can you please help me with question 7c and 3c? thanks

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lemjay | High School Teacher | (Level 3) Senior Educator

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(7c) Find all possible values of m and n, if any, for which


Verify your solutions.


Express the right side in x+yi form.


Then, express each complex number in exponential form `re^(theta i).`

To convert them to exponetial form apply the formulas:


`theta =tan^(-1) (y/x)`

When solving for the values of the angle, consider the signs of x and y values to determine the quadrant in which the angle belongs. Also, the angle should be in radians.

For `sqrt3-i`    , plug-in `x=sqrt3 `  and  `y=-1` to the formulas above.` `



So, the exponetial form of  `sqrt3-i`  is  `2e^((11pi)/6i)` .

For -1+i, plug-in x=-1 and y=1.



Hence, the exponential form of `-1+i`  is  `sqrt2e^((3pi)/4i)` .

For -4+0i, plug-in x=-4 and y=0.



Thus, the exponential form of  `-4+0i`  is `4e^(pi i)` .

So when express the complex numbers are expresses in exponential form, the equation becomes:

`(sqrt3 -i)/(-1+i)=-4+i`   

`((2e^((11pi)/6i))^m)/((sqrt2e^((3pi)/4i))^n)=4e^(pi i)`

Then, apply the exponents rule to simplify the left side.

`(2^me^((11pi)/6mi))/(2^(n/2)e^((3pi)/4ni))=4e^(pi i)`

`2^(m-n/2)e^((11pi)/6mi -(3pi)/4ni)=4e^(pi i)`

In order for the left side and right side to have same base, factor 4.

`2^(m-n/2) e^((11pi)/6mi-(3pi)/4ni)=2^2e^(pi i)`

Then, for same base, set their exponents equal to each other.

For base 2, the resulting equation is:


`2m -n=4`

`n=2m-4`          (Let this be EQ1.)

For base e, the resulting equation is:

`(11pi)/6mi-(3pi)/4ni=pi i`

`(11pi)/6mi-(3pi)/4ni-pi i=0`

`pi i(11/6m-3/4n-1)=0`

`11/6m-3/4n-1=0`              (Let this be EQ2.)

Then , plug-in EQ1 to EQ2.






And, plug-in the value of m to EQ1.



Therefore, the values of m and n that satisfies the given equation are m=-6 and n=-16.