Question Number 151468 by peter frank last updated on 21/Aug/21
$$\mathrm{z}^{\mathrm{2}} −\mathrm{7z}+\mathrm{16}=\mathrm{i}\left(\mathrm{z}−\mathrm{11}\right) \\ $$$$ \\ $$
Answered by peter frank last updated on 21/Aug/21
$$\mathrm{z}^{\mathrm{2}} −\mathrm{7z}−\mathrm{iz}+\mathrm{16}+\mathrm{11i}=\mathrm{0} \\ $$$$\mathrm{z}^{\mathrm{2}} −\left(\mathrm{7}+\mathrm{i}\right)\mathrm{z}+\mathrm{16}+\mathrm{11i}=\mathrm{0} \\ $$$$\mathrm{z}=\frac{\left(\mathrm{7}+\mathrm{i}\right)\pm\sqrt{\left(\mathrm{7}+\mathrm{i}\right)^{\mathrm{2}} −\mathrm{4}\left(\mathrm{1}\right)\left(\mathrm{16}+\mathrm{11i}\right)}}{\mathrm{2}} \\ $$$$…… \\ $$
Commented by MJS_new last updated on 22/Aug/21
$$\mathrm{you}\:\mathrm{get}\:\mathrm{the}\:\mathrm{same} \\ $$$$… \\ $$$${z}=\frac{\left(\mathrm{7}+\mathrm{i}\right)\pm\sqrt{−\mathrm{16}−\mathrm{30i}}}{\mathrm{2}} \\ $$$$\sqrt{−\mathrm{16}−\mathrm{30i}}=\mathrm{3}−\mathrm{5i} \\ $$$$\Rightarrow \\ $$$${z}=\mathrm{2}+\mathrm{3i}\vee{z}=\mathrm{5}−\mathrm{2i} \\ $$
Answered by MJS_new last updated on 21/Aug/21
$${z}^{\mathrm{2}} −\left(\mathrm{7}+\mathrm{i}\right){z}+\mathrm{16}+\mathrm{11i}=\mathrm{0} \\ $$$${z}={t}+\frac{\mathrm{7}+\mathrm{i}}{\mathrm{2}} \\ $$$${t}^{\mathrm{2}} +\mathrm{4}+\frac{\mathrm{15}}{\mathrm{2}}\mathrm{i}=\mathrm{0} \\ $$$${t}=\pm\sqrt{−\mathrm{4}−\frac{\mathrm{15}}{\mathrm{2}}\mathrm{i}}=\pm\left(\frac{\mathrm{3}}{\mathrm{2}}−\frac{\mathrm{5}}{\mathrm{2}}\mathrm{i}\right) \\ $$$${z}=\mathrm{2}+\mathrm{3i}\vee{z}=\mathrm{5}−\mathrm{2i} \\ $$
Commented by peter frank last updated on 22/Aug/21
$$\mathrm{more}\:\mathrm{step}\:\:\mathrm{sir}\:\mathrm{not}\:\mathrm{understood} \\ $$
Commented by MJS_new last updated on 22/Aug/21
![x^2 +px+q=0 let x=t−(p/2) (just another version to get the usual term] t^2 −(p^2 /4)+q=0 t=±(√((p^2 /4)−q)) p=−(7+i)∧q=16+11i ⇒ t=±(√(−4−((15)/2)i)) (a+bi)^2 =−4−((15)/2)i a^2 −b^2 +2abi=−4−((15)/2)i ⇒ a^2 −b^2 =−4∧2ab=−((15)/2) ⇒ a=±(3/2)∧b=∓(5/2) ±(√(−4−((15)/2)i))=±((3/2)−(5/2)i)](https://www.tinkutara.com/question/Q151594.png)
$${x}^{\mathrm{2}} +{px}+{q}=\mathrm{0} \\ $$$$\mathrm{let}\:{x}={t}−\frac{{p}}{\mathrm{2}}\:\:\left(\mathrm{just}\:\mathrm{another}\:\mathrm{version}\:\mathrm{to}\:\mathrm{get}\:\mathrm{the}\:\mathrm{usual}\:\mathrm{term}\right] \\ $$$${t}^{\mathrm{2}} −\frac{{p}^{\mathrm{2}} }{\mathrm{4}}+{q}=\mathrm{0} \\ $$$${t}=\pm\sqrt{\frac{{p}^{\mathrm{2}} }{\mathrm{4}}−{q}} \\ $$$${p}=−\left(\mathrm{7}+\mathrm{i}\right)\wedge{q}=\mathrm{16}+\mathrm{11i} \\ $$$$\Rightarrow \\ $$$${t}=\pm\sqrt{−\mathrm{4}−\frac{\mathrm{15}}{\mathrm{2}}\mathrm{i}} \\ $$$$\left({a}+{b}\mathrm{i}\right)^{\mathrm{2}} =−\mathrm{4}−\frac{\mathrm{15}}{\mathrm{2}}\mathrm{i} \\ $$$${a}^{\mathrm{2}} −{b}^{\mathrm{2}} +\mathrm{2}{ab}\mathrm{i}=−\mathrm{4}−\frac{\mathrm{15}}{\mathrm{2}}\mathrm{i} \\ $$$$\Rightarrow \\ $$$${a}^{\mathrm{2}} −{b}^{\mathrm{2}} =−\mathrm{4}\wedge\mathrm{2}{ab}=−\frac{\mathrm{15}}{\mathrm{2}}\:\Rightarrow\:{a}=\pm\frac{\mathrm{3}}{\mathrm{2}}\wedge{b}=\mp\frac{\mathrm{5}}{\mathrm{2}} \\ $$$$\pm\sqrt{−\mathrm{4}−\frac{\mathrm{15}}{\mathrm{2}}\mathrm{i}}=\pm\left(\frac{\mathrm{3}}{\mathrm{2}}−\frac{\mathrm{5}}{\mathrm{2}}\mathrm{i}\right) \\ $$