+9664 1) Defining cosh z as \(\frac{1}{2}\left(e^z+e^{-z}\right)\), express in the form a + bi
(a) \(\cosh(5i)\)
(b) \(\cosh(2+5i)\)
2) Express in the form a + bi:
\(\ln(3+4i)\)
3) Evaluate \(\displaystyle\int^{\pi/2}_{0}\sin^6x\cos x\mathtt{dx}\)
4) Find \(\displaystyle\int^{a}_{0}(a^2-x^2)^{5/2}\mathtt{dx}\) and \(\displaystyle\int^{a}_{0}x^2\left(a^2-x^2\right)^{5/2}\mathtt{dx}\)
5)Find \(e^t\displaystyle\int^{t}_{0}\dfrac{x^n}{n!}e^{-x}\mathtt{dx}\)
The first 2 questions about complex number, I still have some clue for solving it, I only know that I should use Euler's relation \(e^{i\theta}=\cos \theta + i\sin \theta\), but the last 3 question about integrals...... *sigh* totally no clue in solving those :(
Also can anyone tell me that for a complex number z, what is \(\arg(z)\) and \(|z|\)?
+33616 Here are a couple of hints:
3) What is d(sin7x)/dx ?
4) (a2 - x2)5/2 → a5(1 - (x/a)2)5/2 Let sin y = x/a
arg(z) and |z| are the same thing, namely the absolute value of the complex number. If
z = a + ib then arg(z) = |z| = sqrt(a2 + b2)
+26376 1) Defining cosh z as , express in the form a + bi
(a) \(\cosh(5i)\)
(b) \(\cosh(2+5i)\)
Formula:
\(\begin{array}{|rcll|} \hline \cosh(a+i\cdot b) &=& \cosh(a) \cos(b) + i\cdot \sinh(a) \sin(b) \\ \hline \end{array} \)
(a)
\(\begin{array}{|rcll|} \hline \cosh(5i) \qquad a=0 \qquad b = 5 \\ \cosh(5i) &=& \underbrace{\cosh(0)}_{=1} \cos(5\cdot rad) + i\cdot \underbrace{\sinh(0)}_{=0} \sin(5\cdot rad) \\ \cosh(5i) &=& \cos(5\cdot rad) \\ \cosh(5i) &=& 0.28366218546\dots \\ \hline \end{array} \)
(b)
\(\begin{array}{|rcll|} \hline \cosh(2+5i) \qquad a=2 \qquad b = 5 \\ \cosh(2+5i) &=& \cosh(2)\cos(5\cdot rad) + i \cdot \sinh(2) \sin(5\cdot rad) \\ \cosh(2+5i) &=& 3.76219569108363145956\dots ~ \cdot 0.28366218546\dots \\ &+& i \cdot 3.626860407847018767668\dots (-0.95892427466) \\ \cosh(2+5i) &=& 1.067192651873\dots ~ - i \cdot 3.477884485899\dots \\ \hline \end{array} \)
+26376 2) Express in the form a + bi:
\(\ln(3+4i)\)
Formula:
\(\begin{array}{|rcll|} \hline \ln(a+i\cdot b)=\ln(\sqrt{a^2+b^2}) + i\cdot \arctan(\frac{y}{x}) \\ \hline \end{array}\)
\(\begin{array}{|rcll|} \hline \ln(3+4i) \qquad a=3 \qquad b = 4 \\ \ln(3+4i) &=& \ln(\sqrt{3^2+4^2}) + i\cdot \arctan(\frac{y}{x}) \\ \ln(3+4i) &=& \ln(5) + i\cdot 0.92729521800\dots \\ \ln(3+4i) &=& 1.609437912434100\dots + i\cdot 0.92729521800161\dots \\ \hline \end{array}\)
+26376 Sorry, without mistakes:
2) Express in the form a + bi:
\(\ln(3+4i)\)
Formula:
\(\begin{array}{|rcll|} \hline \ln(a+i\cdot b)=\ln(\sqrt{a^2+b^2}) + i\cdot \arctan(\frac{b}{a}) \\ \hline \end{array}\)
\(\begin{array}{|rcll|} \hline \ln(3+4i) \qquad a=3 \qquad b = 4 \\ \ln(3+4i) &=& \ln(\sqrt{3^2+4^2}) + i\cdot \arctan(\frac{4}{3}) \\ \ln(3+4i) &=& \ln(5) + i\cdot 0.92729521800\dots \\ \ln(3+4i) &=& 1.609437912434100\dots + i\cdot 0.92729521800161\dots \\ \hline \end{array}\)
WOW HOW DO YOU EVEN DO THAT
