Question

Find the polar form of complex number.
$z = \dfrac{(2 \sqrt{3} - 1) + (2 + \sqrt{3} )i}{4}$
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Answer 1

Answer:

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Answer 2

Question:-

Fund the polar form of complex number.

$\sf \large z = \dfrac{(2 \sqrt{3} - 1) + (2 + \sqrt{3} )i}{4}$

Given:-

$\sf \large z = \dfrac{(2 \sqrt{3} - 1) + (2 + \sqrt{3} )i}{4}$

To Find:-

• The polar form of complex number.

Solution:-

The given complex number is,

$\sf \large z = \dfrac{(2 \sqrt{3} - 1) + (2 + \sqrt{3} )i}{4} .$

On comparing with,

z = a + ib.

We have,

$\sf \large a + ib = \frac{(2 \sqrt{3} - 1) }{4} + \frac{(2 + \sqrt{3} )i}{4}$

This gives,

$\sf \large a = \frac{(2 \sqrt{3} -1) }{4} \\ \\ \sf \large b = \frac{(2 + \sqrt{3} )}{4}$

Therefore, we have

$\sf \large r {}^{2} = a {}^{2} + b {}^{2}$

$\sf \large r {}^{2} = \frac{(2 \sqrt{3} - 1) {}^{2} }{4 {}^{2} } + \frac{(2 + \sqrt{3}) {}^{2} }{4 {}^{2} }$

$\sf \large r {}^{2} = \frac{(2 \sqrt{3} ) {}^{2} + (1) {}^{2} - 2 \times 2 \sqrt{3} \times 1 }{4 {}^{2} } + \frac{(2) {}^{2} + ( \sqrt{3}) {}^{2} + 2 \times 2 \times \sqrt{3} }{4 {}^{2} }$

$\sf \large r {}^{2} = \frac{4 \times 3 + 1 - 4 \sqrt{3} }{4 {}^{2} } + \frac{4 + 3 + 4 \sqrt{3} }{4 {}^{2} }$

$\sf \large r {}^{2} = \frac{12 + 1 - 4 \sqrt{3} + 7 + 4 \sqrt{3} }{4 {}^{2} }$

This gives,

$\sf \large r = \sqrt{ \frac{20}{4 {}^{2} } } \\ \\ \sf \large r = \frac{2 \sqrt{5} }{4} \\ \\ \sf \large r = \frac{ \sqrt{5} }{2} = 1.12$

Now, as

$\sf \large \theta = tan {}^{ - 1} ( \frac{b}{a} )$

Substituting the values of a and b, we have

$\sf \large \theta = tan {}^{ - 1} \bigg( \frac{ \frac{(2 + \sqrt{3)} }{4} }{ \frac{(2 \sqrt{3} - 1) }{4} } \bigg)$

$\sf \large \theta = tan {}^{ - 1} \bigg( \frac{(2 + \sqrt{3} )}{(2 \sqrt{3} - 1)} \bigg)$

$\sf \large \theta = tan {}^{ - 1} \bigg( \frac{3.73}{2.46} \bigg)$

$\sf \large \theta = tan {}^{ - 1} (1.5)$

$\sf \large \theta = 0.98$

Thus we can write the given complex number in polar form as,

z = r (cosθ + i sinθ)

Substituting the value of r and theta, we have

$\sf \large z = \frac{ \sqrt{5} }{2} (cos(0.98) + i \: sin(0.98))$

$\sf \large z = 1.12(cos(0.98) + i \: sin(0.98))$

Answer:-

$\sf \large \blue{z = 1.12(cos(0.98) + i \: sin(0.98)).}$

Hope you have satisfied. ⚘