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goldenfox [79]
3 years ago
5

Write the complex number in the form a + bi. square root of six(cos 315° + i sin 315°)

Mathematics
2 answers:
topjm [15]3 years ago
4 0

Answer:

\sqrt{3} -i\sqrt{3}

Step-by-step explanation:

The given complex number is,

\sqrt{6}(\cos 315^{\circ} + i \sin 315^{\circ})

we know that,

\sin 315^{\circ}=-\dfrac{1}{\sqrt2} and

\cos 315^{\circ}=\dfrac{1}{\sqrt2}

Putting the values,

=\sqrt{6}\left(\dfrac{1}{\sqrt2} -i\dfrac{1}{\sqrt2}\right)

=\sqrt{2}\sqrt{3}\left(\dfrac{1}{\sqrt2} -i\dfrac{1}{\sqrt2}\right)

=\sqrt{2}\sqrt{3}\left(\dfrac{1}{\sqrt2}\right) -i\sqrt{2}\sqrt{3}\left(\dfrac{1}{\sqrt2}\right)

=\sqrt{3} -i\sqrt{3}

tamaranim1 [39]3 years ago
3 0
a=\sqrt6\cdot \cos 315=\sqrt 6 \cdot\dfrac{\sqrt2}{2}=\dfrac{\sqrt{12}}{2}=\dfrac{2\sqrt3}{2}=\sqrt3\\
b=\sqrt6\cdot \sin 315=\sqrt 6 \cdot\left(-\dfrac{\sqrt2}{2}\right)=-\dfrac{\sqrt{12}}{2}=-\dfrac{2\sqrt3}{2}=-\sqrt3\\\\
a+bi=\sqrt3-\sqrt3i
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garik1379 [7]

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Step-by-step explanation:

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Hit and trial error method.

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