Complex Numbers and Euler's Identity

[mkematt96]

Homework Statement

exp(z)=-4+3i, find z in x+iy form

Homework Equations

See attached image.

The Attempt at a Solution

See attached image. exp(z)=exp(x+iy)=exp(x)*exp(iy)=exp(x)*[cos(y)+isin(y)] ... y=inv(tan(-3/4)=-.6432 ... mag(-4+3i)=5, x= ln (5)..exp(ln(5))=5 ... 5*[cos(-.6432)+isin(-6.432)] = -4+3i

z=r*exp(i*theta)..z=5exp(-.6432) Did I do this right?

 

[mfb]

Check the value of the inverse tangent function you used. In particular, check the signs of the second last equation where you didn't see the error made earlier.

The last line is wrong. You found z already, what do you do in the last line?

 

[patric44]

you should take the lin of both sides its much simpler
and use the the law for a lin of acomplex numper

 

[Mark44]

you should take the lin of both sides its much simpler
and use the the law for a lin of acomplex numper

The natural log is usually abbreviated as "ln" not "lin".

 

[mkematt96]

Check the value of the inverse tangent function you used. In particular, check the signs of the second last equation where you didn't see the error made earlier.

The last line is wrong. You found z already, what do you do in the last line?

Theta should be the inv(tan(-3/4))= -.6435+pi =2.498 which fixes the sign error when I plug that in.

The thing I am confused on is, is exp(ln(5))* [ cos(2.498)+isin(2.498) ] is that Z ? because that expression is equal to -4 +3i, but exp(-4+3i) ISNT equal to -4+3i

 

[mkematt96]

you should take the lin of both sides its much simpler
and use the the law for a lin of acomplex numper
View attachment 203530

I agree but the professor wants us to solve it using Euler's identities

 

[patric44]

The natural log is usually abbreviated as "ln" not "lin".

i know mark44 thanks . i saw some of my professors write it as lin.

 

[vela]

Theta should be the inv(tan(-3/4))= -.6435+pi =2.498 which fixes the sign error when I plug that in.

The thing I am confused on is, is exp(ln(5))* [ cos(2.498)+isin(2.498) ] is that Z ? because that expression is equal to -4 +3i, but exp(-4+3i) ISNT equal to -4+3i

Hint: You found $e^{\ln 5}e^{2.498i}=-4+3i = e^{z}$. Rewrite the left side in the form $e^{x+iy}$.