Question

Find the square root
0 -47 +8 root 3i ( i=root-1)​

Answer 1

$Square roots of the given quantity are</p><p></p><p>=±√−8+8i(√3)</p><p></p><p>=±√4(−2+2(√3i))</p><p></p><p>=±√4(12+(√3i)2+2⋅1⋅(√3i))</p><p></p><p>=±√22(1+(√3i)2)</p><p></p><p>=±2(1+(√3i)</p><p></p><p>Now square roots of =+2(1+(√3i)</p><p></p><p>=±√2(1+(√3i))</p><p></p><p>=±√2+2√3i</p><p></p><p>=±√(√3)2+i2+2√3i</p><p></p><p>=±√(√3+i)2</p><p></p><p>=±(√3+i)</p><p></p><p>Again square roots of =−2(1+(√3i)</p><p></p><p>=±√−2(1+(√3i))</p><p></p><p>=±i√2+2√3i</p><p></p><p>=±i√(√3)2+i2+2√3i</p><p></p><p>=±i√(√3+i)2</p><p></p><p>=±i(√3+i)</p><p></p><p>=±(√3i+i2)</p><p></p><p>=±(−1+√3i)</p><p></p><p>So four fourth roots of the complex number (−8+8i(√3) are</p><p></p><p>=±(√3+i)</p><p>and</p><p>=±(−1+√3i)</p><p>mark \: it \: as \: branlist \\ follow \: me \: pls</p><p>$