Question

The Values of $\sqrt{7-24i},\sqrt{-7+24i},\sqrt{-7-24i},\sqrt{7+24i}$, in this order are:

• A. $\pm (4-3i),\pm (3+4i),\pm (4+3i),\pm (-3+4i)$
• B. $\pm (4-3i),\pm (3+4i),\pm (3-4i),\pm (4+3i)$
• C. $\pm (4-3i),\pm (3+4i),\pm (3+4i),\pm (4+3i)$
• D. $\pm (4-3i),\pm (4+3i),\pm (-3+4i),\pm (3+4i)$

Answer 1

Answer: (B)

$\pm (4-3i),\pm (3+4i),\pm (3-4i),\pm (4+3i)$

let us custom $\sqrt{7-24i}=a_1+i{b}_1$,

$\sqrt{-7+24i}=a_2+i{b}_2$,

$\sqrt{-7-24i}=a_3+i{b}_3$,

$\sqrt{7+24i}=a_4+i{b}_4$

$(a_1+i{b}_1{)}^2=7-24i;(a_2+i{b}_2{)}^2=-7+24i;(a_3+i{b}_3{)}^2=-7-24i;(a_4+i{b}_4{)}^2=7+24i$

$a_1^2-b_1^2=7$ $a_2^2-b_2^2=-7$ $a_3^2-b_3^2=-7$ $a_4^2-b_4^2=7$

$2a_1{b}_1=-24$ $2a_2{b}_2=24$ $2a_3{b}_3=-24$ $2a_4{b}_4=24$

$a_1^2+b_1^2=25$ $a_2^2+b_2^2=25$

$a_3^2+b_3^2=24$ $a_4^2+b_4^2=25$

$\therefore a_1{=}_{-}^{+}4$ $a_2{=}_{-}^{+}3$ $a_3{=}_{-}^{+}3$ $a_4{=}_{-}^{+}4$

$\therefore b_1{=}_{+}^{-}3$ $b_2{=}_{-}^{+}4$ $b_3{=}_{+}^{-}4$ $b_4{=}_{-}^{+}3$

$\therefore a_1+ib,$ $\therefore a_2+i{b}_2$ $\therefore a_3+i{b}_3$ $\therefore a_4+i{b}_4$

${=}_{-}^{+}(4-3i)$ ${=}_{-}^{+}(3+4i)$ ${=}_{-}^{+}(3-4i)$ ${=}_{-}^{+}(4+3i)$