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## Solutions to TRTR Ch. 4 (Magical complex numbers) 4.1-4.5
### 4.1 (easy)
One just needs to check that $(c+id)(c-id)=c^2+d^2$, or equivalently $\frac{1}{c+id} = \frac{c-id}{c^2+d^2}.$ Multiplying by $a+ib$ on both sides of the equation gives us the desired result.
### 4.2 (medium)
By "checking" he means to write each complex number as the sum of its real and imaginary components and mechanically apply the rules $(a+ib)+(c+id)=(a+c)+i(b+d)$ and $(a+ib)(c+id)=(ac-bd)+i(ad+bc)$ to each of the equations to be checked, and to show that the real and imaginary parts are respectively the same on both sides. This is straightforward but tedious to do, and I will omit it here.
### 4.3 (easy)
We want to find real numbers $u$ and $v$ so that $(u+iv)^2=(u^2-v^2)+i(2uv)=a+ib$, that is $\begin{cases}u^2-v^2=a,\\2uv=b.\end{cases}$ Multiplying the first equation by $4u^2$ on both sides and noticing that $4u^2v^2$ is just $b^2$ because of the second equation, we have $4u^4-4u^2a-b^2=0$. This is a quadratic equation in $u^2$, and can be readily solved to yield $u^2 = (a\pm\sqrt{a^2+b^2})/2$. But $u^2$ is positive, so we can only take the plus sign here. Using the first equation again, we find $v^2 = u^2-a = (-a+\sqrt{a^2+b^2})/2$. Therefore, the square root of $a+ib$ is (plus or minus) the quantity given in the text.
### 4.4 (easy)
Assume that the series $1+x^2+x^4+\cdots$ converges to some function $f(x)$. Then $x^2f(x) = x^2+x^4+\cdots=f(x)-1$. Solving for $f(x)$, we find $f(x) = 1/(1-x^2).$
### 4.5 (easy)
Set $x=iy$ in the previous series, we get $1+(iy)^2+(iy)^4+\cdots=1/(1-(iy)^2).$ But $(iy)^2=-y^2$, $(iy^4)=y^4$, $(iy)^6=-y^6$, etc. Therefore the above equation is the same as saying $1-y^2+y^4-\cdots = 1/(1+y^2).$