{\cal P}$. Then,as above, by property (iii) we have $(-i)^2=-1\in{\cal P}$ which leads to a contradiction as before. Thus it must be that $-i\not\in{\cal P}$.

However (i) says that one of $i$ and $-i$ must be in $\cal P$. This contradiction shows that the set $\cal P$ doesn't exist.