solve complex solutions math

EXAMPLE 1 Find the square roots of the following: (i) $7-24i$ (ii) $5+12i$ SOLUTION Let$\sqrt{7-24i}=x+iy$. Then, $\sqrt{7-24i}=x+iy$ $\Rightarrow\quad7-24i=(x+iy)^{2}$ $\Rightarrow\quad7-24i=(x^{2}-y^{2})+2ixy$ $\Rightarrow\quad x^{2}-y^{2}=7$ ...(i) and,..

naming the complex

Introduction Consider the quadratic equation $\mathrm{x}^{2}+1=0$. Obviously it has no solution in the set of all real numbers because the square of a real number can never be negative. This equation becomes solvable if we introduce new quantities : square roots of negative numbers. The square roots of -ve numbers were..

complex numbers

31. If the complex numbers $\mathrm{z}_{1},\mathrm{z}_{2},\mathrm{z}_{3}$ represent the vertices of an equilateral $\triangle$ such $|z_{1}|=|z_{2}|=|z_{3}|$ . then prove that $\mathrm{z}_{1}+\mathrm{z}_{2}+\mathrm{z}_{3}=0$ (IIT) 32. If $\mathrm{w}$ is an imaginary cube root of 1, find the value of 1. $(2-\mathrm{w})(2-\mathrm{w..

solutions

LEVEL II 1. We know from Napier's analogy that : $\qquad\qquad{\displaystyle \tan\left(\frac{B-C}{2}\right)=\frac{b-c}{b+c}\cot\frac{A}{2}}$ $\qquad\qquad{\displaystyle \tan\left(\frac{A-B}{2}\right)=\frac{a-b}{a+b}\cot\frac{C}{2}}$ $\qquad\qquad{\displaystyle \tan\left(\frac{C-A}{2}\right)=\frac{c-a}{c+a..

Solution

> = . * M [log | sin (x - a) | - log | sin (x - fc) [ ] + c = log + * sm {a - b) sin (a - b) sm (x - b) 11. ^ = f~-dx [For alternative solution see Q. 3 in solved Ex. 21 J1 + cot x J sin x + cos x x 1 = f si" x (cos x - sin x) ^ _ fsin 2x -2sin 2 x ^ J (cos x + sin x) (cos x - sin x) J 2 cos 2x fsin 2x + cos 2x -1 , l'f-2sin2x...

Solution

(0 sin = sin -J = 1 sin"* solved Ex. 1(0 2 2 6 J -1 [ .v^n, tt ..

square roots of a complex number

Let $a+ib$ be a complex number such that $\sqrt{a+ib}=x+iy$, where $x$ and $y$ are real numbers. Then, $\sqrt{a+ib}=x+iy$ $\Rightarrow\quad(a+ib)=(x+iy)^{2}$ $\Rightarrow\quad a+ib=(x^{2}-y^{2})+2ixy$ On equating real and imaginary parts, we get $x^{2}-\mathrm{y}^{2}=a$ ...(i) ..

example of complex conjugates

Type II ON EQUALITY OF COMPLEX NUMBERS Recall that two complex numbers $z_{1}$ and $z_{2}$. are equal iff Re $(\mathrm{z}_{1})={\rm Re}(\mathrm{z}_{2})$ and ${\rm Im}(\mathrm{z}_{1})={\rm Im}(\mathrm{z}_{2})$ EXAMPLE 6 Given that: ${\displaystyle \frac{2\sqrt{3}\cos30^{\mathrm{o}}-2i\sin30^{\mathrm{o}}}{\sq..

complex numbers examples

Find the amplitude of the complex number $\sin\frac{6\pi}{5}+i\left(1-\cos\frac{6\pi}{5}\right)$ [ISC 2005] SOLUTION Let $z={\displaystyle \sin\frac{6\pi}{5}+i\left(1-\cos\frac{6\pi}{5}\right)}$. Then, $z=2\sin\frac{3\pi}{5}\cos\frac{3\pi}{5}+2i\sin^{2}\frac{3\pi}{5}$ Let$\alpha$ be the acute angle gi..

examples of reciprocals complex numbers

Let $\mathrm{z}_{1},\mathrm{z}_{2},z_{3}$ be three complex numbers such that $|z_{1}|=1,|\mathrm{z}_{2}|=2,|z_{3}|=3$ and $|z_{1}+\mathrm{z}_{2}+\mathrm{z}_{3}|=1$. Find $|9\mathrm{z}_{1}\mathrm{z}_{2}+4\mathrm{z}_{1}z_{3}+\mathrm{z}_{2}z_{3}|$. SOLUTION We have, $|\mathrm{z}_{1}+\mathrm{z}_{2}+z_{3}|=1$ $\Rig..

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