Question

if √3+I= r( cos teta +i sin teta) then find the value of teta in radian measure
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Answer 1

Answer:

The value of theta in radian measure is

$\displaystyle{\sf\:\dfrac{\pi}{6}}$

Step-by-step-explanation:

We have given that,

$\displaystyle{\sf\:\sqrt{3}\:+\:i\:=\:r\:(\:\cos\:\theta\:+\:i\:\sin\:\theta\:)}$

We have to find the value of theta in radian measure.

Now,

$\displaystyle{\sf\:z\:=\:\sqrt{3}\:+\:i}$

Comparing with a + ib, we get,

$\displaystyle{\sf\:a\:=\:\sqrt{3}}$

$\displaystyle{\sf\:b\:=\:1}$

Now, we know that,

$\displaystyle{\pink{\sf\:|\:z\:|\:=\:\sqrt{a^2\:+\:b^2}}}$

$\displaystyle{\implies\sf\:|\:z\:|\:=\:\sqrt{(\:\sqrt{3}\:)^2\:+\:1^2}}$

$\displaystyle{\implies\sf\:|\:z\:|\:=\:\sqrt{3\:+\:1}}$

$\displaystyle{\implies\sf\:|\:z\:|\:=\:\sqrt{4}}$

$\displaystyle{\implies\sf\:|\:z\:|\:=\:2}$

$\displaystyle{\therefore\:\sf\:r\:=\:2}$

Now, we know that,

$\displaystyle{\pink{\sf\:\arg\:z\:(\:\theta\:)\:=\:\tan^{-\:1}\:\left(\:\dfrac{b}{a}\:\right)}}$

$\displaystyle{\implies\sf\:\theta\:=\:\tan^{-\:1}\:\left(\:\dfrac{1}{\sqrt{3}}\:\right)}$

$\displaystyle{\implies\underline{\boxed{\red{\sf\:\theta\:=\:\dfrac{\pi}{6}\:}}}}$

∴ The value of theta in radian measure is

$\displaystyle{\sf\:\dfrac{\pi}{6}}$

Answer 2

Step-by-step explanation:

Look at the attachment above.

I hope this helps you.