Complex conjugate of complex number written as fraction

[Qwertyuiop[]]

Hi, I have this question: [math]\frac{2+5i}{1-i} + \frac{2-5i}{1+i}[/math] and we are asked to write it in algebraic form. So i went the usual way by rationalizing the denominator and got the correct answer but the expression got very big. The answer is -3. But our professor told us that we could use the following property :
[math]z + \bar{z} = 2 Re(z)[/math] and using this property we get the answer much quicker. So i have a question: The complex conjugate of a number in fraction is found by just flipping the signs?

 

[Dr.Peterson]

Hi, I have this question: [math]\frac{2+5i}{1-i} + \frac{2-5i}{1+i}[/math] and we are asked to write it in algebraic form. So i went the usual way by rationalizing the denominator and got the correct answer but the expression got very big. The answer is -3. But our professor told us that we could use the following property :
[math]z + \bar{z} = 2 Re(z)[/math] and using this property we get the answer much quicker. So i have a question: The complex conjugate of a number in fraction is found by just flipping the signs?

Yes. One of the properties of the conjugate is that the conjugate of a quotient is the quotient of the conjugates.

Since the numerators, and the denominators, are each conjugate pairs, the sum is a sum of a number and its conjugate.

 

[khansaheb]

Hi, I have this question: [math]\frac{2+5i}{1-i} + \frac{2-5i}{1+i}[/math] and we are asked to write it in algebraic form. So i went the usual way by rationalizing the denominator and got the correct answer but the expression got very big. The answer is -3. But our professor told us that we could use the following property :
[math]z + \bar{z} = 2 Re(z)[/math] and using this property we get the answer much quicker. So i have a question: The complex conjugate of a number in fraction is found by just flipping the signs?

.
To test your hypothesis - the complex conjugate of a number in fraction is found by just flipping the signs - apply it to (a + ib)/(c + id) and see what you get.
.

 

[Qwertyuiop[]]

Yes. One of the properties of the conjugate is that the conjugate of a quotient is the quotient of the conjugates.

Since the numerators, and the denominators, are each conjugate pairs, the sum is a sum of a number and its conjugate.

I have another question which is related to complex numbers:
show that: [math](z_2)^5 + (\bar{z_2})^5=32 \\ z_2 = 2e^{\frac{i\pi }{3}}[/math]I find it's conjugate : [math]\bar{z_2} = 2e^{-\frac{i\pi }{3}}[/math]After factoring out 2^5 I get stuck at this point: [math]2^5\cdot e^{\frac{5\pi i}{3}}+2^5\cdot e^{-\frac{5\pi i}{3}} \\ 2^5(e^{\frac{5\pi i}{3}}+e^{-\frac{5\pi i}{3}})[/math]At this point i don't know what to do next. can you give a hint? Is there a property that we need to apply here?

 

[Dr.Peterson]

I have another question which is related to complex numbers:
show that: [math](z_2)^5 + (\bar{z_2})^5=32 \\ z_2 = 2e^{\frac{i\pi }{3}}[/math]I find it's conjugate : [math]\bar{z_2} = 2e^{-\frac{i\pi }{3}}[/math]After factoring out 2^5 I get stuck at this point: [math]2^5\cdot e^{\frac{5\pi i}{3}}+2^5\cdot e^{-\frac{5\pi i}{3}} \\ 2^5(e^{\frac{5\pi i}{3}}+e^{-\frac{5\pi i}{3}})[/math]At this point i don't know what to do next. can you give a hint? Is there a property that we need to apply here?

I'm not sure what you are being asked to do. Is it this?

Show that [imath](z_2)^5 + (\bar{z_2})^5=32 \;\;\text{ if }\;\; z_2 = 2e^{\frac{i\pi }{3}}[/imath]​

Do you notice that you again have a sum of a number and its conjugate?

Or, just write [imath]e^{\frac{5\pi i}{3}}[/imath] in rectangular form. You don't absolutely have to use a special property.

 

[Qwertyuiop[]]

I'm not sure what you are being asked to do. Is it this?

Show that [imath](z_2)^5 + (\bar{z_2})^5=32 \;\;\text{ if }\;\; z_2 = 2e^{\frac{i\pi }{3}}[/imath]​

Do you notice that you again have a sum of a number and its conjugate?

Or, just write [imath]e^{\frac{5\pi i}{3}}[/imath] in rectangular form. You don't absolutely have to use a special property.

yes , I have to prove this expression equals 32.
The expression inside the brackets is in the form [imath]z + \overline{z_2}[/imath] which is equal to 2Re(z).
Now, is Re(z) the expression [imath]e^{\frac{5\pi i}{3}}[/imath] ?

 

[topsquark]

yes , I have to prove this expression equals 32.
The expression inside the brackets is in the form [imath]z + \overline{z_2}[/imath] which is equal to 2Re(z).
Now, is Re(z) the expression [imath]e^{\frac{5\pi i}{3}}[/imath] ?

[imath]e^{5 \pi i/3}[/imath] is not a real number.

Use the Euler identity:
[imath]e^{5 \pi i/3} = cos \left ( \dfrac{5 \pi }{3} \right ) + i \, sin \left ( \dfrac{5 \pi }{3} \right )[/imath]

So
[imath]\Re \left ( e^{5 \pi i/3} \right ) = cos \left ( \dfrac{5 \pi }{3} \right ) [/imath]

Apply this also to your conjugate and you will get the result.

-Dan