149,946 views
38 votes
38 votes
Write the complex number in trigonometric form express the angle and radiance do not round any intermediate computations round the values in your answer to the nearest hundredth

Write the complex number in trigonometric form express the angle and radiance do not-example-1
User Nathan Tew
by
2.1k points

1 Answer

22 votes
22 votes

Given a complex number in trigonometric form below:


\begin{gathered} x+yi=r(\cos \theta+i\sin \theta),where \\ r=\sqrt[]{x^2+y^2} \\ \theta=\tan ^(-1)((y)/(x)),-\pi<\theta\leq\pi \end{gathered}

Given


z=6+4i
\begin{gathered} x=6 \\ y=4 \\ r=\sqrt[]{6^2+4^2} \\ r=\sqrt[]{36+16} \\ r=\sqrt[]{52} \end{gathered}
\begin{gathered} \theta=\tan ^(-1)((4)/(6)) \\ \theta=\tan ^(-1)(0.6667) \\ \theta=33.69^0 \end{gathered}

Convert 33.69 degrees to radian


33.69^0=0.588rad

Since tan is positive in the first and third quadrants, the value of teetha would be


\begin{gathered} 1st\text{ quadrant,} \\ \theta=33.69^0=0.588rad \\ 4th\text{ quadrant} \\ \theta=180^0+33.69^0 \\ \theta=213.69^0=3.73\text{rad} \end{gathered}

Therefore,


\begin{gathered} z=6+4i \\ z=\sqrt[]{52}(\cos 0.588+i\sin 0.588),or \\ z=\sqrt[]{52}(\cos 3.73+i\sin 3.73) \end{gathered}

Hence, the trigonometric form 6+4i is √52(cos0.588+isin0.588)

User Thomas Kejser
by
2.7k points