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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Here is a little puzzle from the book 100 Geometric Games by Pierre Berloquin. The side of a small square is one meter long and the side of a larger square one and a half meters long. One vertex of the large square is at the center of the small square. The side of the large square cuts two sides of the small square into one- third parts and two-thirds parts. What is the area where the squares overlap?

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