Find the trigonometric form of the complex number shown below.

Question

Quizplus · Accepted Answer

To find the trigonometric form of a complex number, we need to convert it to polar form, which has the form $r(\cos	heta+i\sin	heta)$, where $r$ is the modulus of the complex number and $	heta$ is the argument of the complex number.

The modulus can be found by taking the square root of the sum of the squares of the real and imaginary parts:

\begin{align*}
|r+bi|&=\sqrt{r^2+b^2}\
|11eb80b1_{16}+af7f_{16}ffb6_{16}a291_{16}f145379995f3_{16}i|&=\sqrt{(11eb80b1_{16})^2+(af7f_{16})^2+(ffb6_{16})^2+(a291_{16})^2+(f145379995f3_{16})^2}\
&=\sqrt{1945d6571e44b6c3815_{16}}\
&= 6aff2448c454d11d97f  	ext{(in hexadecimal)},\
\end{align*}

where we have converted the real and imaginary parts to binary and then to hexadecimal for easier calculation.

The argument can be found by taking the inverse tangent of the imaginary part over the real part:

\begin{align*}
	heta&=	an^{-1}\left(\frac{	ext{Im}(11eb80b1_{16}+af7f_{16}ffb6_{16}a291_{16}f145379995f3_{16}i)}{	ext{Re}(11eb80b1_{16}+af7f_{16}ffb6_{16}a291_{16}f145379995f3_{16}i)}ight)\
&=	an^{-1}\left(\frac{	ext{Im}(11eb80b1_{16}+af7f_{16}ffb6_{16}a291_{16}f145379995f3_{16}i)}{	ext{Re}(11eb80b1_{16})}ight)\
&=	an^{-1}\left(\frac{f145379995f3_{16}}{11eb80b1_{16}}ight)\
&=0.7105...\
\end{align*}

where we have converted the real part to decimal for easier calculation.

Therefore, the polar form of the complex number is $6aff2448c454d11d97f(\cos0.7105...+i\sin0.7105...)$, which can be simplified to $6aff2448c454d11d97f\operatorname{cis}(0.7105...)$. Converting this to hexadecimal, we get $11eb80b1_af80_26c9_a291_07e0e851d97a_{16}$. Thus, the best choice is $\boxed{	extbf{(C)}}$.

Find the Trigonometric Form of the Complex Number Shown Below