- #1
BOAS
- 553
- 19
Hello,
The complex numbers [itex]z_{1} = \frac{a}{1 + i}[/itex] and [itex]z_{2} = \frac{b}{1+2i}[/itex] where a and b are real, are such that [itex]z_{1} + z_{2} = 1. Find a and b.
This looked like a time for partial fractions to me, so I went down that road;
[itex] \frac{a}{1 + i} + \frac{b}{1+2i} = 1[/itex]
[itex]\frac{a(1+2i)}{(1 + i)(1+2i)} + \frac{b(1+i)}{(1+2i)(1+i)} = 1[/itex]
[itex]a(1+2i) + b(1+i) = (1+i)(1+2i)[/itex]
Expanding the brackets gives me;
[itex]a(1+2i) + b(1+i) = (1+i)(1+2i)[/itex]
[itex]a(1+2i) + b(1+i) = -1 + 3i[/itex]
∴ [itex]a + b = -1[/itex]
And now I'm stuck...
Is this the right approach? And, how do I move forward?
Thanks!
Homework Statement
The complex numbers [itex]z_{1} = \frac{a}{1 + i}[/itex] and [itex]z_{2} = \frac{b}{1+2i}[/itex] where a and b are real, are such that [itex]z_{1} + z_{2} = 1. Find a and b.
Homework Equations
The Attempt at a Solution
This looked like a time for partial fractions to me, so I went down that road;
[itex] \frac{a}{1 + i} + \frac{b}{1+2i} = 1[/itex]
[itex]\frac{a(1+2i)}{(1 + i)(1+2i)} + \frac{b(1+i)}{(1+2i)(1+i)} = 1[/itex]
[itex]a(1+2i) + b(1+i) = (1+i)(1+2i)[/itex]
Expanding the brackets gives me;
[itex]a(1+2i) + b(1+i) = (1+i)(1+2i)[/itex]
[itex]a(1+2i) + b(1+i) = -1 + 3i[/itex]
∴ [itex]a + b = -1[/itex]
And now I'm stuck...
Is this the right approach? And, how do I move forward?
Thanks!