Notes
Problems
Express in terms of \(a+bi\)
\(\textbf{1)}\) \(\displaystyle e^{\frac{i3\pi}{2}} \)
The answer is \( \displaystyle -i \)
\(\,\,\,\,\,e^{ix}=\cos{x}+i\sin{x}\)
\(\,\,\,\,\,e^{\frac{i3\pi}{2}}=\cos{\frac{3\pi}{2}}+i\sin{\frac{3\pi}{2}}\)
\(\,\,\,\,\,e^{\frac{i3\pi}{2}}=0+i\left(-1\right)\)
\(\,\,\,\,\,\)The answer is \( \displaystyle -i \)
\(\textbf{2)}\) \(e^{\frac{2i\pi}{3}}\)
The answer is \( \frac{-1+\sqrt{3}i}{2} \)
\(\,\,\,\,\,e^{ix}=\cos{x}+i\sin{x}\)
\(\,\,\,\,\,e^{\frac{2i\pi}{3}}=\cos{\frac{2\pi}{3}}+i\sin{\frac{2\pi}{3}}\)
\(\,\,\,\,\,e^{\frac{2i\pi}{3}}=-\frac{1}{2}+i\frac{\sqrt{3}}{2}\)
\(\,\,\,\,\,\)The answer is \( \frac{-1+\sqrt{3}i}{2} \)
\(\textbf{3)}\) \(e^{6i}\)
The answer is \( .96-.28i \)
\(\,\,\,\,\,e^{ix}=\cos{x}+i\sin{x}\)
\(\,\,\,\,\,e^{6i}=\cos{6}+i\sin{6}\)
\(\,\,\,\,\,e^{6i}\approx0.96-0.28i\)
\(\,\,\,\,\,\)The answer is \( .96-.28i \)
\(\textbf{4)}\) \(2e^{4i}\)
The answer is \( -1.31-1.51i \)
\(\,\,\,\,\,e^{ix}=\cos{x}+i\sin{x}\)
\(\,\,\,\,\,2e^{4i}=2\cos{4}+2i\sin{4}\)
\(\,\,\,\,\,2e^{4i}\approx-1.31-1.51i\)
\(\,\,\,\,\,\)The answer is \( -1.31-1.51i \)
\(\textbf{5)}\) \(2.1e^{-3i}\)
The answer is \( -2.08-.30i \)
\(\,\,\,\,\,e^{ix}=\cos{x}+i\sin{x}\)
\(\,\,\,\,\,2.1e^{-3i}=2.1\cos{(-3)}+2.1i\sin{(-3)}\)
\(\,\,\,\,\,2.1e^{-3i}\approx-2.08-0.30i\)
\(\,\,\,\,\,\)The answer is \( -2.08-.30i \)
Express in terms of \(r e^{i\theta}\)
\(\textbf{6)}\) \(5+2i\)
The answer is \( 5.39e^{0.38i} \)
\(\,\,\,\,\,a+bi=\sqrt{a^2+b^2}e^{i\tan^{-1}\left(\frac{b}{a}\right)}\)
\(\,\,\,\,\,5+2i=\sqrt{5^2+2^2}e^{i\tan^{-1}\left(\frac{2}{5}\right)}\)
\(\,\,\,\,\,5+2i\approx5.39e^{0.38i}\)
\(\,\,\,\,\,\)The answer is \( 5.39e^{0.38i} \)
\(\textbf{7)}\) \(-3+4i\)
The answer is \( 5e^{2.21i} \)
\(\,\,\,\,\,a+bi=\sqrt{a^2+b^2}e^{i\tan^{-1}\left(\frac{b}{a}\right)}\)
\(\,\,\,\,\,-3+4i=\sqrt{(-3)^2+4^2}e^{i\tan^{-1}\left(\frac{4}{-3}\right)}\)
\(\,\,\,\,\,-3+4i\approx5e^{2.21i}\)
\(\,\,\,\,\,\)The answer is \( 5e^{2.21i} \)
\(\textbf{8)}\) \(-1-i\)
The answer is \( \sqrt{2}e^{\frac{5\pi}{4}}\) or \(1.41e^{3.92i} \)
\(\,\,\,\,\,a+bi=\sqrt{a^2+b^2}e^{i\tan^{-1}\left(\frac{b}{a}\right)}\)
\(\,\,\,\,\,-1-i=\sqrt{2}e^{i\tan^{-1}\left(\frac{-1}{-1}\right)}\)
\(\,\,\,\,\,-1-i=\sqrt{2}e^{\frac{5\pi}{4}}\)
\(\,\,\,\,\,-1-i\approx1.41e^{3.92i}\)
\(\,\,\,\,\,\)The answer is \( \sqrt{2}e^{\frac{5\pi}{4}}\) or \(1.41e^{3.92i} \)
\(\textbf{9)}\) \(1.5-3i\)
The answer is \( 3.35e^{2.03i} \)
\(\,\,\,\,\,a+bi=\sqrt{a^2+b^2}e^{i\tan^{-1}\left(\frac{b}{a}\right)}\)
\(\,\,\,\,\,1.5-3i=\sqrt{(1.5)^2+(-3)^2}e^{i\tan^{-1}\left(\frac{-3}{1.5}\right)}\)
\(\,\,\,\,\,1.5-3i\approx3.35e^{2.03i}\)
\(\,\,\,\,\,\)The answer is \( 3.35e^{2.03i} \)
