Complex number
- Thread starter Erros
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topsquark
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Let z = a + ib. Then
[imath]\dfrac{z}{z_1} = \dfrac{a + ib}{2 + i}[/imath] Hint: Multiply top and bottom by 2 - i.
[imath]\overline{z} z_1 = (a - ib)(2 + i)[/imath]. Hint: Multiply this out.
You will get two equations in a and b to solve simultaneously.
Let us know what you are able to do.
-Dan
[imath]\dfrac{z}{z_1} = \dfrac{a + ib}{2 + i}[/imath] Hint: Multiply top and bottom by 2 - i.
[imath]\overline{z} z_1 = (a - ib)(2 + i)[/imath]. Hint: Multiply this out.
You will get two equations in a and b to solve simultaneously.
Let us know what you are able to do.
-Dan
But why are the results of Re and Im given?
Harry_the_cat
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So that you have something to equate your expressions involving a and/or b to.
topsquark
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They are just part of the equations that gives a and b. It's like asking for the integer part of a number or telling you to choose the negative value of the square root in an equation.
-Dan
pka
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Here with some change jn notation is an approach.
If [imath]z=a+bi,~w=2+i,~\Re\left(\dfrac{z}{w}\right)=\dfrac{-3}{5},~\&~\Im\left(\overline{~z~}\cdot w\right)=1[/imath].
You should know tht [imath]\dfrac{1}{w}=\dfrac{\overline{~w~}}{|w|^2}[/imath] Thus [imath]\left(\dfrac{z}{w}\right)=\dfrac{z\cdot\overline{~w~}}{5}[/imath]
Then [imath](a+bi)(2-i))=(2a+b)+i(2b-a)[/imath] Now use the given the finish an post your findings.
[imath][/imath][imath][/imath][imath][/imath]