What is (1+i)/(k+2i) ?

The solution to (1+i)/(k+2i) is (k+2)/(k^2+4)+(-2+k)/(k^2+4)i

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simplify (1+i)/(k+2i)

simplify

\frac{1 + i}{k + 2 i}

\frac{k + 2}{k ^{2} + 4} + \frac{- 2 + k}{k ^{2} + 4} i

Solution steps

\frac{1 + i}{k + 2 i}

Multiply by the conjugate

\frac{k - 2 i}{k - 2 i}

= \frac{( 1 + i ) ( k - 2 i )}{( k + 2 i ) ( k - 2 i )}

Simplify

\frac{( 1 + i ) ( k - 2 i )}{( k + 2 i ) ( k - 2 i )} : \frac{k + 2 + ( - 2 + k ) i}{k ^{2} + 4}

= \frac{k + 2 + ( - 2 + k ) i}{k ^{2} + 4}

Rewrite in standard complex form:

\frac{k + 2}{k ^{2} + 4} + \frac{- 2 + k}{k ^{2} + 4} i

= \frac{k + 2}{k ^{2} + 4} + \frac{- 2 + k}{k ^{2} + 4} i

8^{2} - 5^{2}

3 \times \frac{2}{3}

(x^{3} - 2 x^{2} y) - (x y^{2} - 3 y^{3}) - (x^{2} y - 4 x y^{2})

2 + \frac{1}{8}

44 \cdot 3

What is (1+i)/(k+2i) ?
The solution to (1+i)/(k+2i) is (k+2)/(k^2+4)+(-2+k)/(k^2+4)i