752.

z1+z21+z1z2\frac{z_1+z_2}{1+z_1z_2}

TEKST ZADATKA

Za brojeve z1=iz_1=i i z2=1+i2z_2=\frac{1+i}{\sqrt2} odrediti

z1+z21+z1z2\frac{z_1+z_2}{1+z_1z_2}
REŠENJE ZADATKA

Uvrstiti brojeve z1z_1 i z2.z_2.

i+1+i21+i⋅1+i2i22+1+i222+i(1+i)2i2+1+i22+i(1+i)2i2+1+i2+i(1+i)\frac{i+\frac{1+i}{\sqrt2}}{1+i\cdot\frac{1+i}{\sqrt2}} \\ \frac{\frac{i\sqrt2}{\sqrt2}+\frac{1+i}{\sqrt2}}{\frac{\sqrt2}{\sqrt2}+\frac{i(1+i)}{\sqrt2}}\\ \frac{\frac{i\sqrt2+1+i}{\sqrt2}}{\frac{\sqrt2+i(1+i)}{\sqrt2}} \\ \frac{i\sqrt2+1+i}{\sqrt2+i(1+i)}

Pošto je i2=−1i^2=-1 dobija se:

i2+1+i2+i+i2i2+1+i2+i−1i2+1+i(2−1)+i\frac{i\sqrt2+1+i}{\sqrt2+i+i^2} \\ \frac{i\sqrt2+1+i}{\sqrt2+i-1}\\ \frac{i\sqrt2+1+i}{(\sqrt2-1)+i}

Da bi se uklonio imaginarni broj iz imenioca, pomnožiti brojilac i imenilac konjugovanim brojem od (2−1)+i,(\sqrt2-1)+i, što je (2−1)−i.(\sqrt2-1)-i.

i2+1+i(2−1)+i⋅(2−1)−i(2−1)−i(i2+1+i)(2−1−i)(2−1)2−i2(i2+1+i)(2−1−i)(2−1)2−i22i−i2−i22+2−1−i+i2−i−i22−22+1−i2\frac{i\sqrt2+1+i}{(\sqrt2-1)+i} \cdot \frac{(\sqrt2-1)-i}{(\sqrt2-1)-i} \\ \frac{(i\sqrt2+1+i)(\sqrt2-1-i)}{(\sqrt2-1)^2-i^2} \\ \frac{(i\sqrt2+1+i)(\sqrt2-1-i)}{(\sqrt2-1)^2-i^2} \\ \frac{2i-i\sqrt2-i^2\sqrt2+\sqrt2-1-i+i\sqrt2-i-i^2}{2-2\sqrt2+1-i^2}

Pošto je i2=−1i^2=-1 dobija se:

2i−i2+2+2−1−i+i2−i+12−22+1+1\frac{2i-i\sqrt2+\sqrt2+\sqrt2-1-i+i\sqrt2-i+1}{2-2\sqrt2+1+1}

Sabrati realne i imaginarne delove posebno.

i(2−2−1+2−1)+224−22i⋅0+222(2−2)222(2−2)22−2\frac{i(2-\sqrt2-1+\sqrt2-1)+2\sqrt2}{4-2\sqrt2}\\ \frac{i\cdot0+2\sqrt2}{2(2-\sqrt2)} \\ \frac{2\sqrt2}{2(2-\sqrt2)} \\ \frac{\sqrt2}{2-\sqrt2}