Calculate the value of the expression (z_1+z_2)/(1+z_1z_2)…

Calculate the value of the expression $\dfrac{z_1+z_2}{1+z_1z_2}$, if $z_1=i$ and $z_2=\dfrac{1+i}{\sqrt2}$.

Step 1 · Given values

$$ z_1=i, \qquad z_2=\frac{1+i}{\sqrt2} $$

Step 2 · Compute $z_1+z_2$

$$ z_1+z_2 = i+\frac{1+i}{\sqrt2} = \frac{1}{\sqrt2}+i\left(1+\frac{1}{\sqrt2}\right) = \frac{1+i(\sqrt2+1)}{\sqrt2} $$

Step 3 · Compute $z_1z_2$

$$ z_1z_2 = i\cdot\frac{1+i}{\sqrt2} = \frac{i+i^2}{\sqrt2} = \frac{-1+i}{\sqrt2} $$

Step 4 · Compute $1+z_1z_2$

$$ 1+z_1z_2 = 1+\frac{-1+i}{\sqrt2} = \frac{\sqrt2-1+i}{\sqrt2} $$

Step 5 · Form the quotient — the $\sqrt2$ cancels

$$ \frac{z_1+z_2}{1+z_1z_2} = \frac{\dfrac{1+i(\sqrt2+1)}{\sqrt2}}{\dfrac{\sqrt2-1+i}{\sqrt2}} = \frac{1+i(\sqrt2+1)}{\sqrt2-1+i} $$

Step 6 · Multiply by the conjugate of the denominator

The conjugate of $\sqrt2-1+i$ is $\sqrt2-1-i$. Writing the numerator as $1+i+i\sqrt2$:

$$ \frac{(1+i+i\sqrt2)(\sqrt2-1-i)}{(\sqrt2-1+i)(\sqrt2-1-i)} $$

Step 7 · Expand the numerator term by term

$$ 1\cdot(\sqrt2-1-i) = \sqrt2-1-i $$

$$ i\cdot(\sqrt2-1-i) = i\sqrt2-i-i^2 = 1-i+i\sqrt2 $$

$$ i\sqrt2\cdot(\sqrt2-1-i) = 2i-i\sqrt2-i^2\sqrt2 = \sqrt2+2i-i\sqrt2 $$

Adding all three, the real parts give $(\sqrt2-1)+1+\sqrt2=2\sqrt2$, and the imaginary parts give $-i-i+i\sqrt2+2i-i\sqrt2=0$. So the numerator is $2\sqrt2$.

Step 8 · Expand the denominator

$$ (\sqrt2-1+i)(\sqrt2-1-i) = (\sqrt2-1)^2+1 = (2-2\sqrt2+1)+1 = 4-2\sqrt2 = 2(2-\sqrt2) $$

Step 9 · Simplify

$$ \frac{2\sqrt2}{2(2-\sqrt2)} = \frac{\sqrt2}{2-\sqrt2} $$

Step 10 · Rationalize once more

Multiply by $\dfrac{2+\sqrt2}{2+\sqrt2}$:

$$ \frac{\sqrt2(2+\sqrt2)}{(2-\sqrt2)(2+\sqrt2)} = \frac{2\sqrt2+2}{4-2} = \frac{2(\sqrt2+1)}{2} = \sqrt2+1 $$

Answer

$$ \boxed{\frac{z_1+z_2}{1+z_1z_2} = \sqrt2+1 \approx 2.414} $$

Step 11 · Numerical verification

With $z_1=i$ and $z_2\approx0.707+0.707i$:

$$ z_1+z_2\approx0.707+1.707i, \qquad z_1z_2\approx-0.707+0.707i, \qquad 1+z_1z_2\approx0.293+0.707i $$

Taking moduli: $|0.707+1.707i|\approx1.848$ and $|0.293+0.707i|\approx0.765$, so the ratio is $1.848/0.765\approx2.415\approx\sqrt2+1$. ✓ The result is also confirmed to be real, since the imaginary part of the ratio comes out to be $0$.

Step 12 · Remark

In polar form $z_1=e^{i\pi/2}$ and $z_2=e^{i\pi/4}$ (since $|z_2|=1$, $\arg z_2=\frac{\pi}{4}$); the expression $\dfrac{z_1+z_2}{1+z_1z_2}$ does not reduce to a clean product of exponentials, so the direct algebraic route above is the natural one here. It is nonetheless a notable fact that dividing a sum of two non-real unit complex numbers by $1$ plus their product yields a real number in this case.