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# The Euler Formula - Problem 1

###### Norm Prokup

###### Norm Prokup

**Cornell University**

PhD. in Mathematics

Norm was 4th at the 2004 USA Weightlifting Nationals! He still trains and competes occasionally, despite his busy schedule.

We are talking about the Euler Formula which is re to the i theta equals r times the quantity, cosine theta, plus i sine theta.

Now you’ll recognize this on the right hand side as the trig form of the complex number. Well it turns out that the trig formula can be said even more briefly as re to the i theta. It’s the same theta and the same r.

So what we are going to do, is we are going to take a bunch of complex numbers in rectangular form, and convert them to re to the i theta form. So this is a lot more converted to trig from we have to finds the modulus, and we have to find the argument.

So z is equals 1 plus i root 3, what’s the modulus of this number? Well r, the modulus, is going to be the square root of 1², plus root 3². So that’s 1 plus 3, 4, root 4 which is 2. So the modulus is 2. What about the argument?

The cosine of the argument is going to be 1 over 2. The real part over the modulus, and the sine of the argument is going to be root 3 over 2. Now what angle has a cosine of 1/2 and a sine of root 3 over 2? Pi over 3. So that means that z is 2e to the i times pi over 3.

Let’s try another one. Z equals 3 minus 3i. First the modulus r equals 3² plus -3². So that’s going to give me a root 9 plus 9, root 18, which is 3 root 2. Then I have to find the argument. The argument cosine theta is going to be 3 over 3 root 2, and that’s the same as 1 over root 2, and that’s the same as root 2 over 2.And the sine of theta is -3 over 3 root 2, which is exactly the opposite of what the cosine was. So we'll get minus root 2 over 2.

What angle has a cosine of root 2 over 2, and the sine of negative of root 2 over 2? You can draw a little picture if you like. The cosine is positive the sine is negative and it’s got root 2 over 2, which means that the reference angle is going to be pi over 4. So we are looking for 7 pi over 4. So that means theta is 7 pi over 4.

And in re to the i theta form, z is going to be 3 root 2, e to the i times 7 pi over 4. Finally, let’s do z equals -12i. This one is actually going to be pretty easy because you can graph it. -12i goes down here, and the negative part be the imaginary axis. -12i. Its distance from 0 which is the modulus which is 12. So r equals 12, and its argument is 3 pi over 2. So that’s theta 3 pi over 2. And so now I can write z equals r, which is 12, e to the i times 3 pi over 2. That’s it.

So remember all complex numbers can be put in this exponential form z equals re to the i theta.

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###### Norm Prokup

PhD. in Mathematics, University of Rhode Island

B.S. in Mechanical Engineering, Cornell University

He uses really creative examples for explaining tough concepts and illustrates them perfectly on the whiteboard. It's impossible to get lost during his lessons.

Thiswas EXCELLENT! I am a math teacher and have been looking for an easy/logical way to explain the lateral area of a cone to my students and this was incredibly helpful, thank you very much!”

I just learned more In 3 minutes of polygons here than I do in 3 weeks in my math class”

Hahaha, his examples are the same problems of my math HW!”

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