35−j⋅15\sqrt{35-j\cdot 15}

asked by guest
on Mar 16, 2025 at 6:59 pm



You asked:

Evaluate the expression: 35−j⋅15\sqrt{35 - j \cdot 15}

MathBot Answer:

Evaluated



35−j⋅15=35−15j\displaystyle \sqrt{35 - j \cdot 15} = \sqrt{35 - 15 j}

Expanded

35−j⋅15=35−15j\sqrt{35 - j \cdot 15} = \sqrt{35 - 15 j}

Factored

35−j⋅15=57−3j\sqrt{35 - j \cdot 15} = \sqrt{5} \sqrt{7 - 3 j}

Roots

35−j⋅15=i(35−15re⁡(j))2+225(im⁡(j))24sin⁡(atan2⁡(−15im⁡(j),35−15re⁡(j))2)+(35−15re⁡(j))2+225(im⁡(j))24cos⁡(atan2⁡(−15im⁡(j),35−15re⁡(j))2)≈5.91607978309962i((1−0.428571428571429re⁡(j))2+0.183673469387755(im⁡(j))2)0.25sin⁡(atan2⁡(−15im⁡(j),35−15re⁡(j))2)+5.91607978309962((1−0.428571428571429re⁡(j))2+0.183673469387755(im⁡(j))2)0.25cos⁡(atan2⁡(−15im⁡(j),35−15re⁡(j))2)\sqrt{35 - j \cdot 15} = i \sqrt[4]{\left(35 - 15 \operatorname{re}{\left(j\right)}\right)^{2} + 225 \left(\operatorname{im}{\left(j\right)}\right)^{2}} \sin{\left(\frac{\operatorname{atan_{2}}{\left(- 15 \operatorname{im}{\left(j\right)},35 - 15 \operatorname{re}{\left(j\right)} \right)}}{2} \right)} + \sqrt[4]{\left(35 - 15 \operatorname{re}{\left(j\right)}\right)^{2} + 225 \left(\operatorname{im}{\left(j\right)}\right)^{2}} \cos{\left(\frac{\operatorname{atan_{2}}{\left(- 15 \operatorname{im}{\left(j\right)},35 - 15 \operatorname{re}{\left(j\right)} \right)}}{2} \right)} \approx 5.91607978309962 i \left(\left(1 - 0.428571428571429 \operatorname{re}{\left(j\right)}\right)^{2} + 0.183673469387755 \left(\operatorname{im}{\left(j\right)}\right)^{2}\right)^{0.25} \sin{\left(\frac{\operatorname{atan_{2}}{\left(- 15 \operatorname{im}{\left(j\right)},35 - 15 \operatorname{re}{\left(j\right)} \right)}}{2} \right)} + 5.91607978309962 \left(\left(1 - 0.428571428571429 \operatorname{re}{\left(j\right)}\right)^{2} + 0.183673469387755 \left(\operatorname{im}{\left(j\right)}\right)^{2}\right)^{0.25} \cos{\left(\frac{\operatorname{atan_{2}}{\left(- 15 \operatorname{im}{\left(j\right)},35 - 15 \operatorname{re}{\left(j\right)} \right)}}{2} \right)}35−j⋅15=i(−(35−15re⁡(j))2+225(im⁡(j))24sin⁡(atan2⁡(−15im⁡(j),35−15re⁡(j))2))−(35−15re⁡(j))2+225(im⁡(j))24cos⁡(atan2⁡(−15im⁡(j),35−15re⁡(j))2)≈−5.91607978309962i((1−0.428571428571429re⁡(j))2+0.183673469387755(im⁡(j))2)0.25sin⁡(atan2⁡(−15im⁡(j),35−15re⁡(j))2)−5.91607978309962((1−0.428571428571429re⁡(j))2+0.183673469387755(im⁡(j))2)0.25cos⁡(atan2⁡(−15im⁡(j),35−15re⁡(j))2)\sqrt{35 - j \cdot 15} = i \left(- \sqrt[4]{\left(35 - 15 \operatorname{re}{\left(j\right)}\right)^{2} + 225 \left(\operatorname{im}{\left(j\right)}\right)^{2}} \sin{\left(\frac{\operatorname{atan_{2}}{\left(- 15 \operatorname{im}{\left(j\right)},35 - 15 \operatorname{re}{\left(j\right)} \right)}}{2} \right)}\right) - \sqrt[4]{\left(35 - 15 \operatorname{re}{\left(j\right)}\right)^{2} + 225 \left(\operatorname{im}{\left(j\right)}\right)^{2}} \cos{\left(\frac{\operatorname{atan_{2}}{\left(- 15 \operatorname{im}{\left(j\right)},35 - 15 \operatorname{re}{\left(j\right)} \right)}}{2} \right)} \approx - 5.91607978309962 i \left(\left(1 - 0.428571428571429 \operatorname{re}{\left(j\right)}\right)^{2} + 0.183673469387755 \left(\operatorname{im}{\left(j\right)}\right)^{2}\right)^{0.25} \sin{\left(\frac{\operatorname{atan_{2}}{\left(- 15 \operatorname{im}{\left(j\right)},35 - 15 \operatorname{re}{\left(j\right)} \right)}}{2} \right)} - 5.91607978309962 \left(\left(1 - 0.428571428571429 \operatorname{re}{\left(j\right)}\right)^{2} + 0.183673469387755 \left(\operatorname{im}{\left(j\right)}\right)^{2}\right)^{0.25} \cos{\left(\frac{\operatorname{atan_{2}}{\left(- 15 \operatorname{im}{\left(j\right)},35 - 15 \operatorname{re}{\left(j\right)} \right)}}{2} \right)}

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