Solving 5i/(2+i) in Exponential Form | Imaginary Number Homework

  • Thread starter sarahs52
  • Start date
In summary: I think the author of the post you are commenting on wanted you to find a way to do the calculation without having to "remember" that arctan(1/2)=pi/4 and arctan(2)=pi/4+pi/4. I do not have a way to do that, but I think the author wants you to compare the two triangles, one with 1 and 2 as legs and the other with 1/2 and 1 as legs, to see that the difference in the tangents of the angles is the same for both triangles.I know I am not being very clear, but I do not see how to do this without just doing it. Can someone else help with
  • #1
sarahs52
6
0

Homework Statement



Show this by writing the individual factors on the left in exponential form, performing the needed operations, and finally changing back to rectangular coordinates.

Homework Equations



i is an imaginary number.

The Attempt at a Solution



Looking at the numerator, z_1 = 5i where r = |z_1|= 5, theta = pi/2.
Looking at the denominator, z_2 = 2+i where r = |z_2| = sqrt(5), theta = arctan(1/2).

So, in exponential form, 5i/(2+i) becomes 5*e^(i*pi/2) / sqrt(5)*e^(i*arctan(1/2)) =>
sqrt(5)*e^(i*pi/2) / e^(i*arctan(1/2)) = sqrt(5)*e^(i*((pi/2) - arctan(1/2))) but I don't see how this can be turned back into 1+2i since 1+2i in exponential form
is sqrt(5)*e^(i*arctan(2)).

Am I missing an algebra step or did I do something wrong?

Thank you.
 
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  • #2
You have used Euler's Identity and DeMoivre's Theorem for division correctly. The issue now is to use the right trig identity.

So [itex] \frac {5 cis \frac{\pi}{2}}{\sqrt{5} cis \theta} = \sqrt{5} cis ((\frac{\pi}{2}) - \theta) [/itex] . You wish to find [itex]arctan( \frac{\pi}{2} - \theta )[/itex]. Make a picture of a right triangle with one angle being [itex]\theta[/itex] , having [itex]tan \theta = 1/2 [/itex]; then the other angle is the complement [itex]\frac{\pi}{2} - \theta[/itex]. Looking at that angle, what would its tangent be? That will give you the rectangular components you need.
 
  • #3
Use that

tan(alpha) = 1/tan(pi/2-alpha), that is, arctan(1/2)+arctan(2)=pi/2. ehild

Edit: Dynamicsolo beat me...
 
  • #4
sarahs52 said:

Homework Statement



Show this by writing the individual factors on the left in exponential form, performing the needed operations, and finally changing back to rectangular coordinates.

Homework Equations



i is an imaginary number.

The Attempt at a Solution



Looking at the numerator, z_1 = 5i where r = |z_1|= 5, theta = pi/2.
Looking at the denominator, z_2 = 2+i where r = |z_2| = sqrt(5), theta = arctan(1/2).

So, in exponential form, 5i/(2+i) becomes 5*e^(i*pi/2) / sqrt(5)*e^(i*arctan(1/2)) =>
sqrt(5)*e^(i*pi/2) / e^(i*arctan(1/2)) = sqrt(5)*e^(i*((pi/2) - arctan(1/2))) but I don't see how this can be turned back into 1+2i since 1+2i in exponential form
is sqrt(5)*e^(i*arctan(2)).

Am I missing an algebra step or did I do something wrong?

Thank you.

When simplifying complex fractions it is always a good idea to multiply and divide by a factor that makes the denominator real. Your method is WAY too complicated.

RGV
 
  • #5
Ray Vickson said:
When simplifying complex fractions it is always a good idea to multiply and divide by a factor that makes the denominator real. Your method is WAY too complicated.

I don't think anyone would usually do the calculation by other than the "conjugate factor method", but to judge from sarahs52's post, this appears to be a "demonstration problem", where you're asked to show that the solution of the problem by an alternate method does indeed yield the same answer as the "usual" method would...
 
  • #6
Yes, dynamicsolo is correct. As I stated at the beginning of my post, the question asks to write "...the individual factors on the left in exponential form, performing the needed operations, and finally changing back to rectangular coordinates."

Thanks everyone!
 
  • #7
I do not have the trigonometric breakdown, so this may not be very satisfying, but ((pi/2) - arctan(1/2)) is indeed equal to arctan(2). Once you show that equivalence, you can rely on transitivity to take you the rest of the way.
 

What is the equation "Show 5i/(2+i) = 1+ 2i" solving for?

The equation is solving for the value of i, which is the imaginary unit in complex numbers.

What is a complex number?

A complex number is a number that contains both a real part and an imaginary part. It is expressed in the form a + bi, where a is the real part and bi is the imaginary part, with i being the imaginary unit.

How do you simplify a complex number?

To simplify a complex number, you can use the formula (a + bi)/(c + di) = ((a*c + b*d) + (b*c - a*d)i)/(c^2 + d^2). This will simplify the complex number to the form a + bi, where a and b are real numbers.

What is the significance of the complex conjugate in this equation?

The complex conjugate refers to the number with the opposite sign of the imaginary part. In this equation, the complex conjugate of 2+i is 2-i. Multiplying a complex number by its complex conjugate results in a real number, which is why it is used to simplify complex fractions.

How can this equation be used in scientific research or applications?

This equation can be used in various scientific fields that involve complex numbers, such as physics, engineering, and mathematics. It can also be applied in practical applications, such as signal processing and electronic circuit analysis.

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