The equality must be a first return time: an order starting $GR$ or $RG$ would already have stopped at draw two. The only successful four-draw orders are therefore $GGRR$ and $RRGG$. Their probabilities under sampling without replacement add toHere $N=g+r$. Counting all balanced four-draw orders would include games that have already stopped.

Success after two draws occurs on the two mutually exclusive events $GR$ and $RG$. With $N=g+r$, sampling without replacement gives

After one draw the two counts are either $(1,0)$ or $(0,1)$ and cannot agree. ThereforeThis is the elementary parity restriction behind the first return time interpretation of this game.

For the reciprocal Möbius transformation, substitute $z=1/w$, $\bar{z}=1/\bar{w}$ and multiply by $w\bar{w}$, which is nonzero on the stipulated domain. This yieldsConsequentlyFor a nondegenerate original circle, $c=0$ says that it passes through the origin, and its image is a straight line. The restriction $w\ne 0$ must be retained: multiplying through does not authorize adding that excluded point. Degenerate choices can give a constant or an identically zero equation instead of a straight line.

For the translation, expand the products of complex numbers and collect the terms in $w\bar{w}$, $\bar{w}$, $w$ and the constant:It follows thatThe last coefficient is real because $b\bar{\alpha }+\bar{b}\alpha =2\mathrm{Re}(\bar{b}\alpha )$.

For the real positive dilation factor, $\bar{z}=\alpha \bar{w}$ and $z\bar{z}={\alpha }^{2}w\bar{w}$. ThusandThe original PDF specifies a dilation; the differently named transformation in the OCR is an error.

Under the given rotation, $\bar{z}=\bar{\alpha }\bar{w}$ and $z\bar{z}=|\alpha {|}^{2}w\bar{w}=w\bar{w}$. Collecting terms givesSince $\overline{b\bar{\alpha }}=\bar{b}\alpha$, the transformed coefficients are

There is a single point exactly when the radius is zero:Then $|z+b{|}^2=0$ forces this unique value of the complex number $z$.

A squared complex modulus is nonnegative. Conversely, every nonnegative right-hand side of $|z+b{|}^2=|b{|}^2-c$ can be attained. The necessary and sufficient condition is therefore

Since $z\bar{z}=x^2+y^2$, the real quadratic equation isCompleting the square makes its geometry explicit:

Completing the square directly with the complex modulus givesFor $|b{|}^2>c$, this is the circle with centre $-b$ and radius $\sqrt{|b{|}^2-c}$. Equality gives just the point $-b$, and $|b{|}^2<c$ gives the empty locus. In particular, the nondegenerate answer is

Multiplying the complex numbers gives $b\bar{z}+\bar{b}z=2(b_{1}x+b_{2}y)$. Hence the real linear equation is

Using the complex conjugate, $b\bar{z}+\bar{b}z=2\mathrm{Re}(\bar{b}z)$. The locus is therefore the straight lineIn the complex plane, the quantity $\mathrm{Re}(\bar{b}z)/|b|$ is the signed scalar projection of $z$ in the direction $b$. Thus the straight line is perpendicular to $b$ and passes through $-cb/(2|b{|}^2)$. This description uses no separation of $z$ into real coordinates.

Put $D_n=(-1{)}^n{e}^{\pi }-1$. The two linear equations from parts (i) and (ii) give $I_n=D_n+n{J}_n$ and $J_n=-n{I}_n$, so $(1+n^2)I_n=D_n$. ThereforeThese formulas include $n=0$, when $I_0=e^{\pi }-1$ and $J_0=0$.

For $J_n$, take $u=\sin (nx)$ and $dv=e^{x}dx$ in integration by parts. This yieldssince $\sin (n\pi )=\sin 0=0$ for every integer $n$.

Let $I_n$ denote the integral with the cosine factor and $J_n$ the one with the sine factor. In integration by parts, take $u=\cos (nx)$ and $dv=e^{x}dx$. ThenThe integer restriction gives $\cos (n\pi )=(-1{)}^n$.

The reversed product is the dot product, a scalar represented by a $1\times 1$ matrix:

The outer product of the column vector with its matrix transpose is the $2\times 2$ matrix

For $x\ne 0$ and $y\ne 0$, put $v=y/x$. This is the standard substitution for a homogeneous first-order differential equation. Since $y^{\prime }=v+x{v}^{\prime }$, the equation becomesTaking antiderivatives yields $v^2=2\ln |x|+C$. The explicit general solution is thereforeon any interval where $x\ne 0$ and $2\ln |x|+C>0$, with a fixed choice of sign. For a check that does not divide by $y$, set $w=y^2$ to get $w^{\prime }-2w/x=2x$ and hence $w=x^2(2\ln |x|+C)$.

There is no missing zero solution: at $y=0$ with $x\ne 0$, the original equation would require $x^2=0$. Also no real solution on an interval about $x=0$ can arise from this family, since $2\ln |x|+C$ becomes negative near $0$.

Where $\cos x\ne 0$, the separable differential equation is $y^{\prime }=y\tan x$. Separating and taking antiderivatives gives $\ln |y|=-\ln |\cos x|+C$, for nonzero $y$. Including the zero solution, the explicit general solution ison any interval with $\cos x\ne 0$. It can also be obtained without dividing by $y$: the product rule gives $(y\cos x{)}^{\prime }=0$.

A nonzero solution cannot extend through a zero of $\cos x$. If a differentiable solution is required on an interval containing such a zero, the original equation there requires $y=0$, and the only such solution across that zero is $y\equiv 0$.