in polar form.Solution: Given that $Z=\left(i^{25}\right)^{3}$ $\begin{array}{l} =\dot{i}^{75} \\ =\mathrm{i}^{74} \cdot \mathrm{i} \\ =\left(\mathrm{i}^{2}\right)^{37} \cdot \mathrm{i} \\ =(-1)^{37} \cdot...
Solution: It is known to us that the polar form of a complex number $Z=x+i$ iy is given by $Z=|Z|(\cos \theta+i \sin \theta)$ In which, $\begin{array}{l} \left.|Z|=\text { modulus of complex number...
Solution: It is known to us that the polar form of a complex number $Z=x+i$ iy is given by $Z=|Z|(\cos \theta+i \sin \theta)$ In which, $\begin{array}{l} \left.|Z|=\text { modulus of complex number...
Solution: It is known to us that the polar form of a complex number $Z=x+i$ iy is given by $Z=|Z|(\cos \theta+i \sin \theta)$ In which, $\begin{array}{l} \left.|Z|=\text { modulus of complex number...
and 
are two pairs of conjugate complex numbers, prove that arg 
Solution: Given that $\begin{array}{l} z_{1}=\bar{z}_{2} \\ z_{3}=\bar{z}_{4} \end{array}$ It is known that $\arg \left(\mathrm{z}_{1} / \mathrm{z}_{2}\right)=\arg \left(\mathrm{z}_{1}\right)-\arg...
in polar form.Solution: Given that $Z=\sin \pi / 5+i(1-\cos \pi / 5)$ Using the formula, $\begin{array}{l} \sin 2 \theta=2 \sin \theta \cos \theta \\ 1-\cos 2 \theta=2 \sin ^{2} \theta \end{array}$ Therefore,...
Solution: If $b>0, \sqrt{a+i b}=\pm\left[\left(\frac{a+\sqrt{a^{2}+b^{2}}}{2}\right)^{\frac{1}{2}}+i\left(\frac{-a+\sqrt{a^{2}+b^{2}}}{2}\right)^{\frac{1}{2}}\right]$ If $b<0, \sqrt{a+i...
Solution: If $b>0, \sqrt{a+i b}=\pm\left[\left(\frac{a+\sqrt{a^{2}+b^{2}}}{2}\right)^{\frac{1}{2}}+i\left(\frac{-a+\sqrt{a^{2}+b^{2}}}{2}\right)^{\frac{1}{2}}\right]$ If $b<0, \sqrt{a+i...
Solution: If $b>0, \sqrt{a+i b}=\pm\left[\left(\frac{a+\sqrt{a^{2}+b^{2}}}{2}\right)^{\frac{1}{2}}+i\left(\frac{-a+\sqrt{a^{2}+b^{2}}}{2}\right)^{\frac{1}{2}}\right]$ If $b<0, \sqrt{a+i...
Solution: If $b>0, \sqrt{a+i b}=\pm\left[\left(\frac{a+\sqrt{a^{2}+b^{2}}}{2}\right)^{\frac{1}{2}}+i\left(\frac{-a+\sqrt{a^{2}+b^{2}}}{2}\right)^{\frac{1}{2}}\right]$ If $b<0, \sqrt{a+i...
Solution: (i) $(1+2 \mathrm{i})^{-3}$ Simplify and express in the standard form of $(a+i b)$, $\begin{array}{l} (1+2 i)^{-3}=1 /(1+2 i)^{3} \\ =1 /\left(1^{3}+3(1)^{2}(2 i)+2(1)(2 i)^{2}+(2...
Solution: (i) $(2+3 \mathrm{i}) /(4+5 \mathrm{i})$ Simplify and express in the standard form of $(a+i b)$, $(2+3 i) /(4+5 i)=$ [multiply and divide with (4-5i)] $\begin{array}{l} =(2+3 i) /(4+5 i)...
Solution: (i)$(2+i)^{3} /(2+3 i)$ Simplify and express in the standard form of (a +ib), $\begin{array}{l} (2+i)^{3} /(2+3 i)=\left(2^{3}+i^{3}+3(2)^{2}(i)+3(i)^{2}(2)\right) /(2+3 i) \\...
Solution: (i) $(1 + i) (1 + 2i)$ Simplify and express in the standard form of $(a + ib)$, $(1 + i) (1 + 2i) = (1+i)(1+2i)$ $= 1(1+2i)+i(1+2i)$ $= 1+2i+i+2i^2$ $= 1+3i+2(-1) [\text{since}\, i^2 =...
and 
, if 
Solution: (i) $\frac{(1+\mathbf{i}) \mathbf{x}-2 \mathbf{i}}{\mathbf{3}+\mathbf{i}}+\frac{(2-\mathbf{3 i}) \mathbf{y}+\mathbf{i}}{\mathbf{3}-\mathbf{i}}=\mathbf{i}$ Given that $\begin{array}{l}...
and , if 
Solution: (i) $(x+i y)(2-3 i)=4+i$ On simplifying the expression we obtain, $\begin{array}{l} x(2-3 i)+i y(2-3 i)=4+i \\ 2 x-3 x i+2 y i-3 y i^{2}=4+i \\ 2 x+(-3 x+2 y) i-3 y(-1)=4+i\left[\text {...
Solution: (i) $1 /(1+\mathrm{i})$ As the given complex no. is not in the standard form of $(a+i b)$ Convert it to standard form by multiplying and dividing with $(1 - i)$ We obtain, $\begin{aligned}...
Solution: (i) 4 – 5i It is known that the conjugate of a complex number $(a + ib)$ is $(a – ib)$ $\therefore$ $(4 + 5i)$ is the conjugate of $(4 – 5i)$ (ii) 1 / (3 + 5i) As the given complex no. is...
Solution: (i) $4-3 \mathrm{i}$ Given that $4-3 i$ It is known that the multiplicative inverse of a complex number $(Z)$ is $Z^{-1}$ or $1 / Z$ Therefore, $\begin{array}{l} z=4-3 i \\...
Solution: (i) $1-\mathrm{i}$ Given that It is known that the multiplicative inverse of a complex number $(\mathrm{Z})$ is $\mathrm{Z}^{-1}$ or $1 / \mathrm{Z}$ Therefore, $\begin{array}{l}...
, prove that 
Solution: Given that $x+i y=(a+i b) /(a-i b)$ It is known that for a complex number $Z=(a+i b)$ it's magnitude is given by $\left.|z|=\sqrt{(} a^{2}+b^{2}\right)$ $\mathrm{So}$, $|\mathrm{a} /...
Solution: Given that $[(1+i) /(1-i)]-[(1-i) /(1+i)]$ So, $Z=[(1+i) /(1-i)]-[(1-i) /(1+i)]$ $\begin{array}{l} =[(1+i)(1+i)-(1-i)(1-i)] /\left(1^{2}-i^{2}\right) \\...
, find 
Solution: Given that $z_{1}=(2-i)$ and $z_{2}=(-2+i)$ (i) $\mathbf{R e}\left(\frac{\mathbf{z}_{1} \mathbf{z}_{2}}{\overline{\mathbf{z}_{1}}}\right)$ On rationalising the denominator, we get...
, find 
Solution: Given that $\mathrm{z}_{1}=(2-\mathrm{i})$ and $\mathrm{z}_{2}=(1+\mathrm{i})$ It is known that, $|\mathrm{a} / \mathrm{b}|=|\mathrm{a}| /|\mathrm{b}|$ Therefore, $\begin{aligned}...
for which ![[(1+i) /(1-i)]^{n}](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-777ac8b5c711f90d2f7d5b4a4ac74b49_l3.png "Rendered by QuickLaTeX.com")
is real.Solution: Given that $\begin{array}{l} {[(1+i) /(1-i)]^{n}} \\ Z=[(1+i) /(1-i)]^{n} \end{array}$ On multiplying and dividing by $(1+i)$, we get $\begin{array}{l} =\frac{1+i}{1-i} \times...
for which the complex number 
is purely real.Solution: Given that $\begin{array}{l} (1+i \cos \theta) /(1-2 i \cos \theta) \\ Z=(1+i \cos \theta) /(1-2 i \cos \theta) \end{array}$ Multiply and divide by $(1+2 i \cos \theta)$ $\begin{array}{l}...
for which 
is a real number.Solution: Given that $\begin{array}{l} (1+i)^{n} /(1-i)^{n-2} \\ Z=(1+i)^{n} /(1-i)^{n-2} \end{array}$ Multiply and divide by $(1-i)^{2}$ $\begin{array}{l} =\frac{(1+i)^{n}}{(1-i)^{n-2}} \times...
![[(1+i) /(1-i)]^{3}-[(1-i) /(1+i)]^{3}=x+](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-8fd770c4a1e68dbf4628bfa33d738cd9_l3.png "Rendered by QuickLaTeX.com")
iy, find 
Solution: Given that $[(1+i) /(1-i)]^{3}-[(1-i) /(1+i)]^{3}=x+i y$ On rationalizing the denominator, we obtain $\begin{array}{l} \left(\frac{1+i}{1-i} \times...
, find 
Solution: Given that $(1+i)^{2} /(2-i)=x+i y$ On expansion we obtain, $\begin{array}{l} \frac{1^{2}+i^{2}+2(1)(i)}{2-i}=x+i y \\ \frac{1+(-1)+2 i}{2-i}=x+i y \\ \frac{2 i}{2-i}=x+i y \end{array}$ On...
Solution: (i) ${(1 + i)}^6 + {(1 – i)}^3$ Let's simplify, ${(1 + i)}^6 + {(1 – i)}^3 = {(1 + i)^2 }^3 + (1 – i)^2 (1 – i)$ $= {\{1 + i^2 + 2i}\}^3 + (1 + i^2 – 2i)(1 – i)$ $= {\{1 – 1 + 2i}\}^3 + (1...
![\frac{[i^{592} + i^{590} + i^{588} + i^{586} + i^{584}]} {[i^{582} + i^{580} + i^{578} + i^{576} + i^{574}]}](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-825878077790e1cbe2753453ba3095c8_l3.png "Rendered by QuickLaTeX.com")
Solution: (i) $\frac{[i^{592} + i^{590} + i^{588} + i^{586} + i^{584}]} {[i^{582} + i^{580} + i^{578} + i^{576} + i^{574}]}$ Let us simplify we get, $\frac{[i^{592} + i^{590} + i^{588} + i^{586} +...
Solution: (i) $i + i^2 + i^3 + i^4$ Let's simplify, $i + i^2 + i^3 + i^4 = i + i^2 + i^2\times i + i^4$ $= i – 1 + (– 1) \times i + 1 [\text{since}\ i^4 = 1, i^2 = – 1]$ $= i – 1 – i + 1$ $= 0$...
Solution: (i) $i^{49} + i^{68} + i^{89} + i^{110}$ Let's simplify, $i^{49} + i^{68} + i^{89} + i^{110} = i ^{(48 + 1)} + i^{68} + i^{(88 + 1)} + i^{(108 + 2)}$ $= {(i^4)}^{12} \times i + (i^4)^{17}...
is a real number?Solution: Given that: $1 + i^{10} + i^{20} + i^{30} = 1 + i^{(8 + 2)} + i^{20} + i^{(28 + 2)}$ $= 1 + (i^4)^2 \times i^2 + (i^4)^5 + (i^4)^7 \times i^2$ $= 1 – 1 + 1 – 1 [\text{since}\, i^4 = 1, i^2...
Solution: (i) $\mathrm{i}^{30}+\mathrm{i}^{40}+\mathrm{i} 60$ Let's simplify, $\begin{array}{l} \text { i } 30+i 40+i^{60}=j(28+2)+i 40+i 60 \\ =\left(i^{4}\right)^{7}...
![[i^{41} + {1/i}^ {257}]](https://www.learnatnoon.com/s/wp-content/ql-cache/quicklatex.com-0ace01b49760992eb5b6724b39474b9f_l3.png "Rendered by QuickLaTeX.com")
Solution: (i) $\left[\mathrm{i}^{41}+1 / \mathrm{i}^{257}\right]$ Let's simplify, $\begin{array}{l} {[\mathrm{i} 41+1 / \mathrm{i} 257]=\left[\mathrm{i} 40+1+1 / \mathrm{i}^{256+1}\right]} \\...
Solution: (i) $\frac{1} {i^{58}}$ Let's simplify, $\frac{1} {i^{58}} = \frac{1} {i^ {56+2}}$ $= \frac{1}{ i^{56}} \times {i^{2}}$ $= \frac{1} {(i^4)^{14}} \times {i^{2}}$ $= \frac{1} {i^2} [\text...
Solution: Let $A B C$ be the metallic cone, $DECB$ is the required frustum Let the two radii of the frustum be$\mathrm{DO}^{\prime}=\mathrm{r}_{2}$ and $\mathrm{BO}=\mathrm{r}_{1}$From $\triangle...
Given, r1 = 20 cm, r2 = 8 cm and h = 16 cm \[\therefore Volume\text{ }of\text{ }the\text{ }frustum\text{ }=\text{ }\left( \right)\times \pi \times h\left( r12+r22+r1r2 \right)\] It is given that...
Given, For the lower roundabout end, span $(r_1)$ = 10 cm For the upper roundabout end, span $(r_2)$ = 4 cm Inclination tallness (l) of frustum = 15 cm Presently, The space of material to be...
Given, Inclination tallness (l) = 4 cm Perimeter of upper roundabout finish of the frustum = 18 cm \[\therefore 2\pi r_1\text{ }=\text{ }18\] Or on the other hand, $r_1$ = 9/π Likewise, periphery of...
Sweep $(r_1)$ of the upper base = 4/2 = 2 cm Sweep $(r_2)$ of lower the base = 2/2 = 1 cm Tallness = 14 cm Presently, Capacity of glass = Volume of frustum of cone Thus, Capacity of glass = \[\left(...
Think about the accompanying graph Volume of water that streams in t minutes from pipe \[=\text{ }t\times 0.5\pi \text{ }m^3\] Volume of water that streams in t minutes from pipe = \[t\times 0.5\pi...
It is given that the waterway is the state of a cuboid with measurements as: Broadness (b) = 6 m and Height (h) = 1.5 m It is additionally given that The speed of waterway = 10 km/hr Length of...
The outline will be as- Given, Tallness $(h_1)$ of tube shaped piece of the pail = 32 cm Range $(r_1)$ of roundabout finish of the pail = 18 cm Tallness of the cone like pile $(h_2)$ = 24 cm...
It is realized that the coins are round and hollow fit. Along these lines, tallness (h1) of the chamber = 2 mm = 0.2 cm Range (r) of roundabout finish of coins \[=\text{...
Number of cones will be = Volume of chamber/Volume of frozen treat For the chamber part, Range = 12/2 = 6 cm Stature = 15 cm ∴ Volume of chamber \[=\text{ }\mathbf{\pi }\times \mathbf{r2}\times...
The state of the well will be tube shaped as given underneath. Given, Depth $(h_1)$ of well = 14 m Distance across of the roundabout finish of the well =3 m Thus, Radius $(r_1)$ = 3/2 m Width of the...
t is given that the state of the well is looking like a chamber with a width of 7 m In this way, span = 7/2 m Likewise, Depth (h) = 20 m Volume of the earth uncovered will be equivalent to the...
For Sphere 1: Span (r1) = 6 cm ∴ Volume (V1) \[=\text{ }\left( 4/3 \right)\times \pi \times r13\] For Sphere 2: Span (r2) = 8 cm ∴ Volume (V2) \[=\text{ }\left( 4/3 \right)\times \pi \times r23\]...
It is given that range of the circle (R) = 4.2 cm Likewise, Radius of chamber (r) = 6 cm Presently, let stature of chamber = h It is given that the circle is liquefied into a chamber. Along these...
Given, For the chamber part, Height (h) = 8 cm and Radius (R) = (2/2) cm = 1 cm For the circular part, Radius (r) \[=\text{ }\left( 8.5/2 \right)\text{ }=\text{ }4.25\text{ }cm\] Presently, volume...
Here, the volume of water left will be = Volume of chamber – Volume of strong Given, Range of cone = 60 cm, Stature of cone = 120 cm Range of chamber = 60 cm Stature of chamber = 180 cm Range of...
Given, the tallness of the enormous chamber (H) = 220 cm Sweep of the base (R) \[=\text{ }24/12\text{ }=\text{ }12\text{ }cm\] Thus, the volume of the huge chamber = πR2H \[=\text{ }\pi \left( 12...
For the cone, Range = 5 cm, Tallness = 8 cm Too, Range of circle = 0.5 cm The outline will resemble It is realized that, Volume of cone = volume of water in the cone \[=\text{ }\pi r^2h\text{...
Volume of cuboid = length x width x tallness We know the cuboid's measurements as 15 cmx10 cmx3.5 cm Along these lines, the volume of the cuboid \[=\text{ }15\times10\times3.5\text{ }=\text{...
It is realized that the gulab jamuns are like a chamber with two hemispherical finishes. Thus, the absolute stature of a gulab jamun = 5 cm. Measurement = 2.8 cm Thus, range = 1.4 cm ∴ The tallness...
Given, Height of each conical part $=2 cm$ Stature of chamber \[=\text{ }12{-}2{-}{2}\text{ }=\text{ }8\text{ }cm\] Span = 1.5 cm Stature of cone = 2 cm Presently, the all out volume of the air...
Solution: Here r = 1 cm and h = 1 cm. The chart is as per the following. Presently, Volume of strong = Volume of cone like part + Volume of hemispherical part We know the volume of cone $= {⅓ }\pi...
.The outline for the inquiry is as per the following: From the inquiry we know the accompanying: The width of the chamber = measurement of cone shaped hole = 1.4 cm Along these lines, the range of...
It is realized that a tent is a mix of chamber and a cone. From the inquiry we realize that Distance across = 4 m Inclination tallness of the cone (l) = 2.8 m Sweep of the cone (r) = Radius of...
Two half of the globe and one chamber are displayed in the figure given underneath. Here, the measurement of the case = 5 mm \[\therefore Radius\text{ }=\text{ }5/2\text{ }=\text{ }2.5\text{ }mm\]...
The graph is as per the following: Presently, the breadth of side of the equator = Edge of the 3D square = l In this way, the span of half of the globe = l/2 ∴ The all out surface space of strong =...
It is given that each side of 3D square is 7 cm. In this way, the range will be 7/2 cm. We know, The complete surface space of strong (TSA) = surface space of cubical square + CSA of half of the...
The chart is as per the following: Considering that the sweep of the cone and the side of the equator (r) = 3.5 cm or 7/2 cm The complete stature of the toy is given as 15.5 cm. Thus, the stature of...
The chart is as per the following: Presently, the given boundaries are: The measurement of the half of the globe = D = 14 cm The sweep of the half of the globe = r = 7 cm Additionally, the tallness...
The outline is given as: Given, The Volume (V) of each solid shape is = 64 cm3 This suggests that \[a3\text{ }=\text{ }64\text{ }cm3\] ∴ a = 4 cm Presently, the side of the solid shape = a = 4 cm...