MHB Convert 7sqrt5 cis(tan−1 (2)) to Rectangular Form - Yahoo! Answers

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To convert 7√5 cis(tan−1(2)) to rectangular form, the expression is rewritten as 7√5(cos(tan−1(2)) + i sin(tan−1(2))). Using the identity tan(θ) = 2/1, the values for sin(θ) and cos(θ) are determined as sin(θ) = 2/√5 and cos(θ) = 1/√5. Substituting these values into the equation yields z = 7√5(1/√5 + i(2/√5)), simplifying to z = 7 + 14i. The final result is a = 7 and b = 14, confirming the rectangular form of the complex number.
MarkFL
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Hello Meredith,

We are given:

$$z=7\sqrt{5}\text{cis}\left(\tan^{-1}(2) \right)=7\sqrt{5}\left(\cos\left(\tan^{-1}(2) \right)+i\sin\left(\tan^{-1}(2) \right) \right)$$

In order to evaluate the trig functions, consider the following diagram:

https://www.physicsforums.com/attachments/782._xfImport

As you can see:

$$\tan(\theta)=\frac{2}{1}=2\,\therefore\,\theta= \tan^{-1}(2)$$

and so:

$$\sin(\theta)=\frac{2}{\sqrt{5}}$$

$$\cos(\theta)=\frac{1}{\sqrt{5}}$$

and so we have:

$$z=7\sqrt{5}\left(\frac{1}{\sqrt{5}}+i\frac{2}{ \sqrt{5}} \right)=7+14i$$

Hence:

$$a=7,\,b=14$$

To Meredith and any other guests viewing this topic, I invite and encourage you to post other complex number problems in our http://www.mathhelpboards.com/f21/ forum.

Best Regards,

Mark.
 

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