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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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