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Complex Proofs2017-02-11T11:51:35+10:00

## Complex Proofs

$z\in C$ such that $\frac{z}{z-i}$ is real. Show that z is imaginary.

Let $\frac{z}{z-i}=a$, where a is real.

Then $z=a(z-i)$

$z=az-ai$
$ai=az-z$
$ai=z(a-1)$

so $z=\frac{ai}{a-1}$

$z=\frac{a}{a-1}i$

Since a is real, then $\frac{a}{a-1}$ is real, and hence $\frac{a}{a-1}i$ is imaginary

thus z is imaginary

### further examples

#### Solution

= 10x + 15 – 7x + 14 = 3x + 29

### harder examples

#### Solution

u = 1 + √x = 1 + x1/2 du/dx = ½x–1/2 du = $\frac{1}{{2\sqrt x }}dx$ x = 49 u = 8 x = 1 u = 2 = $2\int\limits_2^8 {\frac{1}{{2\sqrt x \sqrt u }}dx}$ = $2\int\limits_2^8 {{u^{ – 1/2}}\,du}$ = $2\left[ {\frac{{{u^{1/2}}}}{{1/2}}} \right]_2^8$ = $4\left[ {\sqrt u } \right]_2^8$ = 4{√8 – √2} = 4(2√2 – √2) = 4 × √2 = 4√2 = √32