Chapter 17

$$\cos (3 i)=\cosh 3=10.0677$$

$$e^{\frac{\pi}{6} i}=\cos \frac{\pi}{6}+i \sin \frac{\pi}{6}=\frac{\sqrt{3}}{2}+\frac{1}{2} i$$

Substituting \(y=2\) into \(u=x^{2}-y^{2}, v=2 x y\) gives the parametric equations \(u=x^{2}-4, v=4 x.\) Using \(x=v / 4\) the first equation gives \(u=v^{2} / 16-4 .\) The graph is the parabola shown.

$$2(\cos 2 \pi+i \sin 2 \pi)$$

$$\sin (-2 i)=i \sinh (-2)=-3.6269 i$$

$$e^{-\frac{\pi}{3} i}=\cos \frac{\pi}{3}-i \sin \frac{\pi}{3}=\frac{1}{2}-\frac{\sqrt{3}}{2} i$$

Substituting \(x=-3\) into \(u=x^{2}-y^{2}, v=2 x y\) gives the parametric equations \(u=9-y^{2},\) \(v=-6 y .\) Using \(y=-v / 6\) the first equation gives \(u=9-v^{2} / 36 .\) The graph is the parabola shown.

$$u=3 x^{2}-3 y^{2}+5 x, \quad v=6 x y+5 y-6 ; \quad \frac{\partial u}{\partial x}=6 x+5=\frac{\partial v}{\partial y}, \quad \frac{\partial u}{\partial y}=-6 y=-\frac{\partial v}{\partial x}$$

$$10(\cos \pi+i \sin \pi)$$

$$\sin \left(\frac{\pi}{4}+i\right)=\sin \frac{\pi}{4} \cosh (1)+i \cos \frac{\pi}{4} \sinh (1)=1.0911+0.8310 i$$

$$e^{-1+\frac{\pi}{4} i}=e^{-1} \cos \frac{\pi}{4}+i e^{-1} \sin \frac{\pi}{4}=e^{-1}\left(\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2} i\right)$$

$$3\left(\cos \frac{3 \pi}{2}+i \sin \frac{3 \pi}{2}\right)$$

$$\cos (2-4 i)=\cos (2) \cosh (-4)-\sin (2) \sinh (-4)=-11.3642-24.8147 i$$

$$e^{2-\frac{\pi}{2} i}=e^{2} \cos \left(-\frac{\pi}{2}\right)+i e^{2} \sin \left(-\frac{\pi}{2}\right)=-e^{2} i$$

$$u=y, \quad v=x ; \quad \frac{\partial u}{\partial x}=0=\frac{\partial v}{\partial y}, \quad \frac{\partial u}{\partial y}=1,-\frac{\partial v}{\partial x}=-1 . \text { since } 1 \neq-1, f \text { is not analytic at any point. }$$

$$6\left(\cos \frac{\pi}{2}+i \sin \frac{\pi}{2}\right)$$

$$i^{11}=i\left(i^{2}\right)^{5}=i(-1)^{5}=-i$$

$$e^{\pi+\pi i}=e^{\pi} \cos \pi+i e^{\pi} \sin \pi=-e^{\pi}$$

$$\sqrt{2}\left(\cos \frac{\pi}{4}+i \sin \frac{\pi}{4}\right)$$

$$7-13 i$$

$$e^{-\pi+\frac{3 \pi}{2} i}=e^{-\pi} \cos \frac{3 \pi}{2}+i e^{-\pi} \sin \frac{3 \pi}{2}=-e^{-\pi} i$$

\(u=x^{2}-y^{2}, \quad v=-2 x y ; \quad \frac{\partial u}{\partial x}=2 x, \quad \frac{\partial v}{\partial y}=-2 x ; \quad \frac{\partial u}{\partial y}=-2 y, \quad-\frac{\partial v}{\partial x}=2 y\) The Cauchy-Riemann equations hold only at \((0,0) .\) since there is no neighborhood about \(z=0\) within which \(f\) is differentiable we conclude \(f\) is nowhere analytic.

$$5 \sqrt{2}\left(\cos \frac{7 \pi}{4}+i \sin \frac{7 \pi}{4}\right)$$

$$\sec (\pi+i)=\frac{1}{\cos (\pi+i)}=\frac{1}{-\cosh (1)}=-0.6481$$

$$e^{1.5+2 i}=e^{1.5} \cos 2+i e^{1.5} \sin 2=-1.8650+4.0752 i$$

\(u=x^{2}+y^{2}, \quad v=0 ; \quad \frac{\partial u}{\partial x}=2 x, \quad \frac{\partial v}{\partial y}=0 ; \quad \frac{\partial u}{\partial y}=2 y, \quad-\frac{\partial v}{\partial x}=0\) The Cauchy-Riemann equations hold only at \((0,0) .\) since there is no neighborhood about \(z=0\) within which \(f\) is differentiable we conclude \(f\) is nowhere analytic.

$$2\left(\cos \frac{5 \pi}{6}+i \sin \frac{5 \pi}{6}\right)$$

$$-7+5 i$$

$$e^{-0.3+0.5 i}=e^{-0.3} \cos 0.5+i e^{-0.3} \sin 0.5=0.6501+0.3552 i$$

\(u=\frac{x}{x^{2}+y^{2}}, v=\frac{y}{x^{2}+y^{2}} ; \quad \frac{\partial u}{\partial x}=\frac{y^{2}-x^{2}}{\left(x^{2}+y^{2}\right)^{2}}, \quad \frac{\partial v}{\partial y}=\frac{x^{2}-y^{2}}{\left(x^{2}+y^{2}\right)^{2}} ; \quad \frac{\partial u}{\partial y}=-\frac{2 x y}{\left(x^{2}+y^{2}\right)^{2}}=\frac{\partial v}{\partial x}\) The Cauchy-Riemann equations hold only at \((0,0) .\) since there is no neighborhood about \(z=0\) within which \(f\) is differentiable, we conclude \(f\) is nowhere analytic.

$$4\left(\cos \frac{4 \pi}{3}+i \sin \frac{4 \pi}{3}\right)$$

$$\cosh (\pi i)=\cos (i(\pi i))=\cos (-\pi)=\cos \pi=-1$$

$$\frac{3 \sqrt{2}}{2}\left(\cos \frac{5 \pi}{4}+i \sin \frac{5 \pi}{4}\right)$$

$$11-10 i$$

$$\tan ^{-1} 3 i=\frac{i}{2} \ln \left(\frac{4 i}{-2 i}\right)=\frac{i}{2} \ln (-2)=-\frac{\pi}{2}-n \pi+i \log _{e} \sqrt{2}, \quad n=0,\pm 1,\pm 2, \ldots$$

$$e^{-0.23-i}=e^{-0.23} \cos (-1)+i e^{-0.23} \sin (-1)=0.4293-0.6686 i$$

$$f(z)=\left(3 x^{2}-3 y^{2}+2 x\right)+i(-6 x y+2 y)$$

$$6\left[\cos \left(-\frac{\pi}{6}\right)+i \sin \left(-\frac{\pi}{6}\right)\right]$$

$$\frac{3}{4}+\frac{2}{3} i$$

$$\sinh \left(1+\frac{\pi}{3} i\right)=\sinh (1) \cos \frac{\pi}{3}+i \cosh (1) \sin \frac{\pi}{3}=0.5876+1.3363 i$$

$$e^{\frac{11 \pi}{12} i}=\cos \frac{11 \pi}{12}+i \sin \frac{11 \pi}{12}=-0.9659+0.2588 i$$

$$\cosh ^{-1} i=\ln [(1+\pm \sqrt{2}) i]=\left\\{\begin{array}{ll} \log _{e}(1+\sqrt{2})+\left(\frac{\pi}{2}+2 n \pi\right) i & n=0,\pm 1,\pm 2, \ldots \\ \log _{e}(\sqrt{2}-1)+\left(-\frac{\pi}{2}+2 n \pi\right) i & \end{array}\right.$$

$$\cosh (2+3 i)=\cosh (2) \cos (3)+i \sinh (2) \sin (3)=-3.7245+0.5118 i$$

$$e^{5+\frac{5 \pi}{2} i}=e^{5} \cos \frac{5 \pi}{2}+i e^{5} \sin \frac{5 \pi}{2}=e^{5} i$$

$$e^{-i z}=e^{y-x i}=e^{y} \cos x-i e^{y} \sin x$$

$$\cos \left(\frac{\pi}{2}+i \ln 2\right)=\cos \frac{\pi}{2} \cosh (\ln 2)-i \sin \frac{\pi}{2} \sinh (\ln 2)=-i \cdot \frac{e^{\ln 2}-e^{\ln 2^{-1}}}{2}=-i \cdot \frac{2-\frac{1}{2}}{2}=-\frac{3}{4} i$$

\(u=\frac{x^{3}+x y^{2}+x}{x^{2}+y^{2}}, v=\frac{x^{2} y+y^{3}-y}{x^{2}+y^{2}} ; \quad \frac{\partial u}{\partial x}=\frac{x^{4}+2 x^{2} y^{2}-x^{2}+y^{2}+y^{4}}{\left(x^{2}+y^{2}\right)^{2}}=\frac{\partial v}{\partial y}, \quad \frac{\partial u}{\partial y}=\frac{-2 x y}{\left(x^{2}+y^{2}\right)^{2}}=-\frac{\partial v}{\partial x}\) \(f\) is analytic in any domain not containing \(z=0\)

$$\frac{i}{1+i} \cdot \frac{1-i}{1-i}=\frac{i+1}{2}=\frac{1}{2}+\frac{1}{2} i$$

\(\frac{e^{i z}-e^{-i z}}{2 i}=2\) gives \(e^{2(i z)}-4 i e^{i z}-1=0 .\) By the quadratic formula, \(e^{i z}=2 i \pm \sqrt{3} i\) and so $$i z=\ln [(2 \pm \sqrt{3}) i]$$ $$z=-i\left[\log _{e}(2 \pm \sqrt{3})+\left(\frac{\pi}{2}+2 n \pi\right) i\right]=\frac{\pi}{2}+2 n \pi-i \log _{e}(2 \pm \sqrt{3}), \quad n=0, \pm 1 .,\pm 2, \ldots$$

$$\frac{2-4 i}{3+5 i} \cdot \frac{3-5 i}{3-5 i}=\frac{-14-22 i}{34}=-\frac{7}{17}-\frac{11}{17} i$$