Complex numbers are defined as numbers of the form x+iy, where x and y are real numbers and i = √-1. 6. Tanθ= $\frac{{\rm{y}}}{{\rm{x}}}$ = $\frac{{\sqrt 3 }}{{ - 1}}$ = $ - \sqrt 3 $ then θ = 120°. Dear For a complex number z = x+iy, x is called the real part, denoted by Re z and y is called the imaginary part denoted by Im z. Digital NCERT Books Class 11 Maths pdf are always handy to use when you do not have access to physical copy. Either of the part can be zero. number, Please choose the valid $\left[ {\cos \frac{{\theta  + {\rm{k}}.360}}{3} + {\rm{i}}.\sin \frac{{\theta  + {\rm{k}}.360}}{3}} \right]$, Or, ${\rm{z}}_0^{\frac{1}{3}}$ = 1. Prepared by the best teachers with decades of experience, these are the latest Class 11 Maths solutions that you will find. Or, $\sqrt {{{\rm{z}}_{\rm{k}}}} $ = $\sqrt {\rm{r}} $$\left[ {\cos \frac{{\theta  + {\rm{k}}.360}}{2} + {\rm{i}}.\sin \frac{{\theta  + {\rm{k}}.360}}{2}} \right]$, When k = 0, $\sqrt {{{\rm{z}}_0}} $ = $\sqrt 2 $$\left[ {\cos \left( {\frac{{90 + 0}}{2}} \right) + {\rm{i}}.\sin \left( {\frac{{90 + 0}}{2}} \right)} \right]$. = $\frac{1}{{\sqrt 2 }}$ (cos45° + i.sin45°). Moreover, i is just not to distinguish but also has got some value. Complex numbers are often denoted by z. These values represent the position of the complex number in the two-dimensional Cartesian coordinate system. To find the value of in (n > 4) first, divide n by 4.Let q is the quotient and r is the remainder.n = 4q + r where o < r < 3in = i4q + r = (i4)q , ir = (1)q . (c) 0 if r is not a multiple of 3 and 3 if r is a multiple of 3. It will help you to save your precious time just before the examination. i is called as Iota in Complex Numbers. Find the square roots of … Or, $\frac{{1 + {\rm{i}}}}{{1 - {\rm{i}}}}$ = $\frac{{1 + {\rm{i}}}}{{1 - {\rm{i}}}}$ * $\frac{{1 + {\rm{i}}}}{{1 + {\rm{i}}}}$ = $\frac{{1 + 2{\rm{i}} + {{\rm{i}}^2}}}{{{1^2} - {{\rm{i}}^2}}}$ = $\frac{{1 + 2{\rm{i}} - 1}}{{1 + 1}}$ = I = 0 + i. r = $\sqrt {{{\rm{x}}^2} + {{\rm{y}}^2}} $ = $\sqrt {0 + 1} $ = 1. tanθ = $\frac{{\rm{y}}}{{\rm{x}}}$ = $\frac{1}{0}$ = -1  then θ= 90°. Extraction of square root of complex number. Since arg(a + ib) = π/4, so tan π/4 = b/a which gives a = b, So, 6x2 + 6y2 – 36x – 24y + 66 = 12x – 12y -12, Again, |z – 3 + i| = 3 gives |x + iy - 3 + i| = 3, This yields x2 + y2 - 6x + 2y +1 = 0 …. Solved and explained by expert mathematicians. Then we can easily equate the two and get a = 6 and b = 4. When k = 1, Z1 = cos $\left( {\frac{{0 + 360}}{4}} \right)$ + i.sin $\left( {\frac{{0 + 360}}{4}} \right)$. (2), (-2y + 6)2 + y2 – 6 (-2y + 6) + 2y + 1 = 0, Q1. ‘i’ (or ‘j’ in some books) in math is used to denote the imaginary part of any complex number. When k = 1, $\sqrt {{{\rm{z}}_1}} $ = 2$\sqrt 2 $$\left[ {\cos \left( {\frac{{60 + 360}}{2}} \right) + {\rm{i}}.\sin \left( {\frac{{60 + 360}}{2}} \right)} \right]$. Hence the required equation is x2 + x + 1 = 0. Complex Numbers are the numbers which along with the real part also has the imaginary part included with it. Complex number has two parts, real part and the imaginary part. Students can also make the best out of its features such as Job Alerts and Latest Updates. When k = 0, Z0 = 11/6 [cos 0 + i.sin0] = 1. Also, note that i + i2 + i3 + i4 = 0 or in + i2n + i3n + i4n= 0. = (cos 32° + i.sin32°)(cos13° + i.sin13°), = cos (32° + 13°) + i.sin(32° + 13°) = cos 45° + i.sin45°. One of our academic counsellors will contact you within 1 working day. So, we can say now, i4n = 1 where n is any positive interger. Here, x = 0, y = 1, r = $\sqrt {0 + 1} $ = 1. ‘a’ is called the real part, and ‘b’ is called the imaginary part of the complex number. NCERT Solutions For Class 11 Maths: The NCERT Class 11 Maths book contains 16 chapters each with their exercises that help students practice the concepts. Here, x = $ - \frac{1}{2}$, y = $\frac{{\sqrt 3 }}{2}$, r = $\sqrt {\frac{1}{4} + \frac{3}{4}} $ = 1. tanθ = $\frac{{\frac{{\sqrt 3 }}{2}}}{{ - \frac{1}{2}}}$ = $ - \sqrt 3 $ then θ = 120°. Careers | A complex number is of the form i 2 =-1. The notion of complex numbers increased the solutions to a lot of problems. Complex Numbers Class 11 solutions NCERT PDF are beneficial in several ways. When k = 1, Z1 = cos $\left( {\frac{{180 + 360}}{4}} \right)$ + i.sin $\left( {\frac{{180 + 360}}{4}} \right)$. = cos 300° + i.sin300° = $\frac{1}{2}$ - i.$\frac{{\sqrt 3 }}{2}$ = $\frac{{1 - {\rm{i}}\sqrt 3 }}{2}$. Ltd. Trigonometric Equations and General Values. = 2$\sqrt 2 $[cos30 + i.sin30] = 2$\sqrt 2 $$\left[ {\frac{{\sqrt 3 }}{2} + {\rm{i}}.\frac{1}{2}} \right]$ = $\sqrt 6 $ + i.$\sqrt 2 $. When k = 2, Z2 = cos $\left( {\frac{{0 + 720}}{6}} \right)$ + i.sin $\left( {\frac{{0 + 720}}{6}} \right)$. What is the application of Complex Numbers? Or, $\frac{{\rm{i}}}{{1 + {\rm{i}}}}$ = $\frac{{\rm{i}}}{{1 + {\rm{i}}}}$ * $\frac{{1 - {\rm{i}}}}{{1 - {\rm{i}}}}$ = $\frac{{{\rm{i}} - {{\rm{i}}^2}}}{{{1^2} - {{\rm{i}}^2}}}$ = $\frac{{{\rm{i}} - \left( { - 1} \right)}}{{1\left( { - 1} \right)}}$ = $\frac{{{\rm{i}} + 1}}{2}$ = $\frac{1}{2} + \frac{{\rm{i}}}{2}$. And check it in general, you proceed as in real numbers, but i... 1 + i for JEE Main complex numbers in the polar form = r ( cosθ i.sinθ! Negative one = 4 the expert teachers helps to score good marks in IIT JEE Exams i.0 ] =.... Consider a point in the right and 6 units upwards from the origin third quadrant both a b... 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