Mathematics - Complex Number Question with Solution | TestHub

(I) $20,\mathrm{b},\mathrm{c}$ are in H.P.

$$\Rightarrow \frac{1}{20}+\frac{1}{c}=\frac{2}{b}$$

This can be written as $(40-b)(c+20)=800$ we must have $20<b<40$ or $0<40-b<20$, consider factorisation of 800 in which one term is less that 20 $(40-b)(c+20)=800=1\times 800=2\times 400=4\times 200$

$$=5\times 160=8\times 100=10\times 80=16\times 50$$

Thus pair $(b,c)=(39,780),(38,380),(36,180),(35,140),(32,80),(30,60),(24,30)$ but be divides c ⇒ Only 5 triples satisfy required requirement

(II) Solution of $\mathrm{m}+\mathrm{n}+\mathrm{p}=10$, are ${}^{3+10-1}{\mathrm{C}}_{3-1}$

(III) $\mathrm{ab}+\mathrm{b}-\mathrm{a}+1=0$

$$\begin{array}{rlrl}& \Rightarrow (\mathrm{a}+1)(\mathrm{b}-1)=-2& \Rightarrow (\mathrm{a}+1,\mathrm{b}-1)\in \{(1,-2),(-1,2),(2,-1),(-2,1)\}& \end{array}$$

$$\Rightarrow (\mathrm{a},\mathrm{b})\in \{(0,-1),(-2,3),(1,0),(-3,2)$$

(IV) Let $\mathrm{z}=\mathrm{x}+\mathrm{iy}$ then $(\mathrm{x}+\mathrm{iy}{)}^{20}=\mathrm{x}-\mathrm{iy}$

$$\begin{array}{rlrl}& \Rightarrow |z{|}^{20}=|z|\Rightarrow |z|\left(|z{|}^{19}-1\right)=0& \Rightarrow |z|=1\text{ or }0& \Rightarrow |z|=0\end{array}$$

∴ Ordered pair $(0,0)$ Let $z\ne 0$ then $(x+iy{)}^{20}(x+iy)=x^2+y^2$ $\Rightarrow z^{21}=|z{|}^2=1\Rightarrow z^{21}-1=0$, which are 21 roots