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

[sarahs52]

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.

 

[dynamicsolo]

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

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

 

[ehild]

Use that

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

Edit: Dynamicsolo beat me...

 

[Ray Vickson]

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

 

[dynamicsolo]

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...

 

[sarahs52]

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!

 

[losingfeeling]

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.