- #1
chwala
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- Homework Statement
- Solve the simultaneous equation for the complex number ##z## and ##w##,
$$(1+i)z+(2-i)w=3+4i$$
$$iz+(3+i)w=-1+5i$$
- Relevant Equations
- Complex numbers
$$(1+i)z+(2-i)w=3+4i$$
$$iz+(3+i)w=-1+5i$$
ok, multiplying the first equation by##(1-i)## and the second equation by ##i##, we get,
$$2z+(1-3i)w=7+i$$
$$-z+(-1+3i)w=-5-i$$
adding the two equations, we get ##z=2##,
We know that, $$iz+(3+i)w=-1+5i$$
$$⇒2i+(3+i)w=-1+5i$$
$$w=\frac {-1+3i}{3+i}$$
$$w=\frac {(-1+3i)(3-i)}{(3+i)(3-i)}$$
$$w=i$$
There may be a different approach from this...
$$iz+(3+i)w=-1+5i$$
ok, multiplying the first equation by##(1-i)## and the second equation by ##i##, we get,
$$2z+(1-3i)w=7+i$$
$$-z+(-1+3i)w=-5-i$$
adding the two equations, we get ##z=2##,
We know that, $$iz+(3+i)w=-1+5i$$
$$⇒2i+(3+i)w=-1+5i$$
$$w=\frac {-1+3i}{3+i}$$
$$w=\frac {(-1+3i)(3-i)}{(3+i)(3-i)}$$
$$w=i$$
There may be a different approach from this...
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