- #1
BOAS
- 546
- 19
Hello,
The complex numbers z_{1} = \frac{a}{1 + i} and z_{2} = \frac{b}{1+2i} where a and b are real, are such that z_{1} + z_{2} = 1. Find a and b.<br /> <br /> <h2>Homework Equations</h2><br /> <br /> <br /> <br /> <h2>The Attempt at a Solution</h2><br /> <br /> This looked like a time for partial fractions to me, so I went down that road;<br /> <br /> \frac{a}{1 + i} + \frac{b}{1+2i} = 1<br /> <br /> \frac{a(1+2i)}{(1 + i)(1+2i)} + \frac{b(1+i)}{(1+2i)(1+i)} = 1<br /> <br /> a(1+2i) + b(1+i) = (1+i)(1+2i)<br /> <br /> Expanding the brackets gives me;<br /> <br /> a(1+2i) + b(1+i) = (1+i)(1+2i)<br /> <br /> a(1+2i) + b(1+i) = -1 + 3i<br /> <br /> ∴ a + b = -1<br /> <br /> And now I'm stuck...<br /> <br /> Is this the right approach? And, how do I move forward?<br /> <br /> Thanks!
Homework Statement
The complex numbers z_{1} = \frac{a}{1 + i} and z_{2} = \frac{b}{1+2i} where a and b are real, are such that z_{1} + z_{2} = 1. Find a and b.<br /> <br /> <h2>Homework Equations</h2><br /> <br /> <br /> <br /> <h2>The Attempt at a Solution</h2><br /> <br /> This looked like a time for partial fractions to me, so I went down that road;<br /> <br /> \frac{a}{1 + i} + \frac{b}{1+2i} = 1<br /> <br /> \frac{a(1+2i)}{(1 + i)(1+2i)} + \frac{b(1+i)}{(1+2i)(1+i)} = 1<br /> <br /> a(1+2i) + b(1+i) = (1+i)(1+2i)<br /> <br /> Expanding the brackets gives me;<br /> <br /> a(1+2i) + b(1+i) = (1+i)(1+2i)<br /> <br /> a(1+2i) + b(1+i) = -1 + 3i<br /> <br /> ∴ a + b = -1<br /> <br /> And now I'm stuck...<br /> <br /> Is this the right approach? And, how do I move forward?<br /> <br /> Thanks!