Euler’s formula connects exponential expressions with complex numbers using \(e^{i\theta}=\cos{\theta}+i\sin{\theta}\). This makes it possible to rewrite complex exponentials in \(a+bi\) form and rewrite complex numbers in polar/exponential form. These problems include exact unit-circle angles, decimal approximations, and conversions between rectangular and exponential form.
Notes
Practice 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 \( -2.08-.30i \)
\(\textbf{6)}\) \(e^{i\pi}\)
The answer is \(-1\)
\(\,\,\,\,\,e^{ix}=\cos{x}+i\sin{x}\)
\(\,\,\,\,\,e^{i\pi}=\cos{\pi}+i\sin{\pi}\)
\(\,\,\,\,\,e^{i\pi}=-1+i(0)\)
\(\,\,\,\,\,\)The answer is \(-1\)
\(\textbf{7)}\) \(3e^{\frac{i\pi}{4}}\)
The answer is \(\displaystyle \frac{3\sqrt{2}}{2}+\frac{3\sqrt{2}}{2}i\)
\(\,\,\,\,\,re^{i\theta}=r(\cos{\theta}+i\sin{\theta})\)
\(\,\,\,\,\,3e^{\frac{i\pi}{4}}=3\left(\cos{\frac{\pi}{4}}+i\sin{\frac{\pi}{4}}\right)\)
\(\,\,\,\,\,3e^{\frac{i\pi}{4}}=3\left(\frac{\sqrt{2}}{2}+i\frac{\sqrt{2}}{2}\right)\)
\(\,\,\,\,\,\)The answer is \(\displaystyle \frac{3\sqrt{2}}{2}+\frac{3\sqrt{2}}{2}i\)
\(\textbf{8)}\) \(4e^{-\frac{i\pi}{6}}\)
The answer is \(2\sqrt{3}-2i\)
\(\,\,\,\,\,re^{i\theta}=r(\cos{\theta}+i\sin{\theta})\)
\(\,\,\,\,\,4e^{-\frac{i\pi}{6}}=4\left(\cos{\left(-\frac{\pi}{6}\right)}+i\sin{\left(-\frac{\pi}{6}\right)}\right)\)
\(\,\,\,\,\,4e^{-\frac{i\pi}{6}}=4\left(\frac{\sqrt{3}}{2}-\frac{1}{2}i\right)\)
\(\,\,\,\,\,\)The answer is \(2\sqrt{3}-2i\)
\(\textbf{9)}\) \(5e^{\frac{5i\pi}{6}}\)
The answer is \(\displaystyle -\frac{5\sqrt{3}}{2}+\frac{5}{2}i\)
\(\,\,\,\,\,re^{i\theta}=r(\cos{\theta}+i\sin{\theta})\)
\(\,\,\,\,\,5e^{\frac{5i\pi}{6}}=5\left(\cos{\frac{5\pi}{6}}+i\sin{\frac{5\pi}{6}}\right)\)
\(\,\,\,\,\,5e^{\frac{5i\pi}{6}}=5\left(-\frac{\sqrt{3}}{2}+\frac{1}{2}i\right)\)
\(\,\,\,\,\,\)The answer is \(\displaystyle -\frac{5\sqrt{3}}{2}+\frac{5}{2}i\)
\(\textbf{10)}\) \(2e^{\frac{3i\pi}{2}}\)
The answer is \(-2i\)
\(\,\,\,\,\,re^{i\theta}=r(\cos{\theta}+i\sin{\theta})\)
\(\,\,\,\,\,2e^{\frac{3i\pi}{2}}=2\left(\cos{\frac{3\pi}{2}}+i\sin{\frac{3\pi}{2}}\right)\)
\(\,\,\,\,\,2e^{\frac{3i\pi}{2}}=2(0-i)\)
\(\,\,\,\,\,\)The answer is \(-2i\)
Express in terms of \(r e^{i\theta}\)
\(\textbf{11)}\) \(5+2i\)
The answer is \( 5.39e^{0.38i} \)
\(\,\,\,\,\,r=\sqrt{a^2+b^2}\)
\(\,\,\,\,\,r=\sqrt{5^2+2^2}=\sqrt{29}\approx5.39\)
\(\,\,\,\,\,\theta=\tan^{-1}\left(\frac{2}{5}\right)\approx0.38\)
\(\,\,\,\,\,5+2i\approx5.39e^{0.38i}\)
\(\,\,\,\,\,\)The answer is \( 5.39e^{0.38i} \)
\(\textbf{12)}\) \(-3+4i\)
The answer is \( 5e^{2.21i} \)
\(\,\,\,\,\,r=\sqrt{a^2+b^2}\)
\(\,\,\,\,\,r=\sqrt{(-3)^2+4^2}=5\)
\(\,\,\,\,\,\text{The point }(-3,4)\text{ is in Quadrant II.}\)
\(\,\,\,\,\,\theta=\pi-\tan^{-1}\left(\frac{4}{3}\right)\approx2.21\)
\(\,\,\,\,\,-3+4i\approx5e^{2.21i}\)
\(\,\,\,\,\,\)The answer is \( 5e^{2.21i} \)
\(\textbf{13)}\) \(-1-i\)
The answer is \( \sqrt{2}e^{\frac{5\pi i}{4}}\) or \(1.41e^{3.93i} \)
\(\,\,\,\,\,r=\sqrt{a^2+b^2}\)
\(\,\,\,\,\,r=\sqrt{(-1)^2+(-1)^2}=\sqrt{2}\)
\(\,\,\,\,\,\text{The point }(-1,-1)\text{ is in Quadrant III.}\)
\(\,\,\,\,\,\theta=\frac{5\pi}{4}\)
\(\,\,\,\,\,-1-i=\sqrt{2}e^{\frac{5\pi i}{4}}\)
\(\,\,\,\,\,-1-i\approx1.41e^{3.93i}\)
\(\,\,\,\,\,\)The answer is \( \sqrt{2}e^{\frac{5\pi i}{4}}\) or \(1.41e^{3.93i} \)
\(\textbf{14)}\) \(1.5-3i\)
The answer is \( 3.35e^{-1.11i} \) or \(3.35e^{5.18i}\)
\(\,\,\,\,\,r=\sqrt{a^2+b^2}\)
\(\,\,\,\,\,r=\sqrt{(1.5)^2+(-3)^2}\approx3.35\)
\(\,\,\,\,\,\text{The point }(1.5,-3)\text{ is in Quadrant IV.}\)
\(\,\,\,\,\,\theta=\tan^{-1}\left(\frac{-3}{1.5}\right)\approx-1.11\)
\(\,\,\,\,\,1.5-3i\approx3.35e^{-1.11i}\)
\(\,\,\,\,\,\)The answer is \( 3.35e^{-1.11i} \) or \(3.35e^{5.18i}\)
\(\textbf{15)}\) \(3+3i\)
The answer is \(3\sqrt{2}e^{\frac{\pi i}{4}}\)
\(\,\,\,\,\,r=\sqrt{a^2+b^2}\)
\(\,\,\,\,\,r=\sqrt{3^2+3^2}=3\sqrt{2}\)
\(\,\,\,\,\,\theta=\tan^{-1}\left(\frac{3}{3}\right)=\frac{\pi}{4}\)
\(\,\,\,\,\,\)The answer is \(3\sqrt{2}e^{\frac{\pi i}{4}}\)
\(\textbf{16)}\) \(-2+2\sqrt{3}i\)
The answer is \(4e^{\frac{2\pi i}{3}}\)
\(\,\,\,\,\,r=\sqrt{a^2+b^2}\)
\(\,\,\,\,\,r=\sqrt{(-2)^2+(2\sqrt{3})^2}\)
\(\,\,\,\,\,r=\sqrt{4+12}=4\)
\(\,\,\,\,\,\text{The point }(-2,2\sqrt{3})\text{ is in Quadrant II.}\)
\(\,\,\,\,\,\theta=\frac{2\pi}{3}\)
\(\,\,\,\,\,\)The answer is \(4e^{\frac{2\pi i}{3}}\)
\(\textbf{17)}\) \(-4i\)
The answer is \(4e^{\frac{3\pi i}{2}}\)
\(\,\,\,\,\,r=4\)
\(\,\,\,\,\,\text{The point }(0,-4)\text{ is on the negative imaginary axis.}\)
\(\,\,\,\,\,\theta=\frac{3\pi}{2}\)
\(\,\,\,\,\,\)The answer is \(4e^{\frac{3\pi i}{2}}\)
\(\textbf{18)}\) \(6\)
The answer is \(6e^{0i}\)
\(\,\,\,\,\,6=6+0i\)
\(\,\,\,\,\,r=6\)
\(\,\,\,\,\,\theta=0\)
\(\,\,\,\,\,\)The answer is \(6e^{0i}\)
\(\textbf{19)}\) \(-5\)
The answer is \(5e^{i\pi}\)
\(\,\,\,\,\,-5=-5+0i\)
\(\,\,\,\,\,r=5\)
\(\,\,\,\,\,\theta=\pi\)
\(\,\,\,\,\,\)The answer is \(5e^{i\pi}\)
\(\textbf{20)}\) \(2-2i\)
The answer is \(2\sqrt{2}e^{-\frac{\pi i}{4}}\) or \(2\sqrt{2}e^{\frac{7\pi i}{4}}\)
\(\,\,\,\,\,r=\sqrt{a^2+b^2}\)
\(\,\,\,\,\,r=\sqrt{2^2+(-2)^2}=2\sqrt{2}\)
\(\,\,\,\,\,\text{The point }(2,-2)\text{ is in Quadrant IV.}\)
\(\,\,\,\,\,\theta=-\frac{\pi}{4}\text{ or }\frac{7\pi}{4}\)
\(\,\,\,\,\,\)The answer is \(2\sqrt{2}e^{-\frac{\pi i}{4}}\) or \(2\sqrt{2}e^{\frac{7\pi i}{4}}\)
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