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- Question 1, Exercise 1.1
- Question 2, Exercise 1.1
- Question 3, Exercise 1.1
- Question 4, Exercise 1.1
- Question 5, Exercise 1.1
- Question 6, Exercise 1.1
- Question 7, Exercise 1.1
- Question 1, Exercise 1.2
- Question 2, Exercise 1.2
- Question 3, Exercise 1.2
- Question 4, Exercise 1.2
- Question 5, Exercise 1.2
- Question 6, Exercise 1.2
- Question 7, Exercise 1.2
- Question 8, Exercise 1.2
- Question 9, Exercise 1.2
- Question 10, Exercise 1.2
- Question 1, Exercise 1.3
- Question 2, Exercise 1.3
- Question 3, Exercise 1.3
- Question 4, Exercise 1.3
- Question 1, Exercise 1.4
- Question 2, Exercise 1.4
- Question 3, Exercise 1.4
- Question 4, Exercise 1.4
- Question 5, Exercise 1.4
- Question 6(i-ix), Exercise 1.4
- Question 6(x-xvii), Exercise 1.4
- Question 7, Exercise 1.4
- Question 8, Exercise 1.4
- Question 9, Exercise 1.4
- Question 10, Exercise 1.4
- Question 1, Review Exercise
- Question 2, Review Exercise
- Question 3, Review Exercise
- Question 4, Review Exercise
- Question 5, Review Exercise
- Question 6, Review Exercise
- Question 7, Review Exercise
- Question 8, Review Exercise
Fulltext results:
- Question 1, Exercise 1.3
- n.** \begin{align} & 3z^2 + 363 \\ = & 3(z^2 - (11i)^2)\\ = & 3(z + 11i)(z - 11i) \end{align} ====Question 1(iv)==== Factorize the polynomial into linear functions: $z^{2}... olynomial into linear functions: $2 z^{3}+9 z^{2}-11 z-30$. **Solution.** Suppose $$P(z)=2z^3 + 9z^2
- Question 7, Review Exercise
- ===== Solve by completing square method $2 z^{2}-11 z+16=0$. ** Solution. ** \begin{align*} &2 z^{2}-11 z+16=0\\ \implies&z^2 - \dfrac{11}{2}z + 8 = 0\\ \implies& z^2 - \dfrac{11}{2}z = -8\\ \implies& z^2 - 2z\dfrac{11}{4}z + \dfrac{121}{1
- Question 7, Exercise 1.1
- ====Question 7(i)==== Find the magnitude of the $11+12 i$. **Solution.** Suppose $$z=11+12i$$ Then \begin{align}|z|&= \sqrt{(11)^2+(12)^2}\\ &=\sqrt{265}\end{align} Hence $|11+12 i|=\sqrt{265}$. GOOD ====Question 7(ii)==== Fin
- Question 4, Exercise 1.1
- \Rightarrow \quad & \dfrac{x}{(2+i)}=\dfrac{(2-5-11i)+(3y-2yi)}{(6+2i^2-4i-3i)}\\ \Rightarrow \quad & \dfrac{x}{(2+i)}=\dfrac{(3y-3)+(-11-2y)i}{(4-7i)}\\ \Rightarrow \quad & x(4-7i)=((3y-3)+(-11-2y)i)(2+i)\\ \Rightarrow \quad & 4x-7xi=(3y-3)(2+i)+(-11i-2yi)(2+i)\\ \Rightarrow \quad & 4x-7xi=6y-6+3yi-
- Question 1, Exercise 1.1
- i^{23} \\ &=-1 \cdot i^{22} \cdot i \\ &=-(i^2)^{11} \cdot i \\ &=-(-1)^{11} \cdot i \\ &=-(-1)\cdot i = i. \end{align} GOOD ====Question 1(iii)==== Eval... }^{-23}} =\dfrac{1}{i{{\left( {{i}^{2}} \right)}^{11}}}\\ &=\dfrac{1}{i{{\left( -1 \right)}^{11}}} =\dfrac{1}{-i}\\ &=\dfrac{1}{-i}\times \dfrac{i}{i}=\df
- Question 10, Exercise 1.2
- i)(1 - 3i) \\ &= -3 + 9i + 2i - 6i^2 \\ &= -3 + 11i + 6\\ &= 3 + 11i. \end{align} Then \[ \overline{z_1 z_2}= 3 - 11i. -- (i)\] Now \[ \overline{z_1} = -3 - 2i, \quad \... 2i)(1 + 3i)\\ & = -3 - 9i - 2i - 6i^2 \\ &= -3 - 11i + 6 \\ \implies \overline{z_1}\,\, \overline{z_
- Question 3, Exercise 1.3
- (iv)==== Solve the quadratic equation: $z^{2}-9 z+11=0$. **Solution.** Given $$z^{2} - 9z + 11 = 0 $$ Using the quadratic formula: $$ z = \dfrac{{-b \pm... }{2a} $$ Where $$ a = 1, \quad b = -9, \quad c = 11 $$ Then \begin{align} z &= \dfrac{{-(-9) \pm \sqrt{{(-9)^2 - 4 \cdot 1 \cdot 11}}}}{2 \cdot 1} \\ &= \dfrac{{9 \pm \sqrt{{81 - 44
- Question 4, Exercise 1.3
- 6i)\omega=3-i+12-10i\\ \implies &(3-14i)\omega=15-11i\\ \end{align} \begin{align} \implies \omega & =\dfrac{15-11i}{3-14i}\\ &= \dfrac{15 - 11i}{3 - 14i} \cdot \dfrac{3 + 14i}{3 + 14i}\\ &= \dfrac{45 + 154 + 210i - ... === <text align="left"><btn type="primary">[[math-11-nbf:sol:unit01:ex1-3-p3|< Question 3]]</btn></tex
- Question 3, Exercise 1.1
- =&\dfrac{(6+4i^2+8i+3i)+(7-4i)}{2+i}\\ =&\dfrac{2+11i+7-4i}{2+i}\\ =&\dfrac{9+7i}{2+i}\\ =&\dfrac{9+7i... == <text align="left"><btn type="primary">[[math-11-nbf:sol:unit01:ex1-1-p2|< Question 2]]</btn></tex... t> <text align="right"><btn type="success">[[math-11-nbf:sol:unit01:ex1-1-p4|Question 4 >]]</btn></tex
- Question 8, Exercise 1.2
- c{x-1}{2}=-5 \\ \implies & x-1=10 \\ \implies & x=11, \end{align} as required. GOOD ====Question 8(vi... === <text align="left"><btn type="primary">[[math-11-nbf:sol:unit01:ex1-2-p7|< Question 7]]</btn></tex... t> <text align="right"><btn type="success">[[math-11-nbf:sol:unit01:ex1-2-p9|Question 9 >]]</btn></tex
- Question 6, Review Exercise
- =& 2^3+3(2^2)i+3(2)i^2+i^3\\ =& 8+12i-6-i\\ =& 2+11i\end{align*} GOOD ====Go to ==== <text align="left"><btn type="primary">[[math-11-nbf:sol:unit01:Re-ex-p5|< Question 5]]</btn></tex... t> <text align="right"><btn type="success">[[math-11-nbf:sol:unit01:Re-ex-p7|Question 7 >]]</btn></tex
- Question 2, Exercise 1.1
- == <text align="left"><btn type="primary">[[math-11-nbf:sol:unit01:ex1-1-p1|< Question 1]]</btn></tex... t> <text align="right"><btn type="success">[[math-11-nbf:sol:unit01:ex1-1-p3|Question 3 >]]</btn></tex
- Question 5, Exercise 1.1
- == <text align="left"><btn type="primary">[[math-11-nbf:sol:unit01:ex1-1-p4|< Question 4]]</btn></tex... t> <text align="right"><btn type="success">[[math-11-nbf:sol:unit01:ex1-1-p6|Question 6 >]]</btn></tex
- Question 6, Exercise 1.1
- == <text align="left"><btn type="primary">[[math-11-nbf:sol:unit01:ex1-1-p5|< Question 5]]</btn></tex... t> <text align="right"><btn type="success">[[math-11-nbf:sol:unit01:ex1-1-p7|Question 7 >]]</btn></tex
- Question 2, Exercise 1.2
- == <text align="left"><btn type="primary">[[math-11-nbf:sol:unit01:ex1-2-p1|< Question 1]]</btn></tex... t> <text align="right"><btn type="success">[[math-11-nbf:sol:unit01:ex1-2-p3|Question 3 >]]</btn></tex