MATH101 Calculus Workshop Exercise 4 Solutions

W4.1(i) We complete the square

$$z^{2}+6z+11=0$$

so

$$(z+3)^{2}+2=0$$

$$(z+3)^{2}=-2$$

hence

$$z+3=\pm i\sqrt{2};$$

finally

$$z=-3\pm i\sqrt{2}.$$

(ii) We complete the square

$$z^{2}+6iz+11=0$$

so

$$(z+3i)^{2}+20=0$$

$$(z+3i)^{2}=-20$$

hence

$$z+3i=\pm i\sqrt{20};$$

finally

$$z=-3i\pm i2\sqrt{5}.$$

W4.2 For $f(x)=e^{x}\sin x$, we compute the successive derivatives in parallel columns

$$\eqalignno{f(x)=e^{x}\sin x,&\qquad f(0)=0\cr f^{\prime}(x)=e^{x}\sin x+e^{x}\cos x,&\qquad f^{\prime}(0)=1\cr f^{\prime\prime}(x)=e^{x}\sin x+e^{x}\cos x+e^{x}\cos x-e^{x}\sin x=2e^{x}\cos x,&\qquad f^{\prime\prime}(0)=2\cr f^{\prime\prime\prime}(x)=2e^{x}\cos x-2e^{x}\sin x,&\qquad f^{\prime\prime\prime}(0)=2;\cr}$$

hence the general formula for the Maclaurin series

$$f(x)=f(0)+f^{\prime}(0)x+{{f^{\prime\prime}(0)}\over{2!}}x^{2}+\dots$$

gives

$$\eqalignno{e^{x}\sin x&=0+x+{{2}\over{2!}}x^{2}+{{2}\over{3!}}x^{3}+\dots\cr&=x+x^{2}+{{x^{3}}\over{3}}+\dots.\cr}$$

W4.3 With $z=1+i$ and $w=2-3i$ we have

$$zw=5-i;\quad z+w=3-2i,\quad z-w=-1+4i,\quad\bar{z}+\bar{w}=1-i+2+3i=3+2i$$

$${{z}\over{w}}={{1+i}\over{2-3i}}={{1+i}\over{2-3i}}{{2+3i}\over{2+3i}}={{-1+5i}\over{13}}.$$

W4.4 (i) In this case the sucessive derivatives are

$$\matrix{f(x)=\sinh 2x,&&f(0)=0;\cr f^{\prime}(x)=2\cosh 2x,&&f^{\prime}(0)=2;\cr f^{\prime\prime}(x)=4\sinh 2x,&&f^{\prime\prime}(0)=0;\cr}$$

where the pattern of derivatives repeats, so the Maclaurin series is

$$\sinh 2x=2x+{{2^{3}x^{3}}\over{3!}}+{{2^{5}x^{5}}\over{5!}}+\dots.$$

(ii) We write

$$\tanh^{-1}(x)={{1}\over{2}}\log{{1+x}\over{1-x}}={{1}\over{2}}\log(1+x)-{{1}\over{2}}\log(1-x)$$

and differentiate to get

$${{d}\over{dx}}\tanh^{-1}x={{1}\over{2}}(1+x)^{-1}+{{1}\over{2}}(1-x)^{-1}.$$

(iii) The Macalurin coefficients are

$$\matrix{f(x)=\tanh^{-1}x,&&f(0)=0;\cr f^{\prime}(x)=2^{-1}(1+x)^{-1}+2^{-1}(1-x)^{-1},&&f^{\prime}(0)=1;\cr f^{\prime\prime}(x)=-2^{-1}(1+x)^{-2}+2^{-1}(1-x)^{-2}&&f^{\prime\prime}(0)=0;\cr f^{\prime\prime\prime}(x)=(1+x)^{-3}+(1-x)^{-3}&&f^{\prime\prime\prime}(0)=2;}$$

so the Maclaurin series is

$$\tanh^{-1}x=x+{{2x^{3}}\over{3!}}+\dots=x+{{x^{3}}\over{3}}+\dots.$$

and generally

$$\tanh^{-1}x=x+{{x^{3}}\over{3}}+{{x^{5}}\over{5}}+\dots+{{x^{2n+1}}\over{2n+1}}+\dots.$$

W4.6 (i) In this case the sucessive derivatives are

$$\matrix{f(x)=\tan^{-1}x,&&f(0)=0;\cr f^{\prime}(x)=(1+x^{2})^{-1},&&f^{\prime}(0)=1;\cr f^{\prime\prime}(x)=-2x(1+x^{2})^{-2},&&f^{\prime\prime}(0)=0;\cr f^{\prime\prime\prime}(x)=-2(1+x^{2})^{-2}+8x^{2}(1+x^{2})^{-3},&&f^{\prime\prime\prime}(0)=-2;\cr}$$

only odd terms are non-zero in the final column, and again the signs alternate. Hence the Maclaurin series begins

$$\tan^{-1}x=x-{{2x^{3}}\over{3!}}+\dots=x-{{x^{3}}\over{3}}+\dots.$$

(ii) Let $x=1/\sqrt{3}$. Then

$${{\pi}\over{6}}=\tan^{-1}{{1}\over{\sqrt{3}}}={{1}\over{\sqrt{3}}}-{{1}\over{3\cdot 3\sqrt{3}}}+{{1}\over{5\cdot 3^{2}\sqrt{3}}}-\dots.$$

so

$$\pi=2\sqrt{3}\Bigl(1-{{1}\over{3\cdot 3}}+{{1}\over{5\cdot 3^{2}}}-{{1}\over{7\cdot 3^{3}}}+\dots\Bigr).$$

W4.7 (i) Let $P_{n}$ be the statement

$$1+z+\dots+z^{n-1}={{1-z^{n}}\over{1-z}}.$$

Basis of induction.$P_{1}$ asserts that $1=1$, which holds.

Induction step Suppose that $P_{n}$ holds for some $n$, and consider $P_{n+1}$; then

$$\eqalignno{\bigl(1+z+\dots+z^{n-1}\bigr)+z^{n}&={{1-z^{n}}\over{1-z}}+z^{n}\cr&={{1-z^{n}+z^{n}-z^{n+1}}\over{1-z}}\cr&={{1-z^{n+1}}\over{1-z}};\cr}$$

hence $P_{n+1}$ holds and hence result by induction.

(ii) The complex conjugate of $1-z=1-re^{i\theta}=1-r\cos\theta-ri\sin\theta$ is $1-r\cos\theta+ir\sin\theta$.

We have

$$\eqalignno{{{1}\over{1-re^{i\theta}}}&={{1}\over{1-r\cos\theta-ri\sin\theta}}{{1-r\cos\theta+ir\sin\theta}\over{1-r\cos\theta+ir\sin\theta}}\cr&={{1-r\cos\theta+ir\sin\theta}\over{(1-r\cos\theta)^{2}+r^{2}\sin^{2}\theta}}\cr&={{1-r\cos\theta+ir\sin\theta}\over{1-2r\cos\theta+r^{2}\cos^{2}\theta+r^{2}\sin^{2}\theta}}\cr&={{1-r\cos\theta+ir\sin\theta}\over{1-2r\cos\theta+r^{2}}}.\cr}$$

W4.8 We have

$$f(x)={{A}\over{x^{2}}}-{{B}\over{x}}\qquad(x>0),$$

so

$$f^{\prime}(x)={{-2A}\over{x^{3}}}+{{B}\over{x^{2}}}\qquad(x>0),$$

$$f^{\prime\prime}(x)={{6A}\over{x^{4}}}-{{2B}\over{x^{3}}}\qquad(x>0);$$

so the stationary points have $0=f^{\prime}(x)$, so

$$Bx-2A=0;$$

note that $x>0$, so $x=0$ does not give a stationary point. Then $x=2A/B$ so

$$f^{\prime\prime}(2A/B)={{6AB^{4}}\over{16A^{4}}}-{{2B^{4}}\over{8A^{3}}}={{B^{4}}\over{8A^{3}}}>0;$$

so $f$ has a local mimimum at $x=2A/B$, where

$$f(2A/B)={{AB^{2}}\over{4A^{2}}}-{{B^{2}}\over{2A}}=-{{B^{2}}\over{4A}};$$

this is the minimum value of $f$ since $f(x)\rightarrow\infty$ as $x\rightarrow 0+$ and $f(x)\rightarrow 0$ as $x\rightarrow\infty$. W4.9. The general binomial theorem asserts that

$$f(x)=(1+x)^{\alpha}=\sum_{n=0}^{\infty}{{\alpha}\choose{n}}x^{n}\qquad(-1<x<1)$$

where the binomial coefficients are

$${{\alpha}\choose{n}}={{\alpha(\alpha-1)(\alpha-2)\dots(\alpha-n+1)}\over{n!}},$$

which involves a product of $n$ terms on the numerator.

With $\alpha=-{{1}\over{2}}$ we have

$$\eqalign{{{-{{1}\over{2}}}\choose{n}}&={{(-{{1}\over{2}})(-{{3}\over{2}})(-{{5}\over{2}})\dots(-{{1}\over{2}}-n+1)}\over{n!}}\cr&=(-1)^{n}{{1.3.5.\dots(2n-1)}\over{2^{n}n!}},\cr}$$

so the coefficients in the binomial series are all positive, and

$$(1-x)^{-1/2}=1+{{x}\over{2}}+{{3x^{2}}\over{2^{2}.2!}}+{{3.5x^{3}}\over{2^{3}3!}}+\dots=1+{{x}\over{2}}+{{3x^{2}}\over{8}}+{{5x^{3}}\over{16}}+\dots.$$

Remark. The product of the first $n$ odd integers divided by the product of the first $n$ even integers equals

$${{1.3.5.\dots(2n-1)}\over{2^{n}n!}}={{1.3.5.\dots(2n-1)}\over{2.4.6.\dots(2n)}}.$$

W4.10 (i) With $w=u+iv$ and $z=x+iy$, we have $w+z=(u+x)+i(v+y)$, so

$$\overline{w+z}=(u+x)-i(v+y)=\overline{w}+\overline{z};$$

also $wz=(ux-vy)+i(uy+vx),$ so

$$\overline{wz}=(ux-vy)-i(uy+vx)=(u-iv)(x-iy)=\overline{w}\overline{z};$$

further, $z/w=z\overline{w}/w\overline{w}$, where $w\overline{w}$ is real, so

$$\overline{(z/w)}=\overline{z}w/w\overline{w}=\overline{z}/\overline{w}.$$

(ii) We expand

$$\eqalign{|wz|^{2}&=|(ux-vy)+i(uy+vx)|^{2}\cr&=(ux-vy)^{2}+(uy+vx)^{2}\cr&=u^{2}x^{2}-2uxvy+v^{2}y^{2}+u^{2}y^{2}+2uyvx+v^{2}x^{2}\cr&=u^{2}x^{2}+v^{2}y^{2}+u^{2}y^{2}+v^{2}x^{2}\cr&=(u^{2}+v^{2})(x^{2}+y^{2})\cr&=|w|^{2}|z|^{2};\cr}$$

so when we take square roots we obtain the nonnegative terms $|wz|=|w||z|.$

W4.11 (i) The differential equation

$${{d^{2}y}\over{dx^{2}}}+4{{dy}\over{dx}}+3y=0$$

has auxiliary equation

$$s^{2}+4s+3=0$$

with roots $s=-3,-1$; so the general solution is

$$a=Ae^{-3x}+Be^{-x}$$

where $A,B$ are arbitrary constants.

(ii) The differential equation

$${{d^{2}y}\over{dx^{2}}}+4{{dy}\over{dx}}+4y=0$$

has auxiliary equation

$$s^{2}+4s+4=0$$

with roots $s=-2,-2$; so the general solution is

$$a=Ae^{-2x}+Bxe^{-2x}$$

where $A,B$ are arbitrary constants.

(iii) The differential equation

$${{d^{2}y}\over{dx^{2}}}+4{{dy}\over{dx}}+5y=0$$

has auxiliary equation

$$s^{2}+4s+5=0$$

with roots $s=-2+i,-2-i$; so the general solution is

$$a=Ae^{-2x}\cos x+Be^{-2x}\sin x$$

where $A,B$ are arbitrary constants.

W4.12 We have

$$\eqalignno{|z-w|^{2}&=(z-w)(\bar{z}-\bar{w})\cr&=(re^{i\theta}-se^{i\phi})(re^{-i\theta}-se^{-i\phi})\cr&=r^{2}+s^{2}-rse^{i(\theta-\phi)}-rse^{-i(\theta-\phi)}\cr&=r^{2}+s^{2}-2rs\cos(\theta-\phi);\cr}$$

which gives the cosine rule of planar trigonometry for the points $0,z,w$.

W4.13. (i) We simplify the fractions

$$w={{1+iz}\over{z+i}}$$

and get

$$\eqalign{u+iv&={{1+ix-y}\over{x+iy+i}}\cr&={{(1+ix-y)}\over{(x+iy+i)}}{{(x-i-iy)}\over{(x-i-iy)}}\cr&={{(1-y)x+x(y+1)+ix^{2}-i(1-y^{2})}\over{x^{2}+(y+1)^{2}}}\cr&={{2x+i(x^{2}+y^{2}-1)}\over{x^{2}+(y+1)^{2}}}\cr}$$

and this is real if and only if $v=0$, or equivalently $x^{2}+y^{2}-1=0$.

(ii) The real values of $w$ give the real axis in the complex plane. The values $z$ such that $x^{2}+y^{2}=1$ lie on the unit circle with centre the origin.

(iii) The unit circle $x^{2}+y^{2}=1$ is mapped to the horizontal line $v=0$.

W4.14 We have $e^{i\theta}=\cos\theta+i\sin\theta$ and $e^{i\theta}=\cos 3\theta+i\sin 3\theta,$ so by De Moivre’s Theorem

$$\eqalignno{\cos 3\theta+i\sin 3\theta&=(\cos\theta+i\sin\theta)^{3}\cr&=\cos^{3}\theta+3i\cos^{2}\theta\sin\theta-3\cos\theta\sin^{2}\theta-i\sin^{3}\theta\cr}$$

so equating real and imaginary parts, we have real part

$$\eqalignno{\cos 3\theta&=\cos^{3}\theta-3\cos\theta\sin^{2}\theta\cr&=\cos^{3}\theta-3\cos\theta(1-\cos^{2}\theta)\cr&=4\cos^{3}\theta-3\cos\theta;\cr}$$

and imaginary part

$$\eqalignno{\sin 3\theta&=3\cos^{2}\theta\sin\theta-\sin^{3}\theta\cr&=3(1-\sin^{2}\theta)\sin\theta-\sin^{3}\theta\cr&=3\sin\theta-4\sin^{3}\theta.\cr}$$

W4.15. We have $2i\sin\theta=e^{i\theta}-e^{-i\theta}$, so from the binomial expansion, we deduce

$$\eqalignno{2^{5}i\sin^{5}\theta&=(e^{i\theta}-e^{-i\theta})^{5}\cr&=e^{5i\theta}-5e^{3i\theta}+10e^{i\theta}-10e^{-i\theta}+5e^{-3i\theta}-e^{-5i\theta}\cr&=(e^{5i\theta}-e^{-5i\theta})-5(e^{3i\theta}-e^{-3i\theta})+10(e^{i\theta}-e^{-i\theta})\cr&=2i\sin 5\theta-10i\sin 3\theta+20i\sin\theta\cr}$$

and hence

$$\sin^{5}\theta={{1}\over{16}}\bigl(\sin 5\theta-5\sin 3\theta+10\sin\theta\Bigr);$$

on integrating, we deduce that

$$\int\sin^{5}\theta\,d\theta=-{{1}\over{80}}\cos 5\theta+{{5}\over{48}}\cos 3\theta-{{5}\over{8}}\cos\theta+C.$$

W4.16 (i) With

$$y={{x^{2}+4x-6}\over{x+3}}\qquad(x\neq-3),$$

the intercepts are at $x=0,y=-2$; and $y=0$ at $x^{2}+4x-6=0$, so $x=-2\pm\sqrt{10}$.

By long division we obtain

$$y=x+1-{{9}\over{x+3}},$$

so $y=x+1$ is an asymptote as $x\rightarrow\pm\infty$. Also, $y\rightarrow\infty$ as $x\rightarrow(-3)-$, while $y\rightarrow-\infty$ as $x\rightarrow(-3)+$; so $x=-3$ is a vertical asymptote.

(ii) We have

$${{dy}\over{dx}}=1+{{9}\over{(x+3)^{2}}}>0$$

so there are no stationary points.

W4.17. We have polar forms

$$z=1+i\sqrt{3}=2e^{i\pi/3},\qquad w=1+i=\sqrt{2}e^{i\pi/4};$$

hence we have

$${{z}\over{w}}={{(1+i\sqrt{3})}\over{(1+i)}}={{(1+i\sqrt{3})}\over{(1+i)}}{{(1-i)}\over{(1-i)}}={{1+\sqrt{3}+i(\sqrt{3}-1)}\over{2}},$$

while the polar form is

$${{z}\over{w}}={{2e^{i\pi/3}}\over{\sqrt{2}e^{i\pi/4}}}=\sqrt{2}e^{i\pi/12};$$

hence by equating real parts we obtain

$$\sqrt{2}\cos{{\pi}\over{12}}={{1+\sqrt{3}}\over{2}};$$

so

$$\cos{{\pi}\over{12}}={{1+\sqrt{3}}\over{2\sqrt{2}}}.$$

W4.18. The point $-1+i$ lies in the second quadrant, hence has polar form $-1+i=\sqrt{2}e^{i3\pi/4}.$ Hence by de Moivre’s theorem

$$(-1+i)^{5}=2^{5/2}e^{i15\pi/4}=2^{5/2}e^{i7\pi/4}=4\sqrt{2}{{1-i}\over{\sqrt{2}}}=4(1-i).$$

W4.19 One can do this by multiplying out, but it is better to use the polar form. We have $(-\sqrt{3}+i)(-\sqrt{3}-i)=4$, and $-\sqrt{3}+i$ lies in the second quadrant, so $-\sqrt{3}+i=2e^{i5\pi/6}$, hence

$$\eqalignno{(-\sqrt{3}+i)^{5}&=2^{5}e^{i25\pi/6}\cr&=32e^{i\pi/6}\cr&=32(\sqrt{3}/2+i/2)\cr&=16\sqrt{3}+i16.\cr}$$

so the real part is $16\sqrt{3}$ and the imaginary part is $16$.

W4.20 The auxiliary equation is

$$as^{2}+bs+c=0,$$

where $b^{2}-4ac>b^{2}$ since $ac<0$, so the roots are real with

$$\alpha={{-b-\sqrt{b^{2}-4ac}}\over{2a}}<0$$

and

$$\beta={{-b+\sqrt{b^{2}-4ac}}\over{2a}}>0.$$

Let $y_{1}(x)=e^{\alpha x}$, which is a solution with $y_{1}(x)>0$ and $y_{1}(x)\rightarrow 0$ as $x\rightarrow\infty$, and $y_{2}(x)=e^{\beta x}$, which is a solution with $y_{2}(x)>0$ and $y_{2}(x)\rightarrow\infty$ as $x\rightarrow\infty$.

W4.21 The sixth complex roots of unity are

$$z=1,\quad e^{2\pi i/6},\quad e^{4\pi i/6},\quad e^{6\pi i/6},\quad e^{8\pi i/6},\quad e^{10\pi i/6};$$

so

$$z=1,\quad\cos(\pi/3)+i\sin(\pi/3),\quad\cos(2\pi/3)+i\sin(2\pi/3),$$

$$\cos\pi+i\sin\pi,\quad\cos(4\pi/3)+i\sin(4\pi/3),\quad\cos(5\pi/3)+i\sin(5\pi/3);$$

so

$$z=1,\quad{{1+i\sqrt{3}}\over{2}},\quad{{-1+i\sqrt{3}}\over{2}},\quad-1,\quad{{-1-i\sqrt{3}}\over{2}},\quad{{1-i\sqrt{3}}\over{2}}.$$

The sixth roots of unity form the vertices of a regular hexagon on the circle with centre $0$ and radius $1$.