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I need to calculate these kind of values in exams in best speedy way.

Convert $1.46 + 3.17j$ to polar form ($r∠θ$)

Is there is any solution to find of the values as quick as possible?

By the way, the answer is $3.5∠65.3^\circ$.

Yogus
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  • How did you manage to find the angle is 65.3 deg ? From where do you know the sin cos and tan of the same angle ? – DiffeoR Mar 04 '14 at 00:29
  • Do you want to do this in your head or something? I mean, to get the angle all you need is $\arctan(3.17/1.46)$ and to get the other number all you need is $\sqrt {1.46^2 + 3.17^2}$. Both of these are easy to get with a calculator. – recursive recursion Mar 04 '14 at 00:46
  • ActuallY i am trying to solve one electrical network theorom. I was able to find it out till the complex number but not able to calculate the polar value. The answer I wrote above was written in the solution of the question. – Yogus Mar 04 '14 at 00:47
  • Ah, I was wondering why you were using $j$ instead of $i$ – recursive recursion Mar 04 '14 at 00:48
  • @recursiverecursion Yeah something in my head as calculators are not allowed . So is there is any fast calculation to get the polar values – Yogus Mar 04 '14 at 00:49
  • @recursiverecursion its 'i' :) – Yogus Mar 04 '14 at 00:49
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    When you think about the definition of polar coordinates, you'll realize that you kind of need $\arctan$ to convert between cartesian and polar. I'm not really seeing a way out of that, though I'll try my best to think of something. – recursive recursion Mar 04 '14 at 00:51
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    @Yogus See my answer here. – Git Gud Mar 04 '14 at 00:58

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The general formula is $r = \sqrt {a^2 + b^2}, \theta = \arctan(b/a)$ for any complex number of the form $a + bi$. Both of these values will be very difficult to find by hand, as there are square roots and trigonometric functions, but there are methods of finding such things by hand as shown here and here. Please note that even though there are methods for estimating the square root and arctan of a number, these methods are by no means fast to do by hand, though it is possible.

recursive recursion
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