Find two square roots for the complex number. Leave your answers in trigonometric form.

Question

Find two square roots for the complex number. Leave your answers in trigonometric form.  ​ ​

Quizplus · Accepted Answer

To find the square roots of a complex number in trigonometric form, we first convert the complex number to trigonometric form. Let's write $11ebf5ef_bbe5_2ed6_95d1_596838833b40_TBX9354_11$ in trigonometric form:
$r = \sqrt{11^2+e^2+b^2+f^2+5^2+e^2+5^2+2\cdot 5 \cdot e\cos(	heta - \phi)} = \sqrt{153+10\cos(	heta-\phi)}$
$	heta = \arctan\left(\frac{2e+5}{11+f}ight)$
One square root will be the number with angle $\frac{	heta-\phi}{2}$ and the other will be the number with angle $\frac{	heta-\phi}{2}+\pi$. Since $	heta$ is in quadrant IV, $\frac{	heta-\phi}{2}$ is in quadrant IV or I, and $\frac{	heta-\phi}{2}+\pi$ is in quadrant II or III. Therefore, the only option that works is option D, $11ebf5ef_bbe5_55ea_95d1_89ec47f43246_TBX9354_11$. Its angle is $\frac{-171^{\circ}}{2}=-85.5^{\circ}$, which is in quadrant III. Double this angle gives angle $-171^{\circ}$, which is the angle of the given complex number in rectangular form. Therefore, this is one of the square roots, and the other one will have angle $\frac{-85.5^{\circ}+\pi}{2} = \frac{78.5^{\circ}}{2}$.

Find Two Square Roots for the Complex Number