Question

The complex number (2i/1+I)^1/6 is equivalent is a)12√2 cos (π/24) b)12√2 cis (25π/24) c)(1+I)^1/6 d) All of the above

Answer 1

Step-by-step explanation:

We have,

$z = \left( \dfrac{2i}{1 + i} \right)^{ \frac{1}{6} }$

$\implies \: z = \left \{\dfrac{2i \left(1 - i \right)}{\left(1 + i \right)\left(1 - i \right)} \right \}^{ \frac{1}{6} }$

$\implies \: z = \left \{\dfrac{2i \left(1 - i \right)}{\left(1 \right)^{2} + \left(1\right)^{2} } \right \}^{ \frac{1}{6} }$

$\implies \: z = \left \{\dfrac{2i \left(1 - i \right)}{2} \right \}^{ \frac{1}{6} }$

$\implies \: z = \left \{i \left(1 - i \right) \right \}^{ \frac{1}{6} }$

$\implies \: z = \left(i - {i}^{2} \right)^{ \frac{1}{6} }$

$\implies \: z = \left(i + 1 \right)^{ \frac{1}{6} }$

$\implies \: z = \left(1 + i \right)^{ \frac{1}{6} }$

$\implies \: z = \left \{ \sqrt{2} \left( \dfrac{1}{ \sqrt{2} } + \dfrac{i}{ \sqrt{2} } \right) \right \}^{ \frac{1}{6} }$

$\implies \: z = \sqrt[12]{2} \left( \dfrac{1}{ \sqrt{2} } + \dfrac{i}{ \sqrt{2} } \right)^{ \frac{1}{6} }$

$\implies \: z = \sqrt[12]{2} \left \{ \cos \left(\dfrac{\pi}{ 4 } \right) + i \sin \left(\dfrac{\pi}{ 4 } \right) \right \}^{ \frac{1}{6} }$

$\implies \: z = \sqrt[12]{2} \left \{ {e}^{i \frac{\pi}{4} } \right \}^{ \frac{1}{6} }$

$\implies \: z = \sqrt[12]{2} \cdot{e}^{i \frac{\pi}{4} \cdot\frac{1}{6} }$

$\implies \: z = \sqrt[12]{2} \cdot{e}^{i \frac{\pi}{24} }$

$\implies \: \tt{ z = \sqrt[12]{2} \cdot cis\left(\frac{\pi}{24} \right) }$